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Mean and standard deviation of a random variable

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52346312
An automobile manufacturer finds that \(8\%\) of its vehicles have a minor paint defect. Assume vehicles are independently classified with this constant defect probability. A sample of \(150\) vehicles is inspected. Let \(X\) be the number with a paint defect. Find the mean and standard deviation of \(X\).

Hints

- Identify \(n\) and \(p\). - Use the binomial mean formula. - Use the binomial standard-deviation formula.

Solution

1. The random variable is binomial with \(n=150\) and \(p=0.08\). 2. The mean is \(\mu=np=150\cdot0.08=12\). 3. The variance is \(np(1-p)=150\cdot0.08\cdot0.92=11.04\). 4. The standard deviation is \(\sigma=\sqrt{11.04}\approx3.32\).

Answer

\(\mu=12\) and \(\sigma\approx3.32\)
52346412
A spinner has \(10\) equal sections, and exactly one section is labeled Grand Prize. The spinner is spun independently \(80\) times. Let \(X\) be the number of Grand Prizes. Find the mean and standard deviation of \(X\). Briefly interpret the mean in context.

Hints

- Find the probability of Grand Prize on one spin. - Use the formulas for the mean and standard deviation of a binomial random variable. - Interpret the mean as a long-run average across repeated sets of spins.

Solution

1. The random variable is binomial with \(n=80\) and \(p=\frac{1}{10}=0.1\). 2. The mean is \(\mu=np=80\cdot0.1=8\). 3. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{80\cdot0.1\cdot0.9}=\sqrt{7.2}\approx2.68\). 4. Over many sets of \(80\) spins, the average number of Grand Prizes would approach \(8\).

Answer

\(\mu=8\) and \(\sigma\approx2.68\). Over many sets of \(80\) spins, the average number of Grand Prizes would be about \(8\).
52347512
A binomial random variable \(X\) has parameters \(n\) and \(p\). Suppose the number of trials is multiplied by \(25\) while the success probability remains unchanged. Describe how the mean \(E(X)\) and standard deviation \(\sigma(X)\) change.

Hints

- Write the formulas for the mean and standard deviation of a binomial random variable. - Identify where \(n\) appears in each formula. - Consider how a factor inside a square root affects the value of the square root. - Compare each new expression with the original one.

Solution

1. The mean of a binomial random variable is \(E(X)=np\). Replacing \(n\) with \(25n\) gives \(E_{\text{new}}=25np=25E(X)\). Therefore, the mean is multiplied by \(25\). 2. The standard deviation is \(\sigma(X)=\sqrt{np(1-p)}\). Replacing \(n\) with \(25n\) gives \(\sigma_{\text{new}}=\sqrt{25np(1-p)}=5\sqrt{np(1-p)}=5\sigma(X)\). Therefore, the standard deviation is multiplied by \(5\).

Answer

The mean is multiplied by \(25\), and the standard deviation is multiplied by \(5\).
52348512
A binomial random variable \(X\) has parameters \(n\) and \(p\), where \(n\ge1\). Suppose \(\operatorname{Var}(X)=0\). Find all possible values of \(p\), and briefly explain what each value means for the possible values of \(X\).

Hints

- Write the variance formula for a binomial random variable. - Determine when a product can equal zero. - Interpret what zero variance means for a random variable. - Consider what happens when the success probability is \(0\%\) or \(100\%\).

Solution

1. The variance of a binomial random variable is \(\operatorname{Var}(X)=np(1-p)\). 2. Setting the variance equal to zero gives \(np(1-p)=0\). 3. Since \(n\ge1\), it follows that \(p(1-p)=0\). 4. By the zero-product property, \(p=0\) or \(1-p=0\), so \(p=0\) or \(p=1\). 5. If \(p=0\), there are no successes, so \(X=0\) every time. If \(p=1\), every trial is a success, so \(X=n\) every time. In either case, \(X\) has no variability.

Answer

\(p=0\) or \(p=1\). If \(p=0\), then \(X=0\) always; if \(p=1\), then \(X=n\) always.
52348912
A binomial random variable \(X\) has success probability \(p=0.25\) and standard deviation \(\sigma=3\). Find the number of trials \(n\), and then find the mean \(\mu\).

Hints

- Write the formula that relates the standard deviation, success probability, and number of trials. - Square both sides to remove the square root. - Substitute all known values before solving for \(n\). - Use the binomial mean formula after finding \(n\).

Solution

1. Use the standard-deviation formula \(\sigma=\sqrt{np(1-p)}\). 2. Square both sides and substitute the known values: \(3^2=n\cdot0.25\cdot(1-0.25)\). 3. Simplify: \(9=0.1875n\). 4. Solve for \(n\): \(n=9\div0.1875=48\). 5. The mean is \(\mu=np=48\cdot0.25=12\).

Answer

\(n=48\) and \(\mu=12\)
52349512
A multiple-choice test has \(60\) questions. Each question has \(4\) answer choices, and exactly one choice is correct. A student selects an answer at random for every question. The random variable \(X\) is the number of correct answers. a) Find the mean \(\mu\) of \(X\), and interpret it in context. b) Find the probability that the student answers exactly the expected number of questions correctly.

Hints

- Identify the number of trials and the probability of a correct answer. - Use the mean formula for a binomial random variable. - For an exact number of successes, use the binomial probability formula.

Solution

1. Here, \(X\) is binomial with \(n=60\) and \(p=0.25\). 2. The mean is \(\mu=np=60\cdot0.25=15\). 3. Over many tests completed by random guessing, the long-run average number of correct answers would be about \(15\) per test. 4. The probability of exactly \(15\) correct answers is \(P(X=15)=\binom{60}{15}(0.25)^{15}(0.75)^{45}\approx0.1182\).

Answer

a) \(\mu=15\). Over many such tests, the student would average about \(15\) correct answers per test. b) \(P(X=15)\approx0.1182\), or about \(11.82\%\).
52672512
The probability distribution of a random variable \(X\) is shown below: <table> <tr> <td>\(x_i\)</td> <td>\(10\)</td> <td>\(20\)</td> <td>\(30\)</td> <td>\(40\)</td> </tr> <tr> <td>\(P(X=x_i)\)</td> <td>\(0.4\)</td> <td>\(0.3\)</td> <td>\(0.2\)</td> <td>\(0.1\)</td> </tr> </table> Find the mean \(\mu\), variance \(\operatorname{Var}(X)\), and standard deviation \(\sigma(X)\).

Hints

- First find the long-run weighted average of the values. - For the variance, square each deviation from the mean and weight it by its probability. - Take the square root of the variance to find the standard deviation.

Solution

1. The mean is \(\mu=10\cdot0.4+20\cdot0.3+30\cdot0.2+40\cdot0.1=4+6+6+4=20\). 2. The variance is \(\operatorname{Var}(X)=(10-20)^2\cdot0.4+(20-20)^2\cdot0.3+(30-20)^2\cdot0.2+(40-20)^2\cdot0.1\). 3. Therefore, \(\operatorname{Var}(X)=100\cdot0.4+0\cdot0.3+100\cdot0.2+400\cdot0.1=40+0+20+40=100\). 4. The standard deviation is \(\sigma(X)=\sqrt{100}=10\).

Answer

\(\mu=20\), \(\operatorname{Var}(X)=100\), and \(\sigma(X)=10\)
52672912
A fair \(12\)-sided die is labeled with the integers \(1\) through \(12\). Let \(X\) be the number rolled. Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\).

Hints

- List the possible values and identify their probabilities. - Find the mean of the equally likely outcomes. - Use the relationship between variance and standard deviation. - You may find the variance from squared deviations or from the second moment.

Solution

1. Each outcome has probability \(\frac{1}{12}\), so \(\mu=E(X)=\frac{1+2+\cdots+12}{12}=\frac{78}{12}=6.5\). 2. Use \(\operatorname{Var}(X)=E(X^2)-\mu^2\). Since \(1^2+2^2+\cdots+12^2=650\), \(E(X^2)=\frac{650}{12}=\frac{325}{6}\). 3. Therefore, \(\operatorname{Var}(X)=\frac{325}{6}-(6.5)^2=\frac{143}{12}\). 4. The standard deviation is \(\sigma=\sqrt{\frac{143}{12}}\approx3.45\).

Answer

\(\mu=6.5\) and \(\sigma\approx3.45\)
52674112
A fair four-sided die has faces labeled \(1\), \(2\), \(2\), and \(5\). The random variable \(X\) is the number showing after one roll. a) Find \(E(X)\). b) Evaluate this statement: “Because every face is labeled with an integer, the expected value must also be an integer.” Justify your answer using part a).

Hints

- The expected value describes a long-run average, not necessarily one individual result. - Combine the two faces labeled \(2\) when finding their probability. - Multiply each possible value by its probability and add.

Solution

1. The outcome probabilities are \(P(X=1)=\frac{1}{4}\), \(P(X=2)=\frac{2}{4}=\frac{1}{2}\), and \(P(X=5)=\frac{1}{4}\). 2. The expected value is \(E(X)=1\cdot\frac{1}{4}+2\cdot\frac{1}{2}+5\cdot\frac{1}{4}=\frac{5}{2}=2.5\). 3. The statement is false. An expected value is a weighted long-run average, so it does not have to be one of the possible outcomes or an integer.

Answer

a) \(E(X)=2.5\) b) The statement is false. The expected value is a long-run average and need not be a possible result; here, \(2.5\) cannot occur on one roll.
52678212
A raffle drum contains tickets with different possible prize values. Let \(X\) be the prize value in dollars on a randomly selected ticket. It is known that \(E(X)=\$5.00\) and \(\sigma(X)=0\). Determine which prize values can appear and find \(P(X=5)\). Explain whether the drum can contain any nonwinning tickets worth \(\$0\).

Hints

- Interpret a standard deviation of zero. - If every value is identical, relate that value to the mean. - Decide whether two different prize values could occur when every squared deviation is zero.

Solution

1. Since \(\sigma(X)=0\), the variance is also \(0\). 2. A random variable with variance \(0\) is constant, so it takes one value with probability \(1\). 3. Because the mean is \(\$5.00\), that constant value must be \(\$5.00\). Therefore, \(P(X=5)=1\). 4. Every other value has probability \(0\), so the drum cannot contain any nonwinning \(\$0\) tickets.

Answer

Every ticket must be worth \(\$5.00\), so \(P(X=5)=1\). The drum cannot contain any \(\$0\) tickets.
52679112
At a carnival game, a player draws one ball from an urn containing \(20\) balls: \(2\) gold, \(5\) silver, and \(13\) white. A gold ball pays \(\$15.00\), a silver ball pays \(\$5.00\), and a white ball pays nothing. What entry fee makes the game mathematically fair?

Hints

- Find the probability of each ball color. - Multiply each payout by its probability and add. - In a fair game, the entry fee equals the expected payout. - A white ball contributes \(\$0\) to the expected payout.

Solution

1. The payout probabilities are \(P(X=15)=\frac{2}{20}=0.10\), \(P(X=5)=\frac{5}{20}=0.25\), and \(P(X=0)=\frac{13}{20}=0.65\). 2. The expected payout is \(E(X)=15\cdot0.10+5\cdot0.25+0\cdot0.65=\$2.75\). 3. A fair entry fee equals the expected payout, so the fee should be \(\$2.75\).

Answer

The fair entry fee is \(\$2.75\).
52680912
A random variable \(Z\) can take the values \(5\) and \(0\), where \(P(Z=5)=p\) and \(P(Z=0)=1-p\). a) Write \(E(Z)\) in terms of \(p\). b) Write \(\operatorname{Var}(Z)\) in terms of \(p\). c) Find \(\sigma(Z)\) when \(p=0.4\).

Hints

- Use the expected-value formula for a discrete random variable. - Relate the variance to the second moment and the square of the mean. - Take the square root of the variance to find the standard deviation.

Solution

1. The mean is \(E(Z)=5p+0\cdot(1-p)=5p\). 2. The second moment is \(E(Z^2)=5^2p+0^2\cdot(1-p)=25p\). 3. Therefore, \(\operatorname{Var}(Z)=E(Z^2)-[E(Z)]^2=25p-(5p)^2=25p(1-p)\). 4. When \(p=0.4\), \(\operatorname{Var}(Z)=25\cdot0.4\cdot0.6=6\). 5. Thus, \(\sigma(Z)=\sqrt{6}\approx2.45\).

Answer

a) \(E(Z)=5p\) b) \(\operatorname{Var}(Z)=25p(1-p)\) c) \(\sigma(Z)=\sqrt{6}\approx2.45\)
52712112
A binomial random variable \(X\) has mean \(\mu=24\) and standard deviation \(\sigma=4\). Find the parameters \(n\) and \(p\).

Hints

- Write the binomial formulas for the mean and variance. - Use the mean equation to simplify the variance equation before solving for \(p\).

Solution

1. For a binomial random variable, \(\mu=np\) and \(\sigma^2=np(1-p)\). 2. Substitute the given values: \(np=24\) and \(16=np(1-p)\). 3. Using \(np=24\), the second equation becomes \(16=24(1-p)\). 4. Thus, \(1-p=\frac{2}{3}\), so \(p=\frac{1}{3}\). 5. Finally, \(24=n\left(\frac{1}{3}\right)\), so \(n=72\).

Answer

\(n=72\) and \(p=\frac{1}{3}\)
52712312
Find the parameters \(n\) and \(p\) of a binomial random variable \(X\) for each set of summary statistics. a) \(\mu=10\) and \(\sigma=\sqrt{8}\) b) \(\mu=48\) and \(\sigma=4\)

Hints

- Write the formulas connecting \(\mu\), \(\sigma\), \(n\), and \(p\). - Express the variance in terms of the mean and \(1-p\). - Find \(p\) before solving for \(n\). - Eliminate one unknown by combining the two formulas.

Solution

1. Use \(\sigma^2=\mu(1-p)\), which follows from \(\mu=np\) and \(\sigma^2=np(1-p)\). 2. For part a), \(8=10(1-p)\), so \(p=0.2\). Then \(10=n\cdot0.2\), giving \(n=50\). 3. For part b), \(16=48(1-p)\), so \(1-p=\frac13\) and \(p=\frac23\). Then \(48=n\left(\frac23\right)\), giving \(n=72\).

Answer

a) \(n=50\) and \(p=0.2\) b) \(n=72\) and \(p=\frac{2}{3}\)
52875912
A gardener knows that \(80\%\) of the seeds of a certain flower variety germinate. The gardener plants \(15\) seeds, and the seeds germinate independently. a) How many seeds are expected to germinate? b) What is the probability that exactly the expected number of seeds germinate?

Hints

- Identify the number of trials and the probability of germination. - Use the mean formula for a binomial random variable. - Use the binomial probability formula for an exact number of germinating seeds.

Solution

1. The number of germinating seeds is binomial with \(n=15\) and \(p=0.8\). 2. The mean is \(E(X)=np=15\cdot0.8=12\). 3. The probability of exactly \(12\) germinating seeds is \(P(X=12)=\binom{15}{12}(0.8)^{12}(0.2)^3\approx0.2501\).

Answer

a) \(12\) seeds b) \(P(X=12)\approx0.2501\), or about \(25.0\%\)
53081112
A long, six-faced die has two square end faces labeled \(1\) and \(2\), and four rectangular side faces labeled \(3\), \(4\), \(5\), and \(6\). Its geometry makes the face probabilities unequal: \(P(1)=P(2)=0.12\) and \(P(3)=P(4)=P(5)=P(6)=0.19\). The random variable \(X\) is the number rolled. a) Find the probability that the number rolled is at least \(4\). b) Find the probability of rolling a prime number. c) Calculate the expected value \(E(X)\).

Hints

- List the outcomes that satisfy “at least \(4\).” - Identify the prime numbers from \(1\) through \(6\). - Use the expected-value formula that weights each possible value by its probability. - Confirm that all six face probabilities add to \(1\).

Solution

1. The outcomes at least \(4\) are \(4\), \(5\), and \(6\), so \(P(X\ge4)=0.19+0.19+0.19=0.57\). 2. The prime outcomes are \(2\), \(3\), and \(5\), so their probability is \(0.12+0.19+0.19=0.50\). 3. The expected value is \(E(X)=\sum xP(X=x)\). Thus, \(E(X)=1\cdot0.12+2\cdot0.12+3\cdot0.19+4\cdot0.19+5\cdot0.19+6\cdot0.19=3.78\).

Answer

a) \(P(X\ge4)=0.57\) b) \(P(\text{prime})=0.50\) c) \(E(X)=3.78\)
53082312
In a manufacturing process, a sensor has probability \(p\) of having a production-related inaccuracy. Find the expected number of sensors with inaccuracies in each situation. 1. \(p=0.025\) and \(n=1200\) sensors produced in one day 2. \(p=0.008\) and \(n=15{,}000\) sensors in one weekly shipment 3. \(p=0.12\) and \(n=450\) sensors in a test run

Hints

- The expected count is the total number of trials multiplied by the success probability. - Treat a sensor with an inaccuracy as a success for this calculation. - The result is a theoretical long-run average.

Solution

Use the binomial mean formula \(E(X)=np\). 1. \(E(X)=1200\cdot0.025=30\). 2. \(E(X)=15{,}000\cdot0.008=120\). 3. \(E(X)=450\cdot0.12=54\).

Answer

1. \(30\) sensors 2. \(120\) sensors 3. \(54\) sensors
53097912
A random variable \(X\) is uniformly distributed over \(M=\{10,20,30,40,50,60,70,80,90,100\}\). Find \(E(X)\).

Hints

- Determine the probability assigned to each value. - A uniform discrete distribution has the same mean as the ordinary average of its values. - Notice the symmetry between the smallest and largest values.

Solution

1. The set \(M\) contains \(10\) values, so each value has probability \(\frac{1}{10}\). 2. The sum of the values is \(10+20+30+40+50+60+70+80+90+100=550\). 3. Therefore, \(E(X)=\frac{1}{10}\cdot550=55\). 4. Equivalently, symmetry gives the midpoint of the smallest and largest values: \(\frac{10+100}{2}=55\).

Answer

\(E(X)=55\)
53104612
The table shows the probability distribution of a binomial random variable \(X\) with \(n=3\). <table border="1"> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.064\)</td><td>\(0.288\)</td><td>\(0.432\)</td><td>\(0.216\)</td></tr> </table> a) Find \(E(X)\) directly from the table using the definition of expected value for a discrete random variable. b) Use \(E(X)\) to determine the success probability \(p\).

Hints

- Multiply each possible value by its probability and add. - Use the expected-value definition for part a). - For a binomial random variable, the mean is related to \(n\) and \(p\). - Solve the mean formula for \(p\).

Solution

1. From the table, \(E(X)=0(0.064)+1(0.288)+2(0.432)+3(0.216)=1.8\). 2. For a binomial random variable, \(E(X)=np\). Therefore, \(1.8=3p\), so \(p=0.6\).

Answer

a) \(E(X)=1.8\) b) \(p=0.6\)
53111412
A multiple-choice test has \(10\) questions. Each question has four answer choices, exactly one of which is correct. A test-taker guesses independently and uniformly at random on every question. Let \(X\) be the number of correct answers. a) Find the mean and variance of \(X\). b) Find the standard deviation and interpret it in context. c) How does the variance change if the test is expanded to \(20\) questions? Justify your answer using the binomial variance formula.

Hints

- Identify the probability of a correct guess. - Use the binomial mean and variance formulas. - Interpret the standard deviation as a typical distance from the mean. - Examine how \(n\) appears in the variance formula.

Solution

1. The random variable is binomial with \(n=10\) and \(p=\frac14=0.25\). 2. The mean is \(E(X)=np=10\cdot0.25=2.5\). 3. The variance is \(\operatorname{Var}(X)=np(1-p)=10\cdot0.25\cdot0.75=1.875\). 4. The standard deviation is \(\sigma=\sqrt{1.875}\approx1.37\). The number of correct answers typically differs from the mean of \(2.5\) by about \(1.37\) answers. 5. With \(20\) questions and the same \(p\), the variance is \(20\cdot0.25\cdot0.75=3.75\). Doubling \(n\) doubles the variance because \(p(1-p)\) is unchanged.

Answer

a) \(E(X)=2.5\) and \(\operatorname{Var}(X)=1.875\) b) \(\sigma\approx1.37\). The count of correct answers typically differs from its mean by about \(1.37\). c) The variance doubles to \(3.75\).
53112512
A binomial random variable \(X\) counts the successes in \(3\) trials with success probability \(p=0.5\). Find the mean \(\mu\), and then calculate the variance directly from the definition \(\operatorname{Var}(X)=\sum_{k=0}^{3}(k-\mu)^2P(X=k)\).

Hints

- Find the expected number of successes first. - List the binomial probability of each possible value. - Square each deviation from the mean. - Weight each squared deviation by its probability.

Solution

1. The mean is \(\mu=np=3\cdot0.5=1.5\). 2. The probabilities are \(P(X=0)=0.125\), \(P(X=1)=0.375\), \(P(X=2)=0.375\), and \(P(X=3)=0.125\). 3. The squared deviations are \((0-1.5)^2=(3-1.5)^2=2.25\) and \((1-1.5)^2=(2-1.5)^2=0.25\). 4. Therefore, \(\operatorname{Var}(X)=2.25\cdot0.125+0.25\cdot0.375+0.25\cdot0.375+2.25\cdot0.125\). 5. Thus, \(\operatorname{Var}(X)=0.28125+0.09375+0.09375+0.28125=0.75\).

Answer

\(\mu=1.5\) and \(\operatorname{Var}(X)=0.75\)
53116512
A school raffle ticket costs \(\$2.50\). The payout \(X\) has the following distribution. <table> <tr> <td>Payout \(x_i\)</td> <td>\(\$0\)</td> <td>\(\$5\)</td> <td>\(\$10\)</td> </tr> <tr> <td>Probability \(P(X=x_i)\)</td> <td>\(0.80\)</td> <td>\(0.15\)</td> <td>\(0.05\)</td> </tr> </table> a) Find the expected net gain per ticket for a player. b) Interpret this expected value for a very large number of purchased tickets. c) What ticket price would make the game fair?

Hints

- First find the expected payout. - Net gain equals payout minus ticket price. - A fair game has expected net gain \(0\). - Interpret expected value as a long-run average.

Solution

1. The expected payout is \(E(X)=0\cdot0.80+5\cdot0.15+10\cdot0.05=\$1.25\). 2. The expected net gain is \(\$1.25-\$2.50=-\$1.25\). 3. Over many purchased tickets, a player would lose an average of about \(\$1.25\) per ticket. 4. A fair ticket price equals the expected payout, so it would be \(\$1.25\).

Answer

a) \(-\$1.25\) b) Over many tickets, the player would lose an average of \(\$1.25\) per ticket. c) \(\$1.25\)
53116912
A professional biathlete hits a standing-shooting target with probability \(p=0.85\) on each shot. Assume the \(n=1200\) shots in a training season are independent. a) Find the expected number of hits. b) Explain the meaning of this value in context.

Hints

- Use the mean formula for a binomial random variable. - Multiply the number of shots by the hit probability. - Expected value is a long-run average, not a guaranteed result.

Solution

1. The number of hits is modeled as binomial with \(n=1200\) and \(p=0.85\). 2. The mean is \(\mu=np=1200\cdot0.85=1020\). 3. Over many comparable training seasons of \(1200\) shots, the average number of hits would approach \(1020\). It is not a guarantee for any one season.

Answer

a) \(1020\) hits b) Over many comparable seasons, the athlete would average about \(1020\) hits per \(1200\) shots.
53126712
A medical study includes \(250{,}000\) participants. The probabilities of four side-effect categories are shown. <table> <tr> <th>Side-effect category</th> <th>Probability \(p\)</th> </tr> <tr> <td>None</td> <td>\(0.9350\)</td> </tr> <tr> <td>Mild</td> <td>\(0.0520\)</td> </tr> <tr> <td>Moderate</td> <td>\(0.0115\)</td> </tr> <tr> <td>Severe</td> <td>\(0.0015\)</td> </tr> </table> Find the expected number of participants in each category. State the theoretical assumption required for these estimates.

Hints

- Multiply each category probability by the total number of participants. - Check that the expected counts add to the total sample size. - Consider what must be true about participants and category probabilities for the model to apply.

Solution

1. For each category, multiply the total number of participants by its probability. 2. None: \(250{,}000\cdot0.9350=233{,}750\). 3. Mild: \(250{,}000\cdot0.0520=13{,}000\). 4. Moderate: \(250{,}000\cdot0.0115=2875\). 5. Severe: \(250{,}000\cdot0.0015=375\). 6. The model assumes that participants’ outcomes are independent and that the same category probabilities apply to every participant. The four counts form a multinomial model, and each individual category count is binomial.

Answer

None: \(233{,}750\) Mild: \(13{,}000\) Moderate: \(2875\) Severe: \(375\) The estimates assume independent participant outcomes with constant category probabilities.
52346912
In an electronics manufacturing process, \(5\%\) of LEDs are defective. Assume LEDs are independently classified with this constant defect probability. A sample of \(20\) LEDs is selected, and \(X\) is the number of defective LEDs. a) Find the probability that at most \(1\) LED is defective. b) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). c) List all possible values of \(X\) in the interval \([\mu-\sigma, \mu+\sigma]\).

Hints

- Identify the binomial parameters. - “At most one” includes \(0\) and \(1\). - Use the binomial mean and standard-deviation formulas. - Remember that \(X\) can take only integer values.

Solution

1. The random variable is binomial with \(n=20\) and \(p=0.05\). 2. \(P(X\le1)=P(X=0)+P(X=1)=(0.95)^{20}+20\cdot0.05\cdot(0.95)^{19}\approx0.7359\). 3. The mean is \(\mu=np=1\). 4. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{0.95}\approx0.9747\). 5. The interval is approximately \([0.0253, 1.9747]\). Since \(X\) is an integer count, the only possible value in this interval is \(1\).

Answer

a) \(P(X\le1)\approx0.7359\) b) \(\mu=1\) and \(\sigma\approx0.9747\) c) \(1\)
52347312
A company manufactures glass bottles. Based on past data, \(4\%\) of the bottles have minor material defects. Assume bottles are independently classified with this constant defect probability. A quality-control sample of \(150\) bottles is selected. Let \(X\) be the number of bottles in the sample that have material defects. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). b) Find the probability that \(X\) is within one standard deviation of its mean.

Hints

- Identify the values of \(n\) and \(p\) for the binomial random variable. - Use the binomial formulas for the mean and standard deviation. - Translate “within one standard deviation” into an interval centered at the mean. - Remember that \(X\) can take only integer values.

Solution

1. The random variable is binomial with \(n=150\) and \(p=0.04\). 2. The mean is \(\mu=np=150\cdot0.04=6\). 3. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{150\cdot0.04\cdot0.96}=\sqrt{5.76}=2.4\). 4. Being within one standard deviation of the mean means \(6-2.4\le X\le6+2.4\), or \(3.6\le X\le8.4\). Since \(X\) is an integer count, this is \(4\le X\le8\). 5. Using the binomial distribution, \(P(4\le X\le8)=P(X\le8)-P(X\le3)\approx0.8515-0.1458=0.7057\).

Answer

a) \(\mu=6\) and \(\sigma=2.4\) b) \(P(4\le X\le8)\approx0.7057\), or about \(70.57\%\)
52347612
For a binomial random variable \(X\) with a fixed success probability \(p\), the number of trials \(n\) is changed so that the standard deviation \(\sigma(X)\) becomes four times as large. Determine how the mean \(E(X)\) changes.

Hints

- Use the formula for the standard deviation of a binomial random variable. - If a square root becomes four times as large, determine the factor by which its radicand changes. - Relate the resulting change in \(n\) to the formula for the mean. - Work backward from the change in the standard deviation.

Solution

1. The standard deviation is \(\sigma=\sqrt{np(1-p)}\). To multiply \(\sigma\) by \(4\), the quantity under the square root must be multiplied by \(16\). 2. Because \(p\) remains fixed, \(p(1-p)\) does not change. Therefore, \(n\) must be multiplied by \(16\). 3. The mean is \(E(X)=np\). With \(p\) fixed and \(n\) multiplied by \(16\), the mean is also multiplied by \(16\).

Answer

The mean is multiplied by \(16\).
52358412
Two archers, Alex and Maya, practice hitting a target. Alex hits the target with probability \(p_A=0.60\) and takes \(n_A=80\) shots. Maya hits the target with probability \(p_M=0.75\) and takes \(n_M=64\) shots. Assume each archer’s shots are independent and that each hit probability remains constant. Show that their expected numbers of hits are equal. Then find each standard deviation and determine whose number of hits varies less around the mean.

Hints

- Find each mean separately and compare the results. - Use the binomial standard-deviation formula for each archer. - A smaller standard deviation indicates less variability around the mean.

Solution

1. Alex's mean is \(\mu_A=n_Ap_A=80\cdot0.60=48\). 2. Maya's mean is \(\mu_M=n_Mp_M=64\cdot0.75=48\). Therefore, their expected numbers of hits are equal. 3. Alex's standard deviation is \(\sigma_A=\sqrt{80\cdot0.60\cdot0.40}=\sqrt{19.2}\approx4.38\). 4. Maya's standard deviation is \(\sigma_M=\sqrt{64\cdot0.75\cdot0.25}=\sqrt{12}\approx3.46\). 5. Since \(\sigma_M<\sigma_A\), Maya's number of hits varies less around the mean.

Answer

Both means are \(48\) hits. Alex has \(\sigma_A\approx4.38\), and Maya has \(\sigma_M\approx3.46\). Maya's number of hits varies less.
52359912
A basketball player makes a free throw with probability \(0.75\). During one practice session, the player takes \(40\) free throws. Let \(X\) be the number of made free throws. a) State the conditions under which \(X\) can be modeled by a binomial distribution, and identify \(n\) and \(p\). b) Find the probability that the player makes at least \(35\) free throws. c) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(X\). d) Find the probability that \(X\) lies in the interval \([\mu-\sigma, \mu+\sigma]\).

Hints

- Recall the four conditions for a binomial setting. - Use a complement to find an “at least” probability from a cumulative binomial probability. - When using the interval around the mean, remember that \(X\) can take only integer values.

Solution

1. The model is binomial if there is a fixed number of trials, each trial has two outcomes, the trials are independent, and the success probability remains constant. Here, \(n=40\) and \(p=0.75\). 2. Using the binomial distribution, \(P(X\ge35)=1-P(X\le34)\approx1-0.9567=0.0433\). 3. The mean is \(\mu=np=40\cdot0.75=30\). 4. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{40\cdot0.75\cdot0.25}=\sqrt{7.5}\approx2.7386\). 5. The interval is approximately \([30-2.7386, 30+2.7386]=[27.2614, 32.7386]\). Since \(X\) is an integer count, this corresponds to \(28\le X\le32\). 6. Therefore, \(P(28\le X\le32)=P(X\le32)-P(X\le27)\approx0.8180-0.1791=0.6389\).

Answer

a) There must be a fixed number of independent trials, two outcomes per trial, and a constant success probability; \(n=40\) and \(p=0.75\). b) \(P(X\ge35)\approx0.0433\) c) \(\mu=30\) and \(\sigma\approx2.7386\) d) \(P(28\le X\le32)\approx0.6389\)
52360512
The random variable \(X\) has a binomial distribution with \(n=10\). The mean \(E(X)\) is known to be an integer. The table shows part of the probability distribution. <table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.3487\)</td> <td>\(0.3874\)</td> <td>\(0.1937\)</td> <td>\(0.0574\)</td> <td>\(0.0112\)</td> </tr> </table> a) Determine the success probability \(p\). b) Find \(P(1\le X\le3)\). c) Find \(P(X\ge1)\).

Hints

- Use the fact that \(E(X)=np\) is an integer to list the possible values of \(p\). - Use the row entry for \(P(X=0)\) to identify which possible value of \(p\) fits. - Add the table entries that correspond to the requested interval. - For “at least one,” consider the complement.

Solution

1. Because \(E(X)=np=10p\) is an integer, \(p\) must be one of \(0,0.1,0.2,\ldots,1\). 2. The table gives \(P(X=0)=(1-p)^{10}\approx0.3487\). Testing the possible values shows that \(p=0.1\), since \((0.9)^{10}\approx0.3487\). 3. Add the listed probabilities: \(P(1\le X\le3)=0.3874+0.1937+0.0574=0.6385\). 4. Use the complement: \(P(X\ge1)=1-P(X=0)=1-0.3487=0.6513\).

Answer

a) \(p=0.1\) b) \(P(1\le X\le3)=0.6385\) c) \(P(X\ge1)=0.6513\)
52360612
A binomial random variable \(X\) has \(n=25\), and its mean is an integer. Part of its probability distribution is shown. <table> <tr> <td>\(k\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.0038\)</td> <td>\(0.0236\)</td> <td>\(0.0708\)</td> <td>\(0.1358\)</td> <td>\(0.1867\)</td> <td>\(0.1960\)</td> </tr> </table> a) Determine the success probability \(p\). b) Find \(P(2\le X\le4)\). c) Find the probability of at least one success.

Hints

- Use the fact that \(E(X)=np\) is an integer to list the possible values of \(p\). - Use \(P(X=0)\) to identify which possible value of \(p\) fits the table. - Add the table entries for \(k=2\), \(k=3\), and \(k=4\). - For “at least one,” consider the complement.

Solution

1. Because \(E(X)=np=25p\) is an integer, \(p\) must be one of \(0,0.04,0.08,\ldots,1\). 2. The table gives \(P(X=0)=(1-p)^{25}\approx0.0038\). Of the possible values, \(p=0.2\) fits because \((0.8)^{25}\approx0.0038\). 3. Add the listed probabilities: \(P(2\le X\le4)=0.0708+0.1358+0.1867=0.3933\). 4. Use the complement: \(P(X\ge1)=1-P(X=0)=1-0.0038=0.9962\).

Answer

a) \(p=0.2\) b) \(P(2\le X\le4)=0.3933\) c) \(P(X\ge1)=0.9962\)
52515212
The lifetime \(T\), in years, of an electronic component has density \(g(t)=kt^2(3-t)\) for \(0\le t\le3\), and \(g(t)=0\) otherwise. a) Show that \(g\) is a probability density function when \(k=\frac{4}{27}\). b) Find the mean lifetime \(E(T)\). c) Find the probability that a component lasts longer than two years.

Hints

- Check both nonnegativity and total area when verifying a density. - Translate “longer than two years” into the limits of an integral. - Check whether your mean is reasonable within the interval \([0, 3]\).

Solution

1. a) On \([0, 3]\), both \(t^2\) and \(3-t\) are nonnegative. Also, \(\int_0^3\frac{4}{27}(3t^2-t^3)\,\text{d}t=\frac{4}{27}\left[t^3-\frac{t^4}{4}\right]_0^3=1\). Therefore, \(g\) is a density. 2. b) \(E(T)=\int_0^3t g(t)\,\text{d}t=\frac{4}{27}\left[\frac{3t^4}{4}-\frac{t^5}{5}\right]_0^3=1.8\) years. 3. c) \(P(T>2)=\int_2^3\frac{4}{27}(3t^2-t^3)\,\text{d}t=\frac{4}{27}\left[t^3-\frac{t^4}{4}\right]_2^3=\frac{11}{27}\approx0.4074\).

Answer

a) \(g(t)\ge0\) and \(\int_0^3g(t)\,\text{d}t=1\) b) \(E(T)=1.8\) years c) \(P(T>2)=\frac{11}{27}\approx0.4074\), or about \(40.74\%\)
52671512
A spinner has four sections numbered \(1\) through \(4\). The probability of landing on a section is proportional to its number, so \(P(X=k)=ck\). The game pays \(\$2.00\) for a \(1\) or \(2\), \(\$5.00\) for a \(3\), and \(\$10.00\) for a \(4\). What entry fee should the game operator charge to earn an average profit of \(\$0.50\) per play over the long run?

Hints

- The probabilities of all four outcomes must add to \(1\). - Find the expected payout by multiplying each payout by its probability and adding. - The operator’s profit is the entry fee minus the payout.

Solution

1. Since the probabilities must add to \(1\), \(c(1+2+3+4)=10c=1\), so \(c=0.1\). 2. Therefore, \(P(X=1)=0.1\), \(P(X=2)=0.2\), \(P(X=3)=0.3\), and \(P(X=4)=0.4\). 3. The expected payout is \(0.1\cdot\$2.00+0.2\cdot\$2.00+0.3\cdot\$5.00+0.4\cdot\$10.00=\$6.10\). 4. To earn an average profit of \(\$0.50\), the operator must charge \(\$6.10+\$0.50=\$6.60\).

Answer

The entry fee should be \(\$6.60\).
52671612
In a game, a fair coin is tossed \(5\) times. The random variable \(X\) is the number of heads. The payout in dollars is \(A=X^2\), and the game costs \(\$8.00\) to play. Determine whether the game is favorable to the player in the long run. Find the expected gain or loss per play.

Hints

- Identify the probability distribution for the number of heads. - For each possible value of \(X\), square it to find the payout. - Compare the expected payout with the entry fee. - A game is favorable to the player when the expected net gain is positive.

Solution

1. The random variable \(X\) is binomial with \(n=5\) and \(p=0.5\). 2. Its probabilities for \(k=0,1,2,3,4,5\) are \(\frac{1}{32},\frac{5}{32},\frac{10}{32},\frac{10}{32},\frac{5}{32},\frac{1}{32}\). 3. The expected payout is \(E(X^2)=0^2\cdot\frac{1}{32}+1^2\cdot\frac{5}{32}+2^2\cdot\frac{10}{32}+3^2\cdot\frac{10}{32}+4^2\cdot\frac{5}{32}+5^2\cdot\frac{1}{32}=\frac{240}{32}=\$7.50\). 4. The expected net gain is \(\$7.50-\$8.00=-\$0.50\). Thus, the player loses an average of \(\$0.50\) per play over the long run.

Answer

The game is not favorable to the player. The expected result is a loss of \(\$0.50\) per play.
52672612
A random variable \(Y\) has the following probability distribution: <table> <tr> <td>\(y_i\)</td> <td>\(-2\)</td> <td>\(-1\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> </tr> <tr> <td>\(P(Y=y_i)\)</td> <td>\(0.15\)</td> <td>\(0.25\)</td> <td>\(0.30\)</td> <td>\(0.20\)</td> <td>\(0.10\)</td> </tr> </table> Find the variance \(\operatorname{Var}(Y)\) and standard deviation \(\sigma(Y)\). Round the standard deviation to the nearest hundredth.

Hints

- Find the mean before calculating the variance. - Be careful with signs when subtracting a negative mean. - Take the square root of the variance to return to the original units.

Solution

1. First find the mean: \(\mu=-2\cdot0.15-1\cdot0.25+0\cdot0.30+1\cdot0.20+2\cdot0.10=-0.15\). 2. The variance is \(\operatorname{Var}(Y)=\sum (y_i-\mu)^2P(Y=y_i)\). 3. Substitute the values: \(\operatorname{Var}(Y)=(-1.85)^2\cdot0.15+(-0.85)^2\cdot0.25+(0.15)^2\cdot0.30+(1.15)^2\cdot0.20+(2.15)^2\cdot0.10\). 4. Therefore, \(\operatorname{Var}(Y)=0.513375+0.180625+0.00675+0.2645+0.46225=1.4275\). 5. The standard deviation is \(\sigma(Y)=\sqrt{1.4275}\approx1.19\).

Answer

\(\operatorname{Var}(Y)=1.4275\) and \(\sigma(Y)\approx1.19\)
52672712
A discrete random variable \(X\) can take the values \(-10\), \(0\), \(10\), and \(20\). Three of the probabilities are shown below: <table> <tr> <td>\(x_i\)</td> <td>\(-10\)</td> <td>\(0\)</td> <td>\(10\)</td> <td>\(20\)</td> </tr> <tr> <td>\(P(X=x_i)\)</td> <td>\(0.1\)</td> <td>\(0.5\)</td> <td></td> <td>\(0.1\)</td> </tr> </table> a) Find \(P(X=10)\). b) Find the mean \(E(X)\). c) Find the standard deviation \(\sigma(X)\).

Hints

- Use the fact that all probabilities must add to \(1\). - Find the probability-weighted average of the possible values. - Use the relationship between variance and standard deviation. - You may calculate the variance from squared deviations or from \(E(X^2)-[E(X)]^2\).

Solution

1. The probabilities must sum to \(1\), so \(P(X=10)=1-(0.1+0.5+0.1)=0.3\). 2. The mean is \(E(X)=-10\cdot0.1+0\cdot0.5+10\cdot0.3+20\cdot0.1=-1+0+3+2=4\). 3. The variance is \(\operatorname{Var}(X)=(-10-4)^2\cdot0.1+(0-4)^2\cdot0.5+(10-4)^2\cdot0.3+(20-4)^2\cdot0.1\). 4. Thus, \(\operatorname{Var}(X)=196\cdot0.1+16\cdot0.5+36\cdot0.3+256\cdot0.1=19.6+8+10.8+25.6=64\). 5. Therefore, \(\sigma(X)=\sqrt{64}=8\).

Answer

a) \(P(X=10)=0.3\) b) \(E(X)=4\) c) \(\sigma(X)=8\)
52672812
The random variable \(Y\) takes the values \(2\), \(5\), and \(k\) with the probabilities shown below: <table> <tr> <td>\(y_i\)</td> <td>\(2\)</td> <td>\(5\)</td> <td>\(k\)</td> </tr> <tr> <td>\(P(Y=y_i)\)</td> <td>\(0.4\)</td> <td>\(0.4\)</td> <td>\(0.2\)</td> </tr> </table> The mean is \(E(Y)=4.4\). a) Find \(k\). b) Find \(\operatorname{Var}(Y)\).

Hints

- Use the mean formula to write an equation containing \(k\). - Complete the distribution after finding the missing value. - Use either squared deviations from the mean or the second-moment formula for the variance.

Solution

1. Use the mean to write \(2\cdot0.4+5\cdot0.4+k\cdot0.2=4.4\). 2. Simplify and solve: \(0.8+2+0.2k=4.4\), so \(0.2k=1.6\) and \(k=8\). 3. Find the second moment: \(E(Y^2)=2^2\cdot0.4+5^2\cdot0.4+8^2\cdot0.2=1.6+10+12.8=24.4\). 4. Therefore, \(\operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2=24.4-(4.4)^2=24.4-19.36=5.04\).

Answer

a) \(k=8\) b) \(\operatorname{Var}(Y)=5.04\)
52673012
A bag contains \(6\) balls: \(4\) green balls labeled \(2\) and \(2\) yellow balls labeled \(5\). Two balls are drawn one at a time with replacement. Let \(S\) be the sum of the two numbers drawn. Find \(E(S)\) and \(\sigma(S)\).

Hints

- List all possible sums from the two draws. - Use a tree diagram or ordered pairs to find the probability of each sum. - With replacement, the color probabilities remain the same on both draws. - Find the mean before calculating the variance.

Solution

1. The possible sums are \(4\), \(7\), and \(10\). 2. Their probabilities are \(P(S=4)=\frac{4}{6}\cdot\frac{4}{6}=\frac{4}{9}\), \(P(S=7)=2\cdot\frac{4}{6}\cdot\frac{2}{6}=\frac{4}{9}\), and \(P(S=10)=\frac{2}{6}\cdot\frac{2}{6}=\frac{1}{9}\). 3. The mean is \(E(S)=4\cdot\frac{4}{9}+7\cdot\frac{4}{9}+10\cdot\frac{1}{9}=6\). 4. The variance is \(\operatorname{Var}(S)=(4-6)^2\cdot\frac{4}{9}+(7-6)^2\cdot\frac{4}{9}+(10-6)^2\cdot\frac{1}{9}=4\). 5. Therefore, \(\sigma(S)=\sqrt{4}=2\).

Answer

\(E(S)=6\) and \(\sigma(S)=2\)
52673712
One of the six words in the sentence “Statistics is a branch of mathematics” is selected at random. Let \(X\) be the number of letters in the selected word, and let \(Y\) be the number of vowels \((a, e, i, o, u)\) in the word. Find the mean and variance of both random variables.

Hints

- List the value of each random variable for all six words. - Each word has probability \(\frac{1}{6}\). - Use the formulas for the mean and variance of a discrete random variable. - The identity \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2\) may simplify the work.

Solution

1. The six equally likely word lengths are \(10,2,1,6,2,11\). 2. The mean of \(X\) is \(E(X)=\frac{10+2+1+6+2+11}{6}=\frac{16}{3}\approx5.33\). 3. Since \(E(X^2)=\frac{10^2+2^2+1^2+6^2+2^2+11^2}{6}=\frac{133}{3}\), the variance is \(\operatorname{Var}(X)=\frac{133}{3}-\left(\frac{16}{3}\right)^2=\frac{143}{9}\approx15.89\). 4. The six vowel counts are \(3,1,1,1,1,4\). 5. The mean of \(Y\) is \(E(Y)=\frac{3+1+1+1+1+4}{6}=\frac{11}{6}\approx1.83\). 6. Since \(E(Y^2)=\frac{3^2+1^2+1^2+1^2+1^2+4^2}{6}=\frac{29}{6}\), the variance is \(\operatorname{Var}(Y)=\frac{29}{6}-\left(\frac{11}{6}\right)^2=\frac{53}{36}\approx1.47\).

Answer

\(E(X)=\frac{16}{3}\approx5.33\) and \(\operatorname{Var}(X)=\frac{143}{9}\approx15.89\) \(E(Y)=\frac{11}{6}\approx1.83\) and \(\operatorname{Var}(Y)=\frac{53}{36}\approx1.47\)
52673812
An inventory system contains five items with identification codes A10, B200, C30, D4000, and E50. One item is selected at random for inspection. Let \(X\) be the number of digits in the selected code, and let \(Y\) be the sum of those digits. Find the mean and variance of \(X\) and \(Y\).

Hints

- Record the values of \(X\) and \(Y\) for each identification code. - Because all items are equally likely, average the five values for each random variable. - For \(Y\), add only the digits in each code. - Use the second moment or squared deviations to calculate each variance.

Solution

1. The five equally likely \((X, Y)\) pairs are A10: \((2, 1)\), B200: \((3, 2)\), C30: \((2, 3)\), D4000: \((4, 4)\), and E50: \((2, 5)\). 2. For \(X\), \(E(X)=\frac{2+3+2+4+2}{5}=2.6\) and \(E(X^2)=\frac{2^2+3^2+2^2+4^2+2^2}{5}=7.4\). 3. Therefore, \(\operatorname{Var}(X)=7.4-(2.6)^2=0.64\). 4. For \(Y\), \(E(Y)=\frac{1+2+3+4+5}{5}=3\) and \(E(Y^2)=\frac{1^2+2^2+3^2+4^2+5^2}{5}=11\). 5. Therefore, \(\operatorname{Var}(Y)=11-3^2=2\).

Answer

\(E(X)=2.6\) and \(\operatorname{Var}(X)=0.64\) \(E(Y)=3\) and \(\operatorname{Var}(Y)=2\)
52674212
A random variable \(Z\) can take only the values \(0\) and \(5\). Its expected value is \(E(Z)=3.5\). a) Determine the probability distribution of \(Z\), including \(P(Z=0)\) and \(P(Z=5)\). b) Interpret \(3.5\) in terms of a very large number of repetitions of the experiment.

Hints

- The probabilities in a distribution must add to \(1\). - Let one probability be a variable and express the other in terms of it. - Use the expected-value formula to form an equation.

Solution

1. Let \(p=P(Z=5)\). Then \(P(Z=0)=1-p\). 2. Use the expected-value equation: \(0(1-p)+5p=3.5\). 3. Solving gives \(5p=3.5\), so \(p=0.7\). 4. Therefore, \(P(Z=5)=0.7\) and \(P(Z=0)=0.3\). 5. Over many repetitions, the average of the observed values will tend to be close to \(3.5\).

Answer

a) \(P(Z=0)=0.3\) and \(P(Z=5)=0.7\) b) Over many repetitions, the average outcome will approach \(3.5\).
52674312
A bakery records the number of muffins purchased by each customer on a Monday morning. Let \(X\) be the number of muffins purchased by a randomly selected customer. The probability distribution is shown below: <table> <tr> <td>\(x_i\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> </tr> <tr> <td>\(P(X=x_i)\)</td> <td>\(0.2\)</td> <td>\(0.5\)</td> <td>\(0.2\)</td> <td>\(0.1\)</td> </tr> </table> a) Find \(E(X)\) and interpret it in context. b) Find \(\sigma(X)\). c) Find the probability that a randomly selected customer buys more muffins than the mean.

Hints

- Find the probability-weighted average of the possible values. - Calculate the variance before taking its square root. - Identify the integer values of \(X\) that are greater than the mean. - Interpret the standard deviation as the typical distance from the mean.

Solution

1. The mean is \(E(X)=0\cdot0.2+1\cdot0.5+2\cdot0.2+3\cdot0.1=1.2\). In the long run, customers buy an average of \(1.2\) muffins each. 2. The variance is \(\operatorname{Var}(X)=(0-1.2)^2\cdot0.2+(1-1.2)^2\cdot0.5+(2-1.2)^2\cdot0.2+(3-1.2)^2\cdot0.1=0.76\). 3. Therefore, \(\sigma(X)=\sqrt{0.76}\approx0.8718\). 4. Since the mean is \(1.2\), buying more than the mean means buying \(2\) or \(3\) muffins. Thus, \(P(X>1.2)=0.2+0.1=0.3\).

Answer

a) \(E(X)=1.2\). Customers buy an average of \(1.2\) muffins each. b) \(\sigma(X)\approx0.8718\) c) \(P(X>1.2)=0.3\)
52674412
During a grocery-store promotion, each participating customer spins a prize wheel and receives a discount coupon. Assume every coupon is redeemed in full. Let \(Y\) be the coupon value in dollars. The probability distribution is partially shown below: <table> <tr> <td>\(y_i\) (dollars)</td> <td>\(0\)</td> <td>\(2\)</td> <td>\(5\)</td> <td>\(10\)</td> </tr> <tr> <td>\(P(Y=y_i)\)</td> <td>\(p\)</td> <td>\(0.4\)</td> <td>\(0.2\)</td> <td>\(0.1\)</td> </tr> </table> a) Find \(p\). b) Find \(E(Y)\). What total coupon cost should the store expect for \(500\) participating customers? c) Find \(\operatorname{Var}(Y)\) and \(\sigma(Y)\).

Hints

- Use the fact that all probabilities in a distribution sum to \(1\). - Multiply the expected value for one customer by the number of customers. - Use \(\operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2\).

Solution

1. The probabilities must sum to \(1\), so \(p=1-(0.4+0.2+0.1)=0.3\). 2. The expected coupon value is \(E(Y)=0\cdot0.3+2\cdot0.4+5\cdot0.2+10\cdot0.1=2.8\), or \(\$2.80\) per customer. 3. For \(500\) customers, the expected total cost is \(500\cdot\$2.80=\$1400\). 4. The second moment is \(E(Y^2)=0^2\cdot0.3+2^2\cdot0.4+5^2\cdot0.2+10^2\cdot0.1=16.6\). 5. Therefore, \(\operatorname{Var}(Y)=16.6-(2.8)^2=8.76\) square dollars, and \(\sigma(Y)=\sqrt{8.76}\approx\$2.96\).

Answer

a) \(p=0.3\) b) \(E(Y)=\$2.80\), and the expected total cost is \(\$1400\). c) \(\operatorname{Var}(Y)=8.76\) square dollars and \(\sigma(Y)\approx\$2.96\)
52675312
A random variable \(X\) can take the values \(0\), \(4\), and \(10\). Its probability distribution is partially shown below: <table> <tr> <td>\(x_i\)</td> <td>\(0\)</td> <td>\(4\)</td> <td>\(10\)</td> </tr> <tr> <td>\(P(X=x_i)\)</td> <td>\(p_1\)</td> <td>\(0.3\)</td> <td>\(p_2\)</td> </tr> </table> The mean is \(E(X)=5.2\). a) Find \(p_1\) and \(p_2\). b) Find \(\operatorname{Var}(X)\).

Hints

- Use the fact that all probabilities must sum to \(1\). - Write an equation from the given mean. - Because one possible value is \(0\), one unknown drops out of the mean equation. - Use either squared deviations or the second-moment formula for the variance.

Solution

1. Since the probabilities sum to \(1\), \(p_1+0.3+p_2=1\), so \(p_1+p_2=0.7\). 2. The mean gives \(0\cdot p_1+4\cdot0.3+10p_2=5.2\). 3. Solving, \(1.2+10p_2=5.2\), so \(p_2=0.4\). 4. Therefore, \(p_1=0.7-0.4=0.3\). 5. The second moment is \(E(X^2)=0^2\cdot0.3+4^2\cdot0.3+10^2\cdot0.4=44.8\). 6. Thus, \(\operatorname{Var}(X)=44.8-(5.2)^2=17.76\).

Answer

a) \(p_1=0.3\) and \(p_2=0.4\) b) \(\operatorname{Var}(X)=17.76\)
52675412
A random variable \(Y\) can take the values \(-2\), \(0\), and \(4\). It is known that \(P(Y=0)=0.4\) and that the mean is \(\mu=-0.6\). a) Find \(P(Y=-2)\) and \(P(Y=4)\). b) Find the standard deviation \(\sigma\) of \(Y\).

Hints

- Write one equation using the sum of the probabilities. - Use the given mean to write a second equation. - Solve the two equations for the unknown probabilities. - Find the variance before taking its square root.

Solution

1. Let \(p_1=P(Y=-2)\) and \(p_2=P(Y=4)\). 2. Since the probabilities sum to \(1\), \(p_1+p_2=0.6\). 3. The mean gives \(-2p_1+4p_2=-0.6\). 4. Solving the system gives \(p_2=0.1\) and \(p_1=0.5\). 5. The second moment is \(E(Y^2)=(-2)^2\cdot0.5+0^2\cdot0.4+4^2\cdot0.1=3.6\). 6. The variance is \(\operatorname{Var}(Y)=3.6-(-0.6)^2=3.24\), so \(\sigma=\sqrt{3.24}=1.8\).

Answer

a) \(P(Y=-2)=0.5\) and \(P(Y=4)=0.1\) b) \(\sigma=1.8\)
52676512
A drawer contains \(5\) USB drives. Two contain an important presentation, and the other three are empty. Drives are selected at random and checked one at a time without replacement. The random variable \(X\) is the number of drives checked before the first drive containing the presentation is found. a) Give the probability distribution of \(X\) in a table. b) Find \(E(X)\).

Hints

- List the possible numbers of drives that might need to be checked. - The fourth checked drive must contain the presentation if the first three are empty. - Multiply conditional probabilities along each without-replacement path. - To find the mean, multiply each value by its probability and add.

Solution

1. Since there are only \(3\) empty drives, the first drive with the presentation must be found by the fourth check. Thus, \(X\in\{1,2,3,4\}\). 2. The probabilities are \(P(X=1)=\frac{2}{5}=0.4\), \(P(X=2)=\frac{3}{5}\cdot\frac{2}{4}=0.3\), \(P(X=3)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=0.2\), \(P(X=4)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3}\cdot\frac{2}{2}=0.1\). 3. Therefore, \(E(X)=1\cdot0.4+2\cdot0.3+3\cdot0.2+4\cdot0.1=2\).

Answer

a) <table border="1"> <tr> <td>\(k\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(0.4\)</td> <td>\(0.3\)</td> <td>\(0.2\)</td> <td>\(0.1\)</td> </tr> </table> b) \(E(X)=2\)
52676612
A technician inspects a shipment of \(7\) components, including \(3\) defective components. The components are selected at random and tested one at a time without replacement until the first defective component is found. The random variable \(X\) is the number of tests required. a) What is the greatest possible number of tests? b) What is the probability that a defective component is found on the first or second test? c) Find the expected number of tests.

Hints

- Consider how many nondefective components could be tested before a defective one is guaranteed. - Find the separate probabilities for success on the first and second tests. - Build the full distribution before calculating the expected value.

Solution

1. In the worst case, all \(4\) nondefective components are tested first, so the fifth component must be defective. The maximum is \(5\) tests. 2. \(P(X=1)=\frac{3}{7}\), and \(P(X=2)=\frac{4}{7}\cdot\frac{3}{6}=\frac{2}{7}\). Therefore, \(P(X\le2)=\frac{3}{7}+\frac{2}{7}=\frac{5}{7}\approx0.714\). 3. The remaining probabilities are \(P(X=3)=\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{3}{5}=\frac{6}{35}\), \(P(X=4)=\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{35}\), and \(P(X=5)=\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{2}{5}\cdot\frac{1}{4}=\frac{1}{35}\). 4. Thus, \(E(X)=1\cdot\frac{15}{35}+2\cdot\frac{10}{35}+3\cdot\frac{6}{35}+4\cdot\frac{3}{35}+5\cdot\frac{1}{35}=2\).

Answer

a) \(5\) tests b) \(P(X\le2)=\frac{5}{7}\approx0.714\) c) \(E(X)=2\) tests
52677712
A discrete random variable \(X\) can take the values \(0\), \(2\), \(4\), and \(6\). Its probability mass function is \(P(X=x)=k(x+1)\), where \(k\) is a positive constant. a) Find \(k\). b) Find \(E(X)\).

Hints

- The probabilities of all possible values must add to \(1\). - Substitute each possible value of \(X\) into the probability formula. - To find the mean, multiply each value by its probability and add.

Solution

1. The probabilities must add to \(1\), so \(k(0+1)+k(2+1)+k(4+1)+k(6+1)=1\). 2. Thus, \(k(1+3+5+7)=16k=1\), giving \(k=\frac{1}{16}=0.0625\). 3. The probabilities are \(\frac{1}{16}\), \(\frac{3}{16}\), \(\frac{5}{16}\), and \(\frac{7}{16}\). 4. Therefore, \(E(X)=0\cdot\frac{1}{16}+2\cdot\frac{3}{16}+4\cdot\frac{5}{16}+6\cdot\frac{7}{16}=\frac{68}{16}=4.25\).

Answer

a) \(k=\frac{1}{16}=0.0625\) b) \(E(X)=4.25\)
52677812
The random variable \(Y\) can take the values \(1\), \(2\), \(3\), and \(4\), with \(P(Y=1)=c\), \(P(Y=2)=2c\), \(P(Y=3)=0.3\), and \(P(Y=4)=0.1\). Find \(c\), and then calculate \(E(Y)\).

Hints

- The probabilities of all possible outcomes must add to \(1\). - Use that fact to write an equation for \(c\). - After finding all probabilities, multiply each value by its probability and add.

Solution

1. The probabilities must add to \(1\): \(c+2c+0.3+0.1=1\). 2. Therefore, \(3c=0.6\), so \(c=0.2\). 3. The complete distribution has probabilities \(0.2\), \(0.4\), \(0.3\), and \(0.1\). 4. Thus, \(E(Y)=1\cdot0.2+2\cdot0.4+3\cdot0.3+4\cdot0.1=2.3\).

Answer

\(c=0.2\) \(E(Y)=2.3\)
52678112
A random variable \(X\) can take the values \(10\) and \(20\). Let \(P(X=10)=p\), where \(0\le p\le1\). a) Write \(\operatorname{Var}(X)\) as a function of \(p\). b) Find the values of \(p\) for which \(\operatorname{Var}(X)=0\). c) Interpret the results from part b) in terms of the distribution of \(X\).

Hints

- Find the mean in terms of \(p\). - Use the definition of variance as a probability-weighted sum of squared deviations. - Determine when the nonnegative product can equal zero. - Interpret what it means for one outcome to have probability \(1\).

Solution

1. Since \(P(X=20)=1-p\), the mean is \(E(X)=10p+20(1-p)=20-10p\). 2. The variance is \(\operatorname{Var}(X)=p[10-(20-10p)]^2+(1-p)[20-(20-10p)]^2\). 3. Simplifying gives \(\operatorname{Var}(X)=100p(1-p)^2+100p^2(1-p)=100p(1-p)\). 4. Setting the variance equal to zero gives \(100p(1-p)=0\), so \(p=0\) or \(p=1\). 5. If \(p=0\), then \(X=20\) with probability \(1\). If \(p=1\), then \(X=10\) with probability \(1\). In either case, the random variable is constant.

Answer

a) \(\operatorname{Var}(X)=100p(1-p)\) b) \(p=0\) or \(p=1\) c) The variance is zero exactly when \(X\) is constant: \(X=20\) always when \(p=0\), or \(X=10\) always when \(p=1\).
52678312
A student claims, “The numerical value of a standard deviation \(\sigma\) can never be greater than the numerical value of its variance \(V\), because \(\sigma=\sqrt{V}\) and taking a square root always makes a number smaller.” a) Evaluate the claim. For \(V>0\), state the condition under which \(\sigma>V\). b) Let \(X\) take the values \(0\) and \(1\), each with probability \(0.5\). Find \(\operatorname{Var}(X)\) and \(\sigma(X)\), and use them as a counterexample.

Hints

- Examine what the square-root function does to numbers between \(0\) and \(1\). - Determine when a positive number is greater than its square. - Use the discrete variance formula for the example. - One counterexample is enough to disprove a universal claim.

Solution

1. Taking a square root makes a number smaller only when the number is greater than \(1\). For \(0<V<1\), \(\sqrt{V}>V\). 2. Since \(\sigma=\sqrt{V}\), the inequality \(\sigma>V\) holds exactly when \(0<V<1\). 3. For the example, \(E(X)=0\cdot0.5+1\cdot0.5=0.5\). 4. The variance is \(\operatorname{Var}(X)=(0-0.5)^2\cdot0.5+(1-0.5)^2\cdot0.5=0.25\). 5. The standard deviation is \(\sigma(X)=\sqrt{0.25}=0.5\). Since \(0.5>0.25\), the example disproves the claim.

Answer

a) The claim is false. The numerical inequality \(\sigma>V\) holds when \(0<V<1\). b) \(\operatorname{Var}(X)=0.25\) and \(\sigma(X)=0.5\), so \(\sigma(X)>\operatorname{Var}(X)\).
52678712
A spinner is divided into four sections labeled with payouts of \(\$0\), \(\$2\), \(\$5\), and \(\$9\). a) Find the entry fee that makes the game fair when all four sections have equal area. b) In a new version, the entry fee is \(\$3.50\). Give one possible set of central angles for the four sections that keeps the game fair. Explain your reasoning.

Hints

- A fair game has an entry fee equal to the expected payout. - A section’s probability is its central angle divided by \(360^\circ\). - In part b, there are several correct answers because there are more unknown probabilities than equations. - Check that the probabilities add to \(1\).

Solution

1. With four equal sections, each payout has probability \(\frac{1}{4}\). The expected payout is \(0\cdot\frac{1}{4}+2\cdot\frac{1}{4}+5\cdot\frac{1}{4}+9\cdot\frac{1}{4}=\$4.00\). Therefore, a fair entry fee is \(\$4.00\). 2. For part b, let the probabilities for \(\$0\), \(\$2\), \(\$5\), and \(\$9\) be \(p_0,p_2,p_5,p_9\). Fairness requires \(2p_2+5p_5+9p_9=3.5\), and the probabilities must add to \(1\). 3. One choice is \(p_9=0.20\) and \(p_5=0.20\). Then \(2p_2+5(0.20)+9(0.20)=3.5\), so \(p_2=0.35\), and \(p_0=0.25\). 4. Multiplying each probability by \(360^\circ\) gives angles of \(90^\circ\), \(126^\circ\), \(72^\circ\), and \(72^\circ\), respectively.

Answer

a) \(\$4.00\) b) One possible set of central angles is \(90^\circ\) for \(\$0\), \(126^\circ\) for \(\$2\), \(72^\circ\) for \(\$5\), and \(72^\circ\) for \(\$9\).
52678812
A prize box contains tickets worth \(\$0\), \(\$10\), or \(\$50\). a) Initially, the probabilities of drawing the three ticket values are \(70\%\), \(20\%\), and \(10\%\), respectively. Find the fair price of one ticket. b) The box is refilled with exactly \(40\) tickets, and the price of one ticket is set at \(\$10.00\). Give one possible number of tickets at each prize value that makes the game fair. There must be at least one ticket of each value.

Hints

- A fair price equals the expected payout. - In part b, use ticket counts divided by \(40\) as probabilities. - Write one equation for the total number of tickets and another for the expected payout. - Ticket counts must be positive integers.

Solution

1. The expected payout in part a is \(0\cdot0.70+10\cdot0.20+50\cdot0.10=\$7.00\). Thus, the fair ticket price is \(\$7.00\). 2. Let \(n_0,n_{10},n_{50}\) be the numbers of tickets at each value. Then \(n_0+n_{10}+n_{50}=40\), and fairness requires \(\frac{10n_{10}+50n_{50}}{40}=10\). 3. The second equation simplifies to \(n_{10}+5n_{50}=40\). Choose \(n_{50}=4\), which gives \(n_{10}=20\). 4. Then \(n_0=40-20-4=16\). All three counts are positive and total \(40\).

Answer

a) \(\$7.00\) b) One possible distribution is \(16\) tickets worth \(\$0\), \(20\) tickets worth \(\$10\), and \(4\) tickets worth \(\$50\).
52679212
A game consists of tossing a fair coin \(3\) times and costs \(\$3.00\) to play. A player receives \(\$x\) for \(3\) heads, \(\$4.00\) for exactly \(2\) heads, and no payout otherwise. Find \(x\) so that the game is fair.

Hints

- Find the probabilities of \(3\) heads and exactly \(2\) heads. - Write the expected payout as a weighted sum. - For a fair game, set the expected payout equal to the entry fee. - Solve the resulting equation for \(x\).

Solution

1. The probability of \(3\) heads is \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\). 2. The probability of exactly \(2\) heads is \(\binom{3}{2}\left(\frac{1}{2}\right)^3=\frac{3}{8}\). 3. The expected payout is \(\frac{x}{8}+4\cdot\frac{3}{8}=\frac{x+12}{8}\). 4. For a fair game, set the expected payout equal to the \(\$3.00\) entry fee: \(\frac{x+12}{8}=3\). 5. Solving gives \(x+12=24\), so \(x=12\).

Answer

\(x=12\), so the payout for \(3\) heads should be \(\$12.00\).
52679312
A company analyzes the possible profit from a new project. Let \(X\) be the profit in thousands of dollars. The probability distribution is shown below: <table> <tr><td>Profit \(x_i\)</td><td>\(-5\)</td><td>\(0\)</td><td>\(10\)</td><td>\(20\)</td></tr> <tr><td>\(P(X=x_i)\)</td><td>\(0.2\)</td><td>\(0.3\)</td><td>\(0.4\)</td><td>\(0.1\)</td></tr> </table> Calculate \(S_1=\sum_{i=1}^{4}x_iP(X=x_i)\) and \(S_2=\sum_{i=1}^{4}(x_i-S_1)^2P(X=x_i)\). Interpret both quantities in the context of the project's profit.

Hints

- Identify the quantity found by multiplying each value by its probability and adding. - Interpret each difference \(x_i-S_1\) as a deviation from the mean. - Recall the name of the probability-weighted average of squared deviations. - Use “long-run average” and “variability” in the interpretation.

Solution

1. \(S_1=-5\cdot0.2+0\cdot0.3+10\cdot0.4+20\cdot0.1=-1+0+4+2=5\) thousand dollars. 2. \(S_2=(-5-5)^2\cdot0.2+(0-5)^2\cdot0.3+(10-5)^2\cdot0.4+(20-5)^2\cdot0.1\). 3. Therefore, \(S_2=100\cdot0.2+25\cdot0.3+25\cdot0.4+225\cdot0.1=60\,\text{(thousand dollars)}^2\). 4. The quantity \(S_1\) is the expected profit, so the company expects an average profit of \(\$5000\) per project in the long run. The quantity \(S_2\) is the variance, which measures how widely the possible profits vary around that mean.

Answer

\(S_1=5\) thousand dollars and \(S_2=60\,\text{(thousand dollars)}^2\). Thus, \(S_1\) is the expected profit of \(\$5000\), and \(S_2\) is the variance, a measure of financial variability.
52679412
A game has three possible payouts: \(\$0\), \(\$5\), and \(\$a\). Their probabilities are \(P(X=0)=0.5\), \(P(X=5)=0.4\), and \(P(X=a)=0.1\). Find \(a\) so that the expected payout is exactly \(\$4\). Then find \(\operatorname{Var}(X)\).

Hints

- Write an expected-value equation using the three payouts and their probabilities. - The probabilities already sum to \(1\). - After finding the missing payout, use squared deviations from the mean. - Weight each squared deviation by its probability.

Solution

1. Use the expected-value equation \(0\cdot0.5+5\cdot0.4+a\cdot0.1=4\). 2. Solving, \(2+0.1a=4\), so \(a=20\). The third payout is \(\$20\). 3. Using the mean \(4\), the variance is \(\operatorname{Var}(X)=(0-4)^2\cdot0.5+(5-4)^2\cdot0.4+(20-4)^2\cdot0.1\). 4. Therefore, \(\operatorname{Var}(X)=16\cdot0.5+1\cdot0.4+256\cdot0.1=34\) square dollars.

Answer

\(a=20\), and \(\operatorname{Var}(X)=34\) square dollars.
52681012
A random variable \(X\) takes the value \(1\) with probability \(p\) and the value \(0\) with probability \(1-p\). Determine the value of \(p\in[0, 1]\) for which the standard deviation \(\sigma(X)\) is greatest.

Hints

- Write the variance as a function of \(p\). - Explain why maximizing the variance also maximizes the standard deviation. - Use a method for finding the maximum of a quadratic function. - Remember that a probability must lie between \(0\) and \(1\).

Solution

1. The variance is \(V(p)=p-p^2=p(1-p)\). 2. Because the square-root function is increasing on nonnegative inputs, \(\sigma(p)=\sqrt{V(p)}\) is greatest at the same value of \(p\) as \(V(p)\). 3. Differentiate: \(V'(p)=1-2p\). Setting \(V'(p)=0\) gives \(p=0.5\). 4. Since \(V''(p)=-2<0\), this critical point is a local maximum. 5. Also, \(V(0)=V(1)=0\) and \(V(0.5)=0.25\), so \(p=0.5\) gives the global maximum on \([0, 1]\).

Answer

\(p=0.5\)
52681512
A discrete random variable \(X\) has possible values \(\{2, 4, 6\}\). It is known that \(E(X)=4.2\) and \(\operatorname{Var}(X)=1.16\). Find \(P(X=2)\), \(P(X=4)\), and \(P(X=6)\).

Hints

- Use the fact that the three probabilities sum to \(1\). - Write one equation from the mean. - Use \(E(X^2)=\operatorname{Var}(X)+[E(X)]^2\) for a third equation. - Three unknown probabilities require three independent equations.

Solution

1. Let \(p_1=P(X=2)\), \(p_2=P(X=4)\), and \(p_3=P(X=6)\). The probabilities satisfy \(p_1+p_2+p_3=1\). 2. The mean gives \(2p_1+4p_2+6p_3=4.2\). 3. Since \(E(X^2)=\operatorname{Var}(X)+[E(X)]^2=1.16+(4.2)^2=18.8\), the second-moment equation is \(4p_1+16p_2+36p_3=18.8\). 4. Solving the system gives \(p_3=0.2\), \(p_2=0.7\), and \(p_1=0.1\). 5. The values are valid probabilities and sum to \(1\).

Answer

\(P(X=2)=0.1\), \(P(X=4)=0.7\), and \(P(X=6)=0.2\)
52681912
A sensor costs a manufacturer \(\$15\) to produce. Historically, \(94\%\) of the sensors work properly. If a customer receives a defective sensor, the company sends a free replacement. Producing and performing an additional quality check on the replacement costs \(\$18\). What selling price gives the company an expected profit of \(\$4\) per sensor sold?

Hints

- Find the company’s cost when the original sensor works. - Find the total cost when a replacement is required. - Use the defect probability to calculate the expected cost. - Selling price minus expected cost equals expected profit.

Solution

1. A sensor works properly with probability \(0.94\) and is defective with probability \(0.06\). 2. If the original sensor works, the cost is \(\$15\). If it is defective, the total cost is \(\$15+\$18=\$33\). 3. The expected cost is \(0.94\cdot\$15+0.06\cdot\$33=\$16.08\). 4. Let the selling price be \(x\). To earn an expected profit of \(\$4\), solve \(x-\$16.08=\$4\). 5. Therefore, \(x=\$20.08\).

Answer

The selling price should be \(\$20.08\).
52682012
An online printing company sells photo books for \(\$14.50\). The variable production cost is \(\$8.00\) per book. When a customer files a valid complaint, the company refunds the full purchase price and also incurs \(\$5.00\) in complaint-handling costs. The current complaint rate is \(4\%\). a) Find the company’s expected profit per order. b) What is the greatest complaint rate that keeps the expected profit from falling below \(\$5.00\) per order?

Hints

- Find the profit in the complaint and no-complaint cases separately. - A refund removes the sales revenue, but the production cost has already been incurred. - For part b, let \(p\) represent the complaint rate and write an inequality. - Remember that one possible profit value is negative.

Solution

1. Without a complaint, the profit is \(\$14.50-\$8.00=\$6.50\). With a complaint, the sale is refunded, but the production cost and complaint-handling cost remain, so the result is \(-\$8.00-\$5.00=-\$13.00\). 2. At a \(4\%\) complaint rate, \(E(G)=0.96\cdot\$6.50+0.04\cdot(-\$13.00)=\$5.72\). 3. Let \(p\) be the complaint rate. Require \((1-p)6.50-13p\ge5\). 4. This simplifies to \(6.50-19.50p\ge5\), so \(p\le\frac{1.50}{19.50}=\frac{1}{13}\approx0.0769\). 5. Therefore, the complaint rate can be at most about \(7.69\%\).

Answer

a) The expected profit is \(\$5.72\) per order. b) The complaint rate can be at most \(\frac{1}{13}\approx7.69\%\).
52684112
A random variable \(X\) can take the values \(1, 2, 3, 4, 5\). Its probability distribution is symmetric about the mean \(E(X)=3\): <table> <tr> <td>\(k\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> </tr> <tr> <td>\(P(X=k)\)</td> <td>\(p\)</td> <td>\(q\)</td> <td>\(0.2\)</td> <td>\(q\)</td> <td>\(p\)</td> </tr> </table> Given that \(\operatorname{Var}(X)=2.48\), find \(p\) and \(q\).

Hints

- Use the fact that all probabilities sum to \(1\). - Relate the variance, mean, and second moment. - Use the symmetric probabilities to simplify the second-moment equation. - Solve the resulting system of two equations.

Solution

1. The probabilities sum to \(1\), so \(2p+2q+0.2=1\). Thus, \(p+q=0.4\). 2. From \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2\), \(E(X^2)=2.48+3^2=11.48\). 3. The second moment is \(E(X^2)=1^2p+2^2q+3^2\cdot0.2+4^2q+5^2p=26p+20q+1.8\). 4. Substitute \(q=0.4-p\): \(26p+20(0.4-p)+1.8=11.48\). 5. Simplifying gives \(6p=1.68\), so \(p=0.28\). 6. Therefore, \(q=0.4-0.28=0.12\).

Answer

\(p=0.28\) and \(q=0.12\)
52684212
A discrete random variable \(Z\) takes the values \(-2\), \(0\), and \(2\), where \(P(Z=-2)=a\), \(P(Z=0)=b\), and \(P(Z=2)=c\). Given that \(E(Z)=0.4\) and \(\operatorname{Var}(Z)=2.24\), find \(a\), \(b\), and \(c\).

Hints

- Three unknown probabilities require three independent equations. - Use the sum of the probabilities, the mean, and the variance. - Rewrite the variance condition using the second moment. - Reduce the system by eliminating one variable at a time.

Solution

1. The probabilities satisfy \(a+b+c=1\). 2. The mean gives \(-2a+2c=0.4\), so \(c-a=0.2\). 3. Since \(\operatorname{Var}(Z)=E(Z^2)-[E(Z)]^2\), \(4a+4c-(0.4)^2=2.24\). 4. Thus, \(4a+4c=2.4\), or \(a+c=0.6\). 5. Solving \(c-a=0.2\) and \(a+c=0.6\) gives \(a=0.2\) and \(c=0.4\). 6. Therefore, \(b=1-0.2-0.4=0.4\).

Answer

\(a=0.2\), \(b=0.4\), and \(c=0.4\)
52686512
An electronics store offers a prize game. First, a player draws one ball from a container holding \(25\) balls, of which \(5\) are marked. Only a player who draws a marked ball may spin a prize wheel with three sections: Gold, Silver, and Bronze. Gold awards a \(\$100\) gift card, Silver awards a \(\$40\) gift card, and Bronze awards a \(\$10\) gift card. The probability of Bronze is \(50\%\). The wheel is designed so that the expected prize value per game entry is \(\$7\). Find the probabilities of landing on Gold and Silver.

Hints

- Treat the game as a two-stage random experiment. - First find the probability that a player earns a spin. - The expected prize per entry includes the probability of reaching the wheel. - The three wheel probabilities must add to \(1\).

Solution

1. The probability of earning a spin is \(\frac{5}{25}=0.2\). 2. Let \(p_G\) and \(p_S\) be the probabilities of Gold and Silver. Since Bronze has probability \(0.5\), \(p_G+p_S=0.5\), so \(p_S=0.5-p_G\). 3. The overall expected prize value is \(0.2(100p_G+40p_S+10\cdot0.5)=7\). 4. Dividing by \(0.2\) gives \(100p_G+40p_S+5=35\), or \(100p_G+40p_S=30\). 5. Substitute \(p_S=0.5-p_G\): \(100p_G+40(0.5-p_G)=30\). Thus, \(60p_G=10\), so \(p_G=\frac{1}{6}\). 6. Then \(p_S=0.5-\frac{1}{6}=\frac{1}{3}\).

Answer

The probability of Gold is \(\frac{1}{6}\approx16.7\%\), and the probability of Silver is \(\frac{1}{3}\approx33.3\%\).
52694112
A bag contains five balls labeled \(1, 1, 2, 3, 3\). Two balls are drawn at the same time without replacement. Let \(X\) be the sum of the two labels. a) Find the probability distribution of \(X\) and display it in a table. b) Find \(E(X)\) and \(\operatorname{Var}(X)\).

Hints

- Count the total number of ways to choose two of the five physical balls. - List the pairs that produce each possible sum, accounting for repeated labels. - Multiply each possible sum by its probability to find the mean. - Use squared deviations from the mean to find the variance.

Solution

1. There are \(\binom{5}{2}=10\) equally likely unordered pairs of balls. 2. A sum of \(2\) occurs once, a sum of \(3\) occurs twice, a sum of \(4\) occurs four times, a sum of \(5\) occurs twice, and a sum of \(6\) occurs once. 3. Therefore, the probabilities are \(0.1, 0.2, 0.4, 0.2, 0.1\) for the sums \(2, 3, 4, 5, 6\), respectively. 4. The mean is \(E(X)=2\cdot0.1+3\cdot0.2+4\cdot0.4+5\cdot0.2+6\cdot0.1=4\). 5. The variance is \(\operatorname{Var}(X)=(2-4)^2\cdot0.1+(3-4)^2\cdot0.2+(4-4)^2\cdot0.4+(5-4)^2\cdot0.2+(6-4)^2\cdot0.1=1.2\).

Answer

a) <table border="1"> <tr> <td>\(x_i\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> <td>\(6\)</td> </tr> <tr> <td>\(P(X=x_i)\)</td> <td>\(0.1\)</td> <td>\(0.2\)</td> <td>\(0.4\)</td> <td>\(0.2\)</td> <td>\(0.1\)</td> </tr> </table> b) \(E(X)=4\) and \(\operatorname{Var}(X)=1.2\)
52712012
A multiple-choice test has \(30\) questions, each with four answer choices and exactly one correct answer. A test-taker guesses independently and uniformly at random on every question. Let \(X\) be the number of correct answers and \(Y\) the number of incorrect answers. a) Find \(E(X)\), \(E(Y)\), \(\sigma(X)\), and \(\sigma(Y)\). b) Find the probability that the test-taker answers at least \(10\) questions correctly.

Hints

- The probability of a correct guess is \(\frac14\). - Use the binomial mean and standard-deviation formulas for each count. - “At least ten” can be found using the complement of “at most nine.”

Solution

1. The success probabilities are \(p_X=0.25\) and \(p_Y=0.75\), with \(n=30\). 2. The means are \(E(X)=30\cdot0.25=7.5\) and \(E(Y)=30\cdot0.75=22.5\). 3. The variances are equal: \(\operatorname{Var}(X)=\operatorname{Var}(Y)=30\cdot0.25\cdot0.75=5.625\). 4. Therefore, \(\sigma(X)=\sigma(Y)=\sqrt{5.625}\approx2.3717\). 5. For part b), \(P(X\ge10)=1-P(X\le9)\approx1-0.8034=0.1966\).

Answer

a) \(E(X)=7.5\), \(E(Y)=22.5\), and \(\sigma(X)=\sigma(Y)\approx2.37\) b) \(P(X\ge10)\approx0.1966\), or about \(19.66\%\)
52712212
Let \(X\sim\operatorname{Bin}(200,p)\). 1. Find the value of \(p\) that maximizes \(\operatorname{Var}(X)\). Then find the standard deviation for that value of \(p\). 2. Now let \(p=0.15\). Find the mean \(\mu\) and standard deviation \(\sigma\). List all integer values \(k\) in the interval \([\mu-\sigma, \mu+\sigma]\).

Hints

- Write the variance as a function of \(p\) and identify the vertex of the quadratic. - Use the binomial mean and standard-deviation formulas. - Find the decimal interval endpoints before listing the integers inside.

Solution

1. The variance is \(V(p)=200p(1-p)\), a downward-opening quadratic with zeros at \(p=0\) and \(p=1\). Its maximum occurs at \(p=0.5\). 2. At \(p=0.5\), \(V=200\cdot0.5\cdot0.5=50\), so \(\sigma=\sqrt{50}\approx7.07\). 3. For \(p=0.15\), the mean is \(\mu=200\cdot0.15=30\). 4. The standard deviation is \(\sigma=\sqrt{200\cdot0.15\cdot0.85}=\sqrt{25.5}\approx5.05\). 5. The interval is approximately \([30-5.05, 30+5.05]=[24.95, 35.05]\), so the integer values are \(25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35\).

Answer

1. \(p=0.5\) and \(\sigma=\sqrt{50}\approx7.07\) 2. \(\mu=30\), \(\sigma\approx5.05\), and \(k\in\{25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35\}\)
52713112
A binomial random variable \(X\) has parameters \(n=50\) and \(p=0.4\). 1. Find the mean \(\mu\) and standard deviation \(\sigma\). 2. List all integer values that satisfy \(|X-\mu|\le\sigma\). 3. Find \(P(|X-\mu|\le\sigma)\).

Hints

- Use the binomial mean and standard-deviation formulas. - Rewrite the absolute-value inequality as an interval. - Identify the integers in the interval. - Use cumulative binomial probabilities to find the probability of the range.

Solution

1. The mean is \(\mu=np=50\cdot0.4=20\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{50\cdot0.4\cdot0.6}=\sqrt{12}\approx3.46\). 3. The inequality \(|X-20|\le3.46\) is equivalent to \(16.54\le X\le23.46\). 4. The integer values are \(17, 18, 19, 20, 21, 22, 23\). 5. Using the binomial distribution, \(P(17\le X\le23)=P(X\le23)-P(X\le16)\approx0.8438-0.1561=0.6877\).

Answer

1. \(\mu=20\) and \(\sigma=\sqrt{12}\approx3.46\) 2. \(k\in\{17, 18, 19, 20, 21, 22, 23\}\) 3. \(P(|X-\mu|\le\sigma)\approx0.6877\)
52713312
A binomial random variable \(X\) has parameters \(n\in\mathbb{N}\), \(n\ge1\), and \(0<p<1\). Its variance is \(60\%\) of its mean. a) Find the success probability \(p\). b) Find \(n\) if the standard deviation is \(\sigma=\sqrt{6}\).

Hints

- Write the formulas for the binomial mean and variance. - Translate “the variance is \(60\%\) of the mean” into an equation. - Use the relationship between variance and standard deviation. - First determine the parameter that follows directly from the ratio of variance to mean.

Solution

1. For a binomial random variable, \(E(X)=np\) and \(\operatorname{Var}(X)=np(1-p)\). 2. The condition gives \(np(1-p)=0.6np\). 3. Since \(np>0\), divide by \(np\): \(1-p=0.6\), so \(p=0.4\). 4. Since \(\sigma=\sqrt{6}\), the variance is \(6\). 5. Using \(6=0.6E(X)\), the mean is \(E(X)=10\). 6. Therefore, \(10=n\cdot0.4\), so \(n=25\).

Answer

a) \(p=0.4\) b) \(n=25\)
52714312
During final inspection at an electronics factory, a sample of \(n\) components is selected. Let \(X\) be the number of defective components, modeled by a binomial distribution. The mean is \(\mu=15\), and the variance is \(\sigma^2=12.75\). a) Find the sample size \(n\) and the probability \(p\) that a component is defective. b) Find the probability that the number of defective components is exactly equal to the mean.

Hints

- Use the formulas connecting \(n\) and \(p\) to the mean and variance. - Substitute the mean expression into the variance equation. - “Exactly equal to the mean” identifies one binomial outcome.

Solution

1. The mean and variance satisfy \(np=15\) and \(np(1-p)=12.75\). 2. Substitute \(np=15\) into the variance equation: \(15(1-p)=12.75\). 3. Thus, \(1-p=0.85\), so \(p=0.15\). 4. Since \(n\cdot0.15=15\), \(n=100\). 5. The probability of exactly \(15\) defective components is \(P(X=15)=\binom{100}{15}(0.15)^{15}(0.85)^{85}\approx0.1111\).

Answer

a) \(n=100\) and \(p=0.15\) b) \(P(X=15)\approx0.1111\), or about \(11.11\%\)
52714412
A spinner has \(k\) equal sections, exactly one of which is a winning section. The spinner is spun \(n\) independent times. Let \(X\) be the number of wins. The binomial random variable has mean \(\mu=5\) and standard deviation \(\sigma=2\). Find \(n\) and \(k\).

Hints

- Convert the standard deviation to a variance. - Write equations using the binomial mean and variance. - Relate the success probability to the number of equal spinner sections.

Solution

1. The variance is \(\sigma^2=2^2=4\). 2. Since \(np=5\) and \(np(1-p)=4\), substitute the mean into the variance equation: \(5(1-p)=4\). 3. Thus, \(1-p=0.8\), so \(p=0.2\). 4. From \(n\cdot0.2=5\), \(n=25\). 5. Since exactly one of the \(k\) equal sections is a winning section, \(p=\frac{1}{k}\). Therefore, \(\frac{1}{k}=0.2\), so \(k=5\).

Answer

\(n=25\) and \(k=5\)
52714512
A binomial random variable \(X\) has parameters \(n\) and \(p\). a) Find \(n\) and \(p\) if \(E(X)=12\) and \(\operatorname{Var}(X)=9\). b) Use \(\operatorname{Var}(X)=np(1-p)\) to explain why, for a fixed \(p\in(0, 1)\), the variance grows proportionally with \(n\). How does the standard deviation grow as \(n\) increases?

Hints

- Use the mean and variance formulas to form two equations. - Identify the factor \(np\) inside the variance formula. - For fixed \(p\), treat \(p(1-p)\) as a constant. - Consider how taking a square root changes the growth rate.

Solution

1. The equations are \(np=12\) and \(np(1-p)=9\). 2. Substitute \(np=12\) into the variance equation: \(12(1-p)=9\). 3. Thus, \(1-p=0.75\), so \(p=0.25\). 4. Since \(n\cdot0.25=12\), \(n=48\). 5. For fixed \(p\), the factor \(p(1-p)\) is a positive constant. Therefore, \(\operatorname{Var}(X)=[p(1-p)]n\), so the variance is directly proportional to \(n\). 6. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{p(1-p)}\sqrt{n}\), so it grows proportionally to \(\sqrt{n}\), not to \(n\).

Answer

a) \(n=48\) and \(p=0.25\) b) For fixed \(p\), \(\operatorname{Var}(X)=[p(1-p)]n\), so the variance is proportional to \(n\). The standard deviation is proportional to \(\sqrt{n}\).
52720212
An experiment with success probability \(p=0.2\) is repeated independently \(80\) times. Let \(X\) be the number of successes. a) Find the interval \([\mu-2\sigma, \mu+2\sigma]\) and list the possible integer success counts in the interval. b) Find the probability that the number of successes is within two standard deviations of the mean.

Hints

- Find the mean and standard deviation first. - Add and subtract twice the standard deviation. - Include only the integer counts that actually lie in the interval.

Solution

1. The mean is \(\mu=80\cdot0.2=16\), and the standard deviation is \(\sigma=\sqrt{80\cdot0.2\cdot0.8}=\sqrt{12.8}\approx3.58\). 2. Thus, \(2\sigma\approx7.16\), and the interval is approximately \([16-7.16, 16+7.16]=[8.84, 23.16]\). 3. The integer success counts in the interval are \(9, 10, \ldots, 23\). 4. Using the binomial distribution, \(P(9\le X\le23)=P(X\le23)-P(X\le8)\approx0.9783-0.0131=0.9652\).

Answer

a) \([8.84, 23.16]\), containing \(9, 10, \ldots, 23\) b) \(P(9\le X\le23)\approx0.9652\)
52721212
In a large screening program, the probability of a positive test result is \(p=0.04\). Let \(Y\) be the number of positive results among \(500\) independently tested people, and assume \(Y\) is binomial. a) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(Y\). b) Find the smallest and largest possible relative frequencies of positive results, given that \(Y\) is within one standard deviation of its mean. c) For any random variable \(Z\) with mean \(\mu\) and standard deviation \(\sigma\), Chebyshev's inequality states that \(P(\mu-k\sigma<Z<\mu+k\sigma)\ge1-\frac{1}{k^2}\). Explain what this says for \(k=3\) in the context of this screening program.

Hints

- Use the binomial formulas for the mean and standard deviation. - Identify the integer values of \(Y\) within the interval. - Divide each endpoint count by the sample size to obtain relative frequencies. - Substitute \(k=3\) into the given inequality and interpret the result.

Solution

1. The mean is \(\mu=np=500\cdot0.04=20\). 2. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{500\cdot0.04\cdot0.96}=\sqrt{19.2}\approx4.38\). 3. Being within one standard deviation gives approximately \([20-4.38, 20+4.38]=[15.62, 24.38]\). Since \(Y\) is an integer, \(16\le Y\le24\). 4. The smallest relative frequency is \(\frac{16}{500}=0.032=3.2\%\), and the largest is \(\frac{24}{500}=0.048=4.8\%\). 5. For \(k=3\), Chebyshev's inequality gives \(1-\frac{1}{3^2}=\frac89\approx0.889\). 6. Therefore, the probability that the number of positive results is less than three standard deviations from the mean is at least \(\frac89\), or about \(88.9\%\).

Answer

a) \(\mu=20\) and \(\sigma\approx4.38\) b) The relative frequency can range from \(3.2\%\) to \(4.8\%\). c) The probability that \(Y\) lies within three standard deviations of its mean is at least \(\frac89\), or about \(88.9\%\).
52876512
In a certain region, \(30\%\) of teenagers play a musical instrument. A school study randomly and independently selects \(50\) teenagers. Would it be considered unusual if \(22\) of them report playing an instrument? Use the two-standard-deviation rule.

Hints

- Find the expected count for the sample size. - Calculate the standard deviation of the binomial count. - Form the interval within two standard deviations of the mean. - Compare the observed count with the interval.

Solution

1. Let \(X\) be the number who play an instrument. Then \(X\) is binomial with \(n=50\) and \(p=0.30\). 2. The mean is \(\mu=np=50\cdot0.30=15\). 3. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{50\cdot0.30\cdot0.70}=\sqrt{10.5}\approx3.24\). 4. The two-standard-deviation interval is approximately \([15-2\cdot3.24, 15+2\cdot3.24]=[8.52, 21.48]\). 5. Since \(22>21.48\), the observed count lies outside this interval and is unusual by the stated rule. This rule alone is not a formal significance test.

Answer

Yes. The count \(22\) lies outside the interval \([8.52, 21.48]\), so it is unusual by the two-standard-deviation rule.
52877112
A survey reports that \(20\%\) of workers in a large city commute by bicycle. A city planner surveys \(100\) randomly and independently selected workers. Use the two-standard-deviation rule to find the range of bicycle-commuter counts that would be considered consistent with the reported population proportion.

Hints

- Model the count with a binomial random variable. - Find the mean and standard deviation. - Add and subtract two standard deviations from the mean. - Include integer endpoint values when they lie in the interval.

Solution

1. Let \(X\) be the number of bicycle commuters. Then \(X\) is binomial with \(n=100\) and \(p=0.20\). 2. The mean is \(\mu=np=100\cdot0.20=20\). 3. The standard deviation is \(\sigma=\sqrt{100\cdot0.20\cdot0.80}=\sqrt{16}=4\). 4. The two-standard-deviation interval is \([\mu-2\sigma, \mu+2\sigma]=[20-8, 20+8]=[12, 28]\). 5. Therefore, counts from \(12\) through \(28\), inclusive, are consistent with the reported proportion under this rule.

Answer

From \(12\) through \(28\) bicycle commuters, inclusive.
52877712
A fair six-sided die is rolled independently \(60\) times, and the number of sixes is recorded. 1. Find the mean \(\mu\) and standard deviation \(\sigma\) of the number of sixes. 2. A result is classified as unusual if it lies outside \([\mu-2\sigma, \mu+2\sigma]\). List all counts of sixes that would raise suspicion that the die is not fair under this rule.

Hints

- Model the number of sixes with a binomial distribution. - Use the binomial mean and standard-deviation formulas. - Determine which integer counts lie outside the interval. - Remember that an unusual result is evidence, not proof, of unfairness.

Solution

1. Let \(X\) be the number of sixes. Then \(X\) is binomial with \(n=60\) and \(p=\frac16\). 2. The mean is \(\mu=60\cdot\frac16=10\). 3. The standard deviation is \(\sigma=\sqrt{60\cdot\frac16\cdot\frac56}=\sqrt{\frac{50}{6}}\approx2.89\). 4. The interval is approximately \([10-2\cdot2.89, 10+2\cdot2.89]=[4.22, 15.78]\). 5. The integer counts inside the interval are \(5, 6, \ldots, 15\). Therefore, the unusual counts are \(0\) through \(4\) and \(16\) through \(60\). The rule may raise suspicion, but it does not prove that the die is unfair.

Answer

1. \(\mu=10\) and \(\sigma\approx2.89\) 2. Counts from \(0\) through \(4\), or from \(16\) through \(60\)
53081212
A game has possible payouts of \(\$0\), \(\$2\), and \(\$10\). Let \(X\) be the payout in dollars. It is known that \(P(X=10)=0.05\) and \(P(X=2)=0.25\). a) Find \(P(X=0)\). b) Find the expected payout \(E(X)\). c) Find the standard deviation \(\sigma(X)\).

Hints

- All probabilities in a distribution must sum to \(1\). - The expected value is the long-run average payout. - Find the variance before taking its square root.

Solution

1. The probabilities sum to \(1\), so \(P(X=0)=1-(0.05+0.25)=0.70\). 2. The expected payout is \(E(X)=0\cdot0.70+2\cdot0.25+10\cdot0.05=1.00\), or \(\$1.00\). 3. The variance is \(\operatorname{Var}(X)=(0-1)^2\cdot0.70+(2-1)^2\cdot0.25+(10-1)^2\cdot0.05=5\) square dollars. 4. Therefore, \(\sigma(X)=\sqrt5\approx\$2.24\).

Answer

a) \(P(X=0)=0.70\) b) \(E(X)=\$1.00\) c) \(\sigma(X)=\sqrt5\approx\$2.24\)
53082412
A prize wheel lands on the Grand Prize section with probability \(p=0.05\). a) How many Grand Prizes are expected in \(800\) spins? b) At an event, the Grand Prize occurred \(35\) times. Suppose this count equals the expected value exactly. How many times was the wheel spun? c) In one trial, the wheel was spun \(2000\) times and landed on Grand Prize \(112\) times. Find the relative frequency and compare it with the theoretical probability.

Hints

- Use the binomial mean formula for part a). - Rearrange the mean formula to solve for the number of spins in part b). - Relative frequency is the observed count divided by the total number of trials. - Compare the two values by subtracting.

Solution

1. For \(800\) spins, the expected count is \(800\cdot0.05=40\). 2. If \(35=n\cdot0.05\), then \(n=\frac{35}{0.05}=700\). 3. The relative frequency is \(\frac{112}{2000}=0.056=5.6\%\). 4. This is \(0.056-0.05=0.006\), or \(0.6\) percentage points, above the theoretical probability.

Answer

a) \(40\) times b) \(700\) spins c) The relative frequency is \(0.056=5.6\%\), which is \(0.006\), or \(0.6\) percentage points, above \(0.05\).
53083712
A four-sided die is rolled \(1000\) times to investigate whether it behaves like a fair die. The observed counts are shown. <table> <thead> <tr> <th>Result \(i\)</th> <th>Observed count</th> </tr> </thead> <tbody> <tr> <td>\(1\)</td> <td>\(210\)</td> </tr> <tr> <td>\(2\)</td> <td>\(290\)</td> </tr> <tr> <td>\(3\)</td> <td>\(240\)</td> </tr> <tr> <td>\(4\)</td> <td>\(260\)</td> </tr> </tbody> </table> 1. Use relative frequencies to estimate the probability distribution of \(X\), the result of one roll. 2. Find the expected value based on the empirical distribution. 3. Find the expected value of an ideal fair four-sided die and compare the two values.

Hints

- Divide each count by \(1000\). - Multiply each outcome by its estimated probability and add. - For a fair die, all four outcomes have equal probability.

Solution

1. Dividing each count by \(1000\) gives \(P(X=1)\approx 0.21\), \(P(X=2)\approx 0.29\), \(P(X=3)\approx 0.24\), and \(P(X=4)\approx 0.26\). 2. The empirical expected value is \(E(X)=1\cdot0.21+2\cdot0.29+3\cdot0.24+4\cdot0.26=2.55\). 3. For a fair die, each result has probability \(\frac{1}{4}\), so \(E(X)=\frac{1+2+3+4}{4}=2.5\). The empirical estimate is \(0.05\) greater.

Answer

1. \(P(X=1)\approx 0.21\), \(P(X=2)\approx 0.29\), \(P(X=3)\approx 0.24\), and \(P(X=4)\approx 0.26\) 2. \(E(X)=2.55\) 3. The fair-die expected value is \(2.5\); the empirical estimate is \(0.05\) greater.
53083812
A quality-control study checks \(4000\) newly produced circuit boards for soldering defects. The random variable \(X\) is the number of defects on one board. <table> <thead> <tr> <th>Number of defects \(k\)</th> <th>Observed count</th> </tr> </thead> <tbody> <tr> <td>\(0\)</td> <td>\(3600\)</td> </tr> <tr> <td>\(1\)</td> <td>\(320\)</td> </tr> <tr> <td>\(2\)</td> <td>\(80\)</td> </tr> </tbody> </table> 1. Use relative frequencies to estimate \(P(X=k)\). 2. Estimate the probability that a randomly selected board has at least one defect. 3. Based on the empirical expected number of defects per board, estimate the total number of defects in a shipment of \(500\) boards.

Hints

- Divide each observed count by \(4000\). - “At least one” includes \(1\) or \(2\) defects. - First find the expected defects per board, then scale to \(500\) boards.

Solution

1. The estimated probabilities are \(P(X=0)\approx\frac{3600}{4000}=0.90\), \(P(X=1)\approx\frac{320}{4000}=0.08\), and \(P(X=2)\approx\frac{80}{4000}=0.02\). 2. \(P(X\ge 1)=P(X=1)+P(X=2)=0.08+0.02=0.10\). 3. The expected number of defects per board is \(E(X)=0\cdot0.90+1\cdot0.08+2\cdot0.02=0.12\). For \(500\) boards, the expected total is \(500\cdot0.12=60\).

Answer

1. \(P(X=0)\approx 0.90\), \(P(X=1)\approx 0.08\), and \(P(X=2)\approx 0.02\) 2. \(0.10\) 3. Approximately \(60\) defects
53098012
A container holds \(n\) tickets numbered \(1,2,3,\ldots,n\). The sum of all numbers on the tickets is \(1275\). Find the expected number on a randomly selected ticket.

Hints

- Use the formula for the sum of the first \(n\) positive integers. - First determine how many tickets are in the container. - For equally likely values, the expected value is the ordinary average.

Solution

1. The sum of the first \(n\) positive integers is \(\frac{n(n+1)}{2}\). 2. Set \(\frac{n(n+1)}{2}=1275\), which gives \(n^2+n-2550=0\). 3. The positive solution is \(n=50\). 4. Since the ticket numbers are equally likely, the expected value is their average: \(E(X)=\frac{1+50}{2}=25.5\).

Answer

\(E(X)=25.5\)
53098112
A school carnival offers a game using a spinner with \(20\) equal sections. The game costs \(\$2.50\) to play. Two sections pay \(\$12.00\), five sections pay \(\$4.00\), and the remaining sections pay nothing. Find the expected net gain \(X\) for a player. Decide whether the game is fair, and find the entry fee that would make it fair.

Hints

- Distinguish between the payout and the player’s net gain. - Find the probability of each payout from the number of spinner sections. - A fair game has expected net gain \(0\). - The fair entry fee equals the expected payout.

Solution

1. Let \(A\) be the payout. Then \(P(A=12)=\frac{2}{20}=0.10\), \(P(A=4)=\frac{5}{20}=0.25\), and \(P(A=0)=\frac{13}{20}=0.65\). 2. The expected payout is \(E(A)=12\cdot0.10+4\cdot0.25=\$2.20\). 3. The expected net gain is \(E(X)=\$2.20-\$2.50=-\$0.30\). 4. The game is not fair because the expected net gain is not \(0\). A fair entry fee equals the expected payout, so it would be \(\$2.20\).

Answer

The player’s expected net gain is \(-\$0.30\), so the game is not fair. A fair entry fee would be \(\$2.20\).
53098212
An urn contains \(25\) tickets. One gold ticket pays \(\$20.00\), four silver tickets each pay \(\$5.00\), and the remaining tickets pay nothing. The game costs \(\$3.00\) per draw, and the original ticket distribution is restored after every draw. Find the player’s expected gain or loss per game. What total profit can the operator expect from \(200\) games?

Hints

- Write the payout distribution for one draw. - Find the expected payout before subtracting the entry fee. - The operator’s expected gain is the negative of the player’s expected net gain. - Use linearity of expectation for \(200\) games.

Solution

1. The payout probabilities are \(P(A=20)=\frac{1}{25}=0.04\), \(P(A=5)=\frac{4}{25}=0.16\), and \(P(A=0)=\frac{20}{25}=0.80\). 2. The expected payout is \(20\cdot0.04+5\cdot0.16=\$1.60\). 3. The player’s expected net gain is \(\$1.60-\$3.00=-\$1.40\), so the player expects to lose \(\$1.40\) per game. 4. The operator expects to gain \(\$1.40\) per game. Over \(200\) games, the expected profit is \(200\cdot\$1.40=\$280.00\).

Answer

The player’s expected loss is \(\$1.40\) per game. The operator’s expected profit over \(200\) games is \(\$280.00\).
53098312
A school carnival game has the following prize schedule. The random variable \(X\) is the payout in dollars. <table> <tr> <th>Payout</th> <td>\(\$0\)</td> <td>\(\$2\)</td> <td>\(\$10\)</td> <td>\(\$50\)</td> </tr> <tr> <th>Probability</th> <td>\(85\%\)</td> <td>\(10\%\)</td> <td>\(4\%\)</td> <td>\(1\%\)</td> </tr> </table> a) Find \(E(X)\). b) The game costs \(\$2.00\) to play. Find the operator’s expected profit per game. c) A new rule requires the expected payout to be at least \(45\%\) of the entry fee. What is the greatest entry fee the operator may charge while keeping the same prize schedule? Give the answer to the nearest cent without exceeding the limit.

Hints

- Multiply each payout by its probability and add. - The operator’s profit is the entry fee minus the payout. - For part c, write an inequality comparing expected payout with \(45\%\) of the fee. - The final fee must not exceed the calculated upper bound.

Solution

1. The expected payout is \(E(X)=0\cdot0.85+2\cdot0.10+10\cdot0.04+50\cdot0.01=\$1.10\). 2. At an entry fee of \(\$2.00\), the operator’s expected profit is \(\$2.00-\$1.10=\$0.90\). 3. Let the entry fee be \(e\). The rule requires \(1.10\ge0.45e\). 4. Thus, \(e\le\frac{1.10}{0.45}\approx2.4444\). The greatest fee to the nearest cent that does not exceed the limit is \(\$2.44\).

Answer

a) \(E(X)=\$1.10\) b) \(\$0.90\) c) \(\$2.44\)
53098412
An arcade machine pays prizes according to the table. <table> <tr> <th>Payout</th> <td>\(\$0.10\)</td> <td>\(\$0.50\)</td> <td>\(\$2.00\)</td> <td>\(\$5.00\)</td> <td>\(\$H\)</td> </tr> <tr> <th>Probability</th> <td>\(\frac{1}{5}\)</td> <td>\(\frac{1}{20}\)</td> <td>\(\frac{1}{50}\)</td> <td>\(\frac{1}{100}\)</td> <td>\(\frac{1}{500}\)</td> </tr> </table> All other outcomes pay nothing. a) Find the probability that a play has no payout. b) The machine costs \(\$0.50\) per play. Find the jackpot \(H\) so that the expected payout is exactly \(60\%\) of the entry fee.

Hints

- The probabilities of all outcomes must add to \(1\). - First find the probability that any positive payout occurs. - Determine the target expected payout from the entry fee. - Write an expected-value equation with \(H\) as the unknown.

Solution

1. The probability of a positive payout is \(\frac{1}{5}+\frac{1}{20}+\frac{1}{50}+\frac{1}{100}+\frac{1}{500}=\frac{141}{500}=0.282\). 2. Therefore, the probability of no payout is \(1-0.282=0.718\). 3. The target expected payout is \(0.60\cdot\$0.50=\$0.30\). 4. Set up the expected-value equation: \(0.10\cdot\frac{1}{5}+0.50\cdot\frac{1}{20}+2\cdot\frac{1}{50}+5\cdot\frac{1}{100}+H\cdot\frac{1}{500}=0.30\). 5. The known terms total \(0.135\), so \(0.135+\frac{H}{500}=0.30\). 6. Thus, \(\frac{H}{500}=0.165\), giving \(H=82.50\).

Answer

a) \(0.718\), or \(71.8\%\) b) \(H=82.50\), so the jackpot is \(\$82.50\).
53102412
A spinner has four equal sections, and one section is labeled Win. The spinner is spun repeatedly. a) Find the probability of event \(A\): exactly \(1\) win in \(4\) spins, and event \(B\): exactly \(2\) wins in \(8\) spins. Are the probabilities equal? b) For \(4\) spins, the expected number of wins is \(1\). Evaluate this claim: “Because \(E(X)=1\), event \(C\), at least one win in \(4\) spins, is certain.” Find \(P(C)\) as part of your justification.

Hints

- A certain event has probability \(1\). - Use the binomial probability formula for part a). - For “at least one,” consider the complement of no wins. - An expected value is an average, not a guaranteed outcome.

Solution

1. The success probability is \(p=0.25\). 2. \(P(A)=\binom{4}{1}(0.25)(0.75)^3=0.421875\). 3. \(P(B)=\binom{8}{2}(0.25)^2(0.75)^6\approx0.3115\). The probabilities are not equal. 4. An expected value describes a long-run average; it does not guarantee a win in every set of four spins. 5. Using the complement, \(P(C)=1-P(X=0)=1-(0.75)^4=0.68359375\approx0.6836\). Since this is less than \(1\), event \(C\) is not certain.

Answer

a) \(P(A)=0.421875\) and \(P(B)\approx0.3115\). They are not equal. b) The claim is false. \(P(C)\approx0.6836\), not \(1\).
53104512
A random variable \(X\) has a binomial distribution with \(n=5\) and \(p=0.4\). a) Find \(P(X=k)\) for every possible value of \(k\), and display the probability distribution in a table. b) Use only the probabilities in your table to find \(E(X)\) from \(E(X)=\sum_{k=0}^{5}kP(X=k)\).

Hints

- Use the binomial probability formula for each value of \(k\). - Organize the values and probabilities in a table. - The expected value is the weighted average of all possible values. - Multiply each value \(k\) by its probability and add.

Solution

1. Using \(P(X=k)=\binom{5}{k}(0.4)^k(0.6)^{5-k}\), the probabilities are: \(P(X=0)=0.07776\), \(P(X=1)=0.2592\), \(P(X=2)=0.3456\), \(P(X=3)=0.2304\), \(P(X=4)=0.0768\), and \(P(X=5)=0.01024\). 2. The expected value from the distribution is \(E(X)=0(0.07776)+1(0.2592)+2(0.3456)+3(0.2304)+4(0.0768)+5(0.01024)=2\).

Answer

a) <table border="1"> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(0.07776\)</td><td>\(0.2592\)</td><td>\(0.3456\)</td><td>\(0.2304\)</td><td>\(0.0768\)</td><td>\(0.01024\)</td></tr> </table> b) \(E(X)=2\)
53111312
Consider two independent Bernoulli trials. Each trial is a success with probability \(p\) and a failure with probability \(q=1-p\). Let \(X\) be the number of successes. a) Write the probability distribution of \(X\). b) Find \(E(X)\) in terms of \(p\). c) Use the definition \(\operatorname{Var}(X)=\sum_{i=0}^{2}[x_i-E(X)]^2P(X=x_i)\) to show that \(\operatorname{Var}(X)=2p(1-p)\).

Hints

- List the possible success counts and obtain their probabilities from the two trials. - Use the expected-value formula for part b). - Substitute the mean into the variance definition. - Factor the resulting expression before expanding everything.

Solution

1. The distribution is \(P(X=0)=q^2=(1-p)^2\), \(P(X=1)=2pq=2p(1-p)\), and \(P(X=2)=p^2\). 2. The mean is \(E(X)=0\cdot q^2+1\cdot2pq+2\cdot p^2=2p(q+p)=2p\). 3. Using the variance definition, \(\operatorname{Var}(X)=(0-2p)^2q^2+(1-2p)^2(2pq)+(2-2p)^2p^2\). 4. Since \(1-2p=q-p\) and \(2-2p=2q\), this becomes \(4p^2q^2+2pq(q-p)^2+4p^2q^2\). 5. Factor: \(\operatorname{Var}(X)=2pq[4pq+(q-p)^2]\). 6. Because \(4pq+(q-p)^2=(p+q)^2=1\), \(\operatorname{Var}(X)=2pq=2p(1-p)\).

Answer

a) \(P(X=0)=(1-p)^2\), \(P(X=1)=2p(1-p)\), and \(P(X=2)=p^2\) b) \(E(X)=2p\) c) \(\operatorname{Var}(X)=2p(1-p)\)
53112412
A binomial random variable \(X\) has \(n=250\) and variance \(\operatorname{Var}(X)=40\). a) Find the two possible values of \(p\). b) Find the mean for each value of \(p\). c) Find the greatest possible variance for a binomial random variable with \(n=250\).

Hints

- Substitute the given variance into the binomial variance formula. - Solve the resulting quadratic equation. - Use \(E(X)=np\) for each solution. - Determine when \(p(1-p)\) is greatest.

Solution

1. The variance equation is \(250p(1-p)=40\). 2. Divide by \(250\): \(p-p^2=0.16\), so \(p^2-p+0.16=0\). 3. Solving the quadratic gives \(p=\frac{1\pm\sqrt{1-0.64}}{2}=\frac{1\pm0.6}{2}\), so \(p=0.8\) or \(p=0.2\). 4. The corresponding means are \(250\cdot0.8=200\) and \(250\cdot0.2=50\). 5. The factor \(p(1-p)\) is greatest at \(p=0.5\). Thus, the maximum variance is \(250\cdot0.5\cdot0.5=62.5\).

Answer

a) \(p=0.8\) or \(p=0.2\) b) The corresponding means are \(200\) and \(50\). c) The maximum variance is \(62.5\).
53112612
A binomial random variable \(X\) has parameters \(n=5\) and \(p=0.5\). Find \(\operatorname{Var}(X)\) using \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2\). Calculate \(E(X)\) and \(E(X^2)\) explicitly from the probability distribution.

Hints

- Distinguish \(E(X^2)\) from \([E(X)]^2\). - Write the complete binomial distribution for \(n=5\). - Use the symmetry of the probabilities when \(p=0.5\). - Substitute the two moments into the variance identity.

Solution

1. The mean is \(E(X)=np=5\cdot0.5=2.5\). 2. The probabilities for \(X=0, 1, 2, 3, 4, 5\) are \(\frac{1}{32}, \frac{5}{32}, \frac{10}{32}, \frac{10}{32}, \frac{5}{32}, \frac{1}{32}\), respectively. 3. The second moment is \(E(X^2)=\frac{0^2\cdot1+1^2\cdot5+2^2\cdot10+3^2\cdot10+4^2\cdot5+5^2\cdot1}{32}\). 4. Therefore, \(E(X^2)=\frac{240}{32}=7.5\). 5. Finally, \(\operatorname{Var}(X)=7.5-(2.5)^2=1.25\).

Answer

\(E(X)=2.5\), \(E(X^2)=7.5\), and \(\operatorname{Var}(X)=1.25\)
53114712
A manufacturer performs continuous-use tests on specialized batteries. All batteries fail by \(30\) hours. The table shows the proportion still operating at each time \(t\). <table> <tr> <td>Time \(t\) in hours</td> <td>\(0\)</td> <td>\(5\)</td> <td>\(10\)</td> <td>\(15\)</td> <td>\(20\)</td> <td>\(25\)</td> <td>\(30\)</td> </tr> <tr> <td>Proportion operating</td> <td>\(100\%\)</td> <td>\(95\%\)</td> <td>\(85\%\)</td> <td>\(60\%\)</td> <td>\(30\%\)</td> <td>\(5\%\)</td> <td>\(0\%\)</td> </tr> </table> Estimate the mean battery life by assuming failures are uniformly distributed within each \(5\)-hour interval and using interval midpoints.

Hints

- Find the proportion that fails within each time interval by subtracting consecutive survival proportions. - Use each interval midpoint as its representative lifetime. - Compute a weighted average using the interval failure probabilities.

Solution

1. Subtract successive operating proportions to find the failure probabilities in the six intervals: \(0.05,0.10,0.25,0.30,0.25,0.05\). 2. The interval midpoints are \(2.5,7.5,12.5,17.5,22.5,27.5\) hours. 3. The estimated mean is \(E(X)=0.05\cdot2.5+0.10\cdot7.5+0.25\cdot12.5+0.30\cdot17.5+0.25\cdot22.5+0.05\cdot27.5=16.25\).

Answer

The estimated mean battery life is \(16.25\) hours.
53116612
A discrete random variable \(X\) can take the values \(1\), \(2\), and \(3\), with \(P(X=1)=a\), \(P(X=2)=0.4\), and \(P(X=3)=0.6-a\), where \(0\le a\le0.6\). a) Find \(a\) when \(E(X)=2.3\). b) Without further calculation, explain whether the expected value increases or decreases as \(a\) increases.

Hints

- Write the expected value as a weighted sum containing \(a\). - Simplify before solving the equation. - Track which outcome gains probability and which loses probability as \(a\) increases. - Shifting probability toward smaller values lowers the mean.

Solution

1. Write the expected value as \(E(X)=a+2\cdot0.4+3(0.6-a)\). 2. Simplifying gives \(E(X)=2.6-2a\). 3. Set \(2.6-2a=2.3\). Then \(2a=0.3\), so \(a=0.15\). 4. As \(a\) increases, probability shifts from the largest value, \(3\), to the smallest value, \(1\), while the probability of \(2\) stays fixed. Therefore, the expected value decreases.

Answer

a) \(a=0.15\) b) The expected value decreases as \(a\) increases.
53117012
A quality-control process finds that \(2.5\%\) of LED lights are defective. a) How many defective lights are expected in a batch of \(600\)? b) After production improvements, only \(1\) in every \(100\) lights is defective. How large must a batch be to have the same expected number of defective lights as in part a)?

Hints

- Use the mean formula for a binomial count. - Convert “one in every hundred” to a probability. - In part b, keep the expected number fixed and solve for the number of trials.

Solution

1. For part a), \(E(X)=np=600\cdot0.025=15\). 2. The new defect probability is \(p=\frac{1}{100}=0.01\). 3. Let the new batch size be \(n\). Set \(n\cdot0.01=15\). 4. Solving gives \(n=1500\).

Answer

a) \(15\) defective lights b) \(1500\) lights
53118012
A binomial random variable \(X\) has mean \(\mu=12\) and variance \(\operatorname{Var}(X)=7.2\). a) Find the parameters \(n\) and \(p\). b) Find \(P(X=12)\), the probability that \(X\) is exactly equal to its mean. Round to four decimal places.

Hints

- Use the binomial mean and variance formulas. - Substitute the mean into the variance equation to eliminate one unknown. - Use the binomial probability formula after finding the parameters.

Solution

1. The mean and variance equations are \(np=12\) and \(np(1-p)=7.2\). 2. Substitute \(np=12\) into the variance equation: \(12(1-p)=7.2\). 3. Thus, \(1-p=0.6\), so \(p=0.4\). 4. Since \(n\cdot0.4=12\), \(n=30\). 5. The required probability is \(P(X=12)=\binom{30}{12}(0.4)^{12}(0.6)^{18}\approx0.1474\).

Answer

a) \(n=30\) and \(p=0.4\) b) \(P(X=12)\approx0.1474\)
53126812
A digital prize wheel at a carnival is played about \(12{,}000\) times during a weekend. The prize probabilities and values are shown. <table> <tr> <th>Prize category</th> <th>Probability</th> <th>Prize value</th> </tr> <tr> <td>Grand prize</td> <td>\(\frac{1}{1000}\)</td> <td>\(\$250\)</td> </tr> <tr> <td>Small prize</td> <td>\(\frac{3}{100}\)</td> <td>\(\$10\)</td> </tr> <tr> <td>Consolation prize</td> <td>\(\frac{1}{10}\)</td> <td>\(\$2\)</td> </tr> <tr> <td>No prize</td> <td>\(\frac{869}{1000}\)</td> <td>\(\$0\)</td> </tr> </table> a) Find the expected number of winners in each of the three prize categories. b) Find the expected total value of all prizes awarded during the weekend.

Hints

- Expected count equals the number of games multiplied by the category probability. - For part b, either multiply each expected count by its prize value or first find the expected prize per game. - Check that all prize probabilities, including no prize, add to \(1\).

Solution

1. Multiply \(12{,}000\) by each winning probability. 2. Grand prize: \(12{,}000\cdot\frac{1}{1000}=12\). 3. Small prize: \(12{,}000\cdot\frac{3}{100}=360\). 4. Consolation prize: \(12{,}000\cdot\frac{1}{10}=1200\). 5. The expected total prize value is \(12\cdot250+360\cdot10+1200\cdot2=\$9000\). 6. Equivalently, the expected prize per game is \(\frac{1}{1000}\cdot250+\frac{3}{100}\cdot10+\frac{1}{10}\cdot2=\$0.75\), and \(12{,}000\cdot0.75=\$9000\).

Answer

a) Grand prizes: \(12\) Small prizes: \(360\) Consolation prizes: \(1200\) b) \(\$9000\)
53209012
A spinner has \(12\) equal sections labeled \(0\), \(1\), \(2\), and \(5\), as shown. The spinner is spun once, and the random variable \(X\) is the payout in dollars. a) Complete the probability distribution. <table border="1" cellpadding="5" style="border-collapse: collapse; text-align: center; margin: 10px 0;"> <thead> <tr> <th style="background-color: #f2f2f2; padding: 8px;">Payout \(x\)</th> <th style="padding: 8px;">\(\$0\)</th> <th style="padding: 8px;">\(\$1\)</th> <th style="padding: 8px;">\(\$2\)</th> <th style="padding: 8px;">\(\$5\)</th> </tr> </thead> <tbody> <tr> <th style="background-color: #f2f2f2; padding: 8px;">\(P(X=x)\)</th> <td style="padding: 8px;">\(\frac{1}{3}\)</td> <td style="padding: 8px;"></td> <td style="padding: 8px;">\(\frac{1}{4}\)</td> <td style="padding: 8px;"></td> </tr> </tbody> </table> b) Find \(E(X)\). c) The game costs \(\$1.50\) to play. Is the game fair to the player in the long run? Justify your answer.
Figure for problem 532090

Hints

- Count how many of the \(12\) equal sections have each label. - Divide each count by \(12\) to get its probability. - Find the expected payout as a weighted average. - Compare the expected payout with the entry fee. - A fair game has expected net gain \(0\).

Solution

1. Four sections are labeled \(1\), so \(P(X=1)=\frac{4}{12}=\frac{1}{3}\). One section is labeled \(5\), so \(P(X=5)=\frac{1}{12}\). 2. The expected payout is \(E(X)=0\cdot\frac{1}{3}+1\cdot\frac{1}{3}+2\cdot\frac{1}{4}+5\cdot\frac{1}{12}=\frac{15}{12}=\$1.25\). 3. The expected net gain is \(\$1.25-\$1.50=-\$0.25\). Therefore, the game is not fair to the player; the player loses an average of \(\$0.25\) per play.

Answer

a) \(P(X=1)=\frac{1}{3}\) and \(P(X=5)=\frac{1}{12}\) b) \(E(X)=\$1.25\) c) The game is not fair. The player’s expected net gain is \(-\$0.25\) per play.
53210312
A carnival game uses an urn containing \(4\) blue balls, \(3\) red balls, and \(1\) green ball, as shown. A player pays \(\$1.50\) and draws \(2\) balls without replacement. The payouts are: - Two red balls: \(\$5.00\) - Two blue balls: \(\$2.00\) - The green ball and any other ball: \(\$1.00\) - Any other result: \(\$0.00\) The random variable \(X\) is the player’s net gain, payout minus entry fee. a) Determine the probability distribution of \(X\). b) Calculate \(E(X)\) and decide whether the game is fair.
Figure for problem 532103

Hints

- First determine every possible net gain or loss. - Count the total unordered selections of \(2\) balls. - Count favorable color combinations for each net value. - Weight each net value by its probability to find the expected value. - A fair game has expected net gain \(0\).

Solution

1. The possible net gains are \(\$3.50\) for two red balls, \(\$0.50\) for two blue balls, \(-\$0.50\) when one ball is green, and \(-\$1.50\) for one red and one blue ball. 2. There are \(\binom{8}{2}=28\) equally likely unordered selections. Thus, \(P(X=3.50)=\frac{\binom{3}{2}}{28}=\frac{3}{28}\), \(P(X=0.50)=\frac{\binom{4}{2}}{28}=\frac{3}{14}\), \(P(X=-0.50)=\frac{7}{28}=\frac{1}{4}\), and \(P(X=-1.50)=\frac{3\cdot4}{28}=\frac{3}{7}\). 3. The expected value is \(E(X)=-1.50\left(\frac{3}{7}\right)-0.50\left(\frac{1}{4}\right)+0.50\left(\frac{3}{14}\right)+3.50\left(\frac{3}{28}\right)=-\frac{2}{7}\approx-0.29\). 4. Because the expected net gain is negative, the game is not fair to the player.

Answer

a) \(P(X=-1.50)=\frac{3}{7}\), \(P(X=-0.50)=\frac{1}{4}\), \(P(X=0.50)=\frac{3}{14}\), and \(P(X=3.50)=\frac{3}{28}\). b) \(E(X)=-\frac{2}{7}\text{ dollars}\approx-\$0.29\). The game is not fair.
53214312
A carnival game uses the spinner shown. It has \(8\) equal sections, and each number is the payout in dollars. a) Find the probability distribution of the payout \(X\), and calculate \(E(X)\). b) What entry fee makes the game fair to the operator, meaning the operator’s expected profit is \(\$0\) per play? c) What entry fee gives the operator an expected profit of \(\$0.50\) per play?
Figure for problem 532143

Hints

- Count the sections for each payout. - Divide each count by \(8\) to find its probability. - The expected payout is a weighted average. - A fair fee equals the expected payout. - Add the desired operator profit to the expected payout for part c).

Solution

1. Counting the equal sections gives \(P(X=0)=\frac{3}{8}\), \(P(X=2)=\frac{2}{8}=\frac{1}{4}\), \(P(X=5)=\frac{2}{8}=\frac{1}{4}\), and \(P(X=10)=\frac{1}{8}\). 2. The expected payout is \(E(X)=0\cdot\frac{3}{8}+2\cdot\frac{2}{8}+5\cdot\frac{2}{8}+10\cdot\frac{1}{8}=\$3.00\). 3. A fair entry fee equals the expected payout, so the fair fee is \(\$3.00\). 4. To earn an expected profit of \(\$0.50\), the operator must charge \(\$3.00+\$0.50=\$3.50\).

Answer

a) <table> <thead> <tr><th>Payout \(x\)</th><th>\(\$0\)</th><th>\(\$2\)</th><th>\(\$5\)</th><th>\(\$10\)</th></tr> </thead> <tbody> <tr><th>\(P(X=x)\)</th><td>\(\frac{3}{8}\)</td><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{8}\)</td></tr> </tbody> </table> \(E(X)=\$3.00\) b) \(\$3.00\) c) \(\$3.50\)
53215112
A carnival game uses the spinner shown, which has \(6\) equal sections. The game costs \(\$3\) to play. A red section labeled \(5\) pays \(\$10\), a blue section labeled \(2\) pays \(\$4\), and a yellow section labeled \(1\) pays nothing. The random variable \(X\) is the player’s net gain, equal to payout minus entry fee. a) Find the probability distribution of \(X\) and display it in a table. b) Find \(E(X)\) and determine whether the game is fair.
Figure for problem 532151

Hints

- Find the net gain for each possible payout. - Count how many sections produce each net gain. - Divide by the total of \(6\) sections. - Compute the expected net gain as a weighted average. - A fair game has expected net gain \(0\).

Solution

1. The possible net gains are \(\$10-\$3=\$7\), \(\$4-\$3=\$1\), and \(\$0-\$3=-\$3\). 2. Three of the six sections are yellow, two are blue, and one is red. Therefore, \(P(X=-3)=\frac{1}{2}\), \(P(X=1)=\frac{1}{3}\), and \(P(X=7)=\frac{1}{6}\). 3. The expected net gain is \(E(X)=-3\cdot\frac{1}{2}+1\cdot\frac{1}{3}+7\cdot\frac{1}{6}=0\). 4. Since the expected net gain is \(\$0\), the game is fair.

Answer

a) <table> <tr><th>Net gain \(x\)</th><td>\(-\$3\)</td><td>\(\$1\)</td><td>\(\$7\)</td></tr> <tr><th>\(P(X=x)\)</th><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{6}\)</td></tr> </table> b) \(E(X)=\$0\), so the game is fair.
53215712
At a carnival game, a spinner with \(8\) equal sections is spun once. The amounts shown are the player's net gain or loss in dollars. Let \(X\) be the player's net result from one play. a) Display the probability distribution of \(X\) in a table. b) Find \(E(X)\). Is the game fair? Explain. c) Find \(\sigma(X)\), rounded to the nearest hundredth.
Figure for problem 532157

Hints

- Count how many of the eight equal sections show each outcome. - Multiply each value by its probability and add to find the expected value. - A fair game has expected net gain \(\$0\) for the player. - Find the variance before taking its square root.

Solution

1. Four of the \(8\) sections show \(-\$2\), three show \(+\$1\), and one shows \(+\$5\). Therefore, \(P(X=-2)=\frac48=0.5\), \(P(X=1)=\frac38=0.375\), and \(P(X=5)=\frac18=0.125\). 2. The expected value is \(E(X)=-2\cdot0.5+1\cdot0.375+5\cdot0.125=0\) dollars. 3. The game is fair because the expected net result for the player is \(\$0\). 4. Since \(E(X)=0\), \(\operatorname{Var}(X)=E(X^2)=(-2)^2\cdot0.5+1^2\cdot0.375+5^2\cdot0.125=5.5\). 5. Therefore, \(\sigma(X)=\sqrt{5.5}\approx\$2.35\).

Answer

a) <table border="1" style="border-collapse: collapse; text-align: center;"> <tr> <th style="padding: 5px;">\(x_i\) (dollars)</th> <td style="padding: 5px;">\(-2\)</td> <td style="padding: 5px;">\(1\)</td> <td style="padding: 5px;">\(5\)</td> </tr> <tr> <th style="padding: 5px;">\(P(X=x_i)\)</th> <td style="padding: 5px;">\(0.5\)</td> <td style="padding: 5px;">\(0.375\)</td> <td style="padding: 5px;">\(0.125\)</td> </tr> </table> b) \(E(X)=\$0\). The game is fair because the expected net result is zero. c) \(\sigma(X)\approx\$2.35\)
53274312
The length of time visitors spend in an art gallery, in hours, is modeled by a continuous random variable \(X\) with density function \(f(x)=\frac{4}{9}x^2-\frac{4}{27}x^3\) for \(0\le x\le 3\), and \(f(x)=0\) otherwise. The graph of \(f\) is shown. a) Show algebraically that \(f\) is a valid probability density function. b) Find the probability that a randomly selected visitor spends at most \(1.5\) hours in the gallery. c) Find a formula for the cumulative distribution function \(F(x)\) on \([0, 3]\). d) Find the mean time visitors spend in the gallery. Give the result in hours and also in hours and minutes.
Figure for problem 532743

Hints

- What two conditions must a probability density function satisfy? - An interval probability is an integral of the density. - The cumulative distribution function is the accumulated area from the left endpoint to \(x\). - For a continuous random variable, the mean is \(\int x f(x)\,dx\).

Solution

1. a) On \([0, 3]\), \(f(x)=\frac{4}{27}x^2(3-x)\ge 0\), and it is \(0\) outside that interval. Also, \(\int_0^3\left(\frac{4}{9}x^2-\frac{4}{27}x^3\right)\,dx=\left[\frac{4}{27}x^3-\frac{1}{27}x^4\right]_0^3=4-3=1\). Thus, \(f\) is a valid density. 2. b) \(P(X\le 1.5)=\left[\frac{4}{27}x^3-\frac{1}{27}x^4\right]_0^{3/2}=\frac{5}{16}=0.3125\). 3. c) For \(0\le x\le 3\), \(F(x)=\int_0^x f(t)\,dt=\frac{4}{27}x^3-\frac{1}{27}x^4\). 4. d) \(E(X)=\int_0^3 x f(x)\,dx=\left[\frac{1}{9}x^4-\frac{4}{135}x^5\right]_0^3=1.8\) hours. Since \(0.8\cdot 60=48\), this is \(1\) hour \(48\) minutes.

Answer

a) \(f(x)\ge 0\) and \(\int_{-\infty}^{\infty}f(x)\,dx=1\). b) \(P(X\le 1.5)=\frac{5}{16}=0.3125\), or \(31.25\%\) c) \(F(x)=\frac{4}{27}x^3-\frac{1}{27}x^4\) for \(0\le x\le 3\) d) \(E(X)=1.8\) hours, or \(1\) hour \(48\) minutes
53477612
Customer wait time \(X\), in minutes, at a service hotline is modeled by the density function \(f(x)=0.2e^{-0.2x}\) for \(x\ge 0\), and \(f(x)=0\) for \(x<0\). The graph of \(f\) is shown on \([0, 15]\). a) Describe the graph's intercept, monotonic behavior, and end behavior. b) Find the probability that a customer waits at most \(5\) minutes. c) Find the cumulative distribution function \(F\). d) Find the mean \(\mu=\int_0^\infty x f(x)\,\mathrm{d}x\). Hint: An antiderivative of \(xe^{ax}\) is \(\frac{e^{ax}}{a^2}(ax-1)\).
Figure for problem 534776

Hints

- An interval probability is area under the density curve. - The cumulative distribution function gives \(P(X\le x)\). - Treat the mean as an improper integral and use the supplied antiderivative hint. - Evaluate \(f(0)\) and consider what happens as \(x\) becomes large.

Solution

1. a) The graph begins at \(f(0)=0.2\), decreases for \(x\ge 0\), and approaches the x-axis as \(x\to\infty\). For reference, \(f(5)\approx 0.0736\), \(f(10)\approx 0.0271\), and \(f(15)\approx 0.0100\). 2. b) \(P(X\le 5)=\int_0^5 0.2e^{-0.2x}\,dx=[-e^{-0.2x}]_0^5=1-e^{-1}\approx 0.6321\). 3. c) \(F(x)=0\) for \(x<0\). For \(x\ge 0\), \(F(x)=\int_0^x0.2e^{-0.2t}\,dt=1-e^{-0.2x}\). 4. d) An antiderivative of \(0.2xe^{-0.2x}\) is \(-e^{-0.2x}(x+5)\). Therefore, \(\mu=\left[-e^{-0.2x}(x+5)\right]_0^\infty=5\) minutes.

Answer

a) The graph starts at \((0, 0.2)\), decreases, and approaches the x-axis. b) \(P(X\le 5)=1-e^{-1}\approx 0.6321\), or about \(63.21\%\) c) \(F(x)=\begin{cases}0 & x<0\\1-e^{-0.2x} & x\ge 0\end{cases}\) d) \(\mu=5\) minutes
53477712
Let \(X\) represent the fraction of daylight at a location that is obscured by clouds. The density function is \(f(x)=6x(1-x)\) for \(0\le x\le 1\), and \(f(x)=0\) otherwise. a) Show that \(f\) is a probability density function. b) Find the mean \(\mu\), and interpret it in context. c) Find the probability that the obscured fraction is between \(25\%\) and \(75\%\). d) Find the cumulative distribution function \(F(x)\) on \([0, 1]\).
Figure for problem 534777

Hints

- A probability density must be nonnegative and integrate to \(1\). - For a continuous random variable, the mean is found using \(\int x f(x)\,dx\). - Convert percentages to decimals before using them as integration limits. - The cumulative distribution function is an antiderivative of the density with value \(0\) at the left endpoint.

Solution

1. a) The function is nonnegative on \([0, 1]\), and \(\int_0^1(6x-6x^2)\,dx=[3x^2-2x^3]_0^1=1\). Thus, it is a density. 2. b) \(E(X)=\int_0^1x(6x-6x^2)\,dx=[2x^3-1.5x^4]_0^1=0.5\). On average, one-half, or \(50\%\), of the daylight is obscured by clouds. 3. c) \(P(0.25\le X\le 0.75)=[3x^2-2x^3]_{0.25}^{0.75}=0.84375-0.15625=0.6875\). 4. d) For \(0\le x\le 1\), \(F(x)=\int_0^x(6t-6t^2)\,dt=3x^2-2x^3\).

Answer

a) \(f(x)\ge 0\) and \(\int_0^1f(x)\,dx=1\) b) \(\mu=0.5\). On average, \(50\%\) of the daylight is obscured. c) \(P(0.25\le X\le 0.75)=0.6875\), or \(68.75\%\) d) \(F(x)=3x^2-2x^3\) for \(0\le x\le 1\)
53478012
Two bus companies, SprintBus and SteadyLine, serve the same route. Passenger wait time, in minutes, is modeled by density \(f\) for SprintBus and density \(g\) for SteadyLine: \(f(x)=\frac{3}{8}x^2\) for \(0\le x\le 2\) \(g(x)=\frac{3}{8}(x-2)^2\) for \(0\le x\le 2\). Both functions are \(0\) outside \([0, 2]\), and their graphs are shown. a) Show algebraically that both functions are valid probability densities. b) For each company, find the probability that a passenger waits longer than \(1\) minute. c) Find the mean wait time for each model. Which company would you prefer when minimizing expected wait time?
Figure for problem 534780

Hints

- A probability density must be nonnegative and integrate to \(1\). - Find each interval probability by integrating the corresponding density. - For a continuous random variable, \(E(X)=\int x f(x)\,dx\). - The graph's area concentration can help you anticipate which model has the smaller mean.

Solution

1. a) Both functions are nonnegative on \([0, 2]\). Also, \(\int_0^2\frac{3}{8}x^2\,dx=1\) and \(\int_0^2\frac{3}{8}(x-2)^2\,dx=1\), so both are valid densities. 2. b) For SprintBus, \(P(X>1)=1-\int_0^1\frac{3}{8}x^2\,dx=1-\frac{1}{8}=\frac{7}{8}=0.875\). For SteadyLine, \(P(X>1)=\int_1^2\frac{3}{8}(x-2)^2\,dx=\frac{1}{8}=0.125\). 3. c) For SprintBus, \(E(X)=\int_0^2x\frac{3}{8}x^2\,dx=1.5\) minutes. For SteadyLine, \(E(X)=\int_0^2x\frac{3}{8}(x-2)^2\,dx=0.5\) minute. SteadyLine has the smaller mean wait time.

Answer

a) Both functions are nonnegative and integrate to \(1\) on \([0, 2]\). b) SprintBus: \(0.875\), or \(87.5\%\); SteadyLine: \(0.125\), or \(12.5\%\) c) SprintBus: \(1.5\) minutes; SteadyLine: \(0.5\) minute. SteadyLine is preferable when minimizing expected wait time.
53604412
A carnival game uses the spinner shown. It has \(10\) equal sections labeled with payout amounts in dollars. The random variable \(X\) is the payout for one play. a) Create a probability distribution table for \(X\). b) Find \(E(X)\). What entry fee makes the game fair? c) The operator wants to charge \(\$5.00\). To keep the game fair, the highest payout, \(\$24.00\), will be replaced by \(\$a\). Find \(a\).
Figure for problem 536044

Hints

- Each of the \(10\) equal sections has probability \(0.1\). - Count how many sections display each payout. - A fair entry fee equals the expected payout. - In part c, replace \(24\) with \(a\) in the expected-value equation. - Solve the resulting equation for \(a\).

Solution

1. Counting the sections gives \(P(X=0)=0.5\), \(P(X=2)=0.3\), \(P(X=10)=0.1\), and \(P(X=24)=0.1\). 2. The expected payout is \(E(X)=0\cdot0.5+2\cdot0.3+10\cdot0.1+24\cdot0.1=\$4.00\). 3. Therefore, a fair entry fee is \(\$4.00\). 4. For the new game, require \(0\cdot0.5+2\cdot0.3+10\cdot0.1+a\cdot0.1=5\). 5. This gives \(1.6+0.1a=5\), so \(a=34\).

Answer

a) <table> <tr><td>Payout \(x\)</td><td>\(\$0\)</td><td>\(\$2\)</td><td>\(\$10\)</td><td>\(\$24\)</td></tr> <tr><td>\(P(X=x)\)</td><td>\(0.5\)</td><td>\(0.3\)</td><td>\(0.1\)</td><td>\(0.1\)</td></tr> </table> b) \(E(X)=\$4.00\), so the fair entry fee is \(\$4.00\). c) \(a=34\), so the new highest payout is \(\$34.00\).
53746012
Two tennis players, A and B, play a best-of-three match. The first player to win \(2\) sets wins the match. Player A wins each set independently with probability \(p=0.6\). a) Use the tree diagram to give the probabilities of all possible match sequences. b) Let \(X\) be the number of sets played. Find the probability distribution of \(X\). c) Find \(E(X)\).
Figure for problem 537460

Hints

- Identify the sequences that end the match after two sets. - Stop a tree path as soon as one player has won two sets. - Combine sequence probabilities that produce the same match length. - Find the expected value from the distribution of \(X\).

Solution

1. A match ends in \(2\) sets with sequences \(AA\) or \(BB\): \(P(AA)=0.6^2=0.36\) and \(P(BB)=0.4^2=0.16\). 2. A match lasts \(3\) sets when the first two sets are split. The sequence probabilities are \(P(ABA)=0.6\cdot0.4\cdot0.6=0.144\), \(P(ABB)=0.6\cdot0.4\cdot0.4=0.096\), \(P(BAA)=0.4\cdot0.6\cdot0.6=0.144\), and \(P(BAB)=0.4\cdot0.6\cdot0.4=0.096\). 3. Therefore, \(P(X=2)=0.36+0.16=0.52\), and \(P(X=3)=0.48\). 4. The expected match length is \(E(X)=2\cdot0.52+3\cdot0.48=2.48\) sets.

Answer

a) \(P(AA)=0.36\), \(P(BB)=0.16\), \(P(ABA)=0.144\), \(P(ABB)=0.096\), \(P(BAA)=0.144\), and \(P(BAB)=0.096\) b) \(P(X=2)=0.52\) and \(P(X=3)=0.48\) c) \(E(X)=2.48\) sets
54725612
A warranty pays \(\$0\) with probability \(0.94\), \(\$200\) with probability \(0.05\), and \(\$1000\) with probability \(0.01\). a) Find the expected payout per warranty. b) Find the variance of the payout. c) Explain why the expected payout is not a prediction of the amount paid on any one warranty.

Hints

- Weight each payout by its probability. - Variance can be found from the second moment and the square of the mean. - Compare the expected value with the actual support of the distribution.

Solution

1. \(E(X)=0\cdot0.94+200\cdot0.05+1000\cdot0.01=20\). 2. \(E(X^2)=0+200^2\cdot0.05+1000^2\cdot0.01=12{,}000\). 3. \(\operatorname{Var}(X)=E(X^2)-[E(X)]^2=12{,}000-400=11{,}600\). 4. The expectation is a long-run average across many warranties; individual payouts are only \(0\), \(200\), or \(1000\).

Answer

a) Expected payout: \(\$20\). b) Variance: \(11{,}600\) dollars squared. c) The expectation is a long-run average, not a possible payout required on one warranty.
54736112
Four event indicators \(I_1,I_2,I_3,I_4\) have event probabilities \(0.20,0.35,0.50,\) and \(0.80\). Let \(N=I_1+I_2+I_3+I_4\) be the number of events that occur. Find \(E(N)\). Does this calculation require the events to be independent? Explain.

Hints

- Express the count as a sum of indicator variables. - Recall the expected value of an event indicator. - Separate the calculation of a mean from the calculation of a joint distribution.

Solution

1. Linearity of expectation gives \(E(N)=E(I_1)+E(I_2)+E(I_3)+E(I_4)\). 2. For an indicator, \(E(I_i)=P(I_i=1)\). 3. Thus \(E(N)=0.20+0.35+0.50+0.80=1.85\). 4. Linearity of expectation does not require independence; dependence affects the distribution of \(N\), but not this mean calculation.

Answer

\(E(N)=1.85\). Independence is not required.
54848812
A company knows that \(16\%\) of the parts produced by a certain machine need rework. A quality engineer will inspect a random sample of \(250\) parts and treat the sampled outcomes as independent. Let \(X\) be the number that need rework and let \(\hat p=X/250\). a) Find the mean and standard deviation of \(X\). b) Find the mean and standard deviation of \(\hat p\). c) Explain the numerical relationship between the two standard deviations.

Hints

- Keep track of whether the random variable is a count or a proportion. - Think about how multiplying a random variable by a constant changes its center and spread. - Check that each standard deviation is expressed in the same type of units as its random variable.

Solution

1. For \(X\), the mean is \(np=250(0.16)=40\), and the standard deviation is \(\sqrt{np(1-p)}=\sqrt{250(0.16)(0.84)}\approx 5.797\). 2. For \(\hat p\), the mean is \(p=0.16\), and the standard deviation is \(\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{(0.16)(0.84)}{250}}\approx 0.02319\). 3. Because \(X=250\hat p\), multiplying \(\hat p\) by \(250\) multiplies its standard deviation by \(250\). Indeed, \(250(0.02319)\approx 5.797\).

Answer

a) Mean \(=40\); standard deviation \(\approx 5.797\) parts. b) Mean \(=0.16\); standard deviation \(\approx 0.02319\). c) The count standard deviation is \(250\) times the sample-proportion standard deviation because \(X=250\hat p\).
52347012
A spinner has three equal sections colored red, blue, and yellow. It is spun independently \(12\) times. Let \(Y\) be the number of times it lands on red. a) Find \(P(Y=4)\). b) Find the mean \(\mu\) and standard deviation \(\sigma\) of \(Y\). c) Find the probability that \(Y\) differs from its mean by more than one standard deviation.

Hints

- The probability of red on one spin is \(\frac{1}{3}\). - Use the binomial probability formula for part a). - Find the numerical endpoints of \([\mu-\sigma, \mu+\sigma]\). - Add the probabilities of the integer counts outside that interval.

Solution

1. The random variable is binomial with \(n=12\) and \(p=\frac{1}{3}\). 2. \(P(Y=4)=\binom{12}{4}\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^8\approx0.2384\). 3. The mean is \(\mu=12\cdot\frac{1}{3}=4\), and the standard deviation is \(\sigma=\sqrt{12\cdot\frac{1}{3}\cdot\frac{2}{3}}=\sqrt{\frac{8}{3}}\approx1.633\). 4. More than one standard deviation from the mean means \(Y<4-1.633\) or \(Y>4+1.633\). Since \(Y\) is an integer, this is \(Y\le2\) or \(Y\ge6\). 5. Therefore, \(P(|Y-\mu|>\sigma)=P(Y\le2)+P(Y\ge6)\approx0.1811+0.1777=0.3588\).

Answer

a) \(P(Y=4)\approx0.2384\) b) \(\mu=4\) and \(\sigma\approx1.633\) c) \(P(|Y-\mu|>\sigma)\approx0.3588\)
52347412
In a carnival game, a player wins each round independently with probability \(p=0.25\). The player completes \(80\) rounds. Find the probability that the number of wins differs from its mean by more than two standard deviations.

Hints

- Find the mean and standard deviation of the binomial random variable first. - Determine which integer values lie outside \([\mu-2\sigma, \mu+2\sigma]\). - Use cumulative binomial probabilities for the two tails. - Pay attention to the strict inequality in “more than.”

Solution

1. Let \(X\) be the number of wins. Then \(X\) is binomial with \(n=80\) and \(p=0.25\). 2. The mean is \(\mu=np=80\cdot0.25=20\). 3. The standard deviation is \(\sigma=\sqrt{np(1-p)}=\sqrt{80\cdot0.25\cdot0.75}=\sqrt{15}\approx3.873\), so \(2\sigma\approx7.746\). 4. More than two standard deviations from the mean means \(|X-20|>7.746\). Thus, \(X<12.254\) or \(X>27.746\). 5. Since \(X\) is an integer, the event is \(X\le12\) or \(X\ge28\). 6. Using the binomial distribution, \(P(X\le12)\approx0.0221\) and \(P(X\ge28)\approx0.0295\). 7. Therefore, \(P(|X-20|>2\sigma)\approx0.0221+0.0295=0.0516\).

Answer

\(P(|X-20|>2\sigma)\approx0.0516\), or about \(5.16\%\)
52515112
In an automated drilling process, the magnitude of the error \(X\), in millimeters, has density \(f(x)=c(4-x^2)\) for \(0\le x\le2\), and \(f(x)=0\) otherwise. a) Find the constant \(c\). b) Find the probability that the error is at most \(1.5\,\text{mm}\). c) Find the mean \(E(X)\) and the standard deviation \(\sigma(X)\).

Hints

- What must the integral of a density over its full support equal? - An interval probability is the integral of the density over that interval. - Recall the formulas for the mean and variance of a continuous random variable. - The density is zero outside \([0,2]\).

Solution

1. a) A density must integrate to \(1\): \(c\int_0^2(4-x^2)\,\text{d}x=c\left[4x-\frac{x^3}{3}\right]_0^2=c\frac{16}{3}=1\). Thus, \(c=\frac{3}{16}\). 2. b) \(P(X\le1.5)=\int_0^{1.5}\frac{3}{16}(4-x^2)\,\text{d}x=\frac{3}{16}\left[4x-\frac{x^3}{3}\right]_0^{1.5}=\frac{117}{128}\approx0.9141\). 3. c) \(E(X)=\int_0^2x f(x)\,\text{d}x=\frac{3}{16}\left[2x^2-\frac{x^4}{4}\right]_0^2=0.75\,\text{mm}\). 4. Also, \(E(X^2)=\frac{3}{16}\int_0^2(4x^2-x^4)\,\text{d}x=0.8\,\text{mm}^2\). Therefore, \(\operatorname{Var}(X)=0.8-0.75^2=0.2375\,\text{mm}^2\), and \(\sigma(X)=\sqrt{0.2375}\approx0.4873\,\text{mm}\).

Answer

a) \(c=\frac{3}{16}=0.1875\) b) \(P(X\le1.5)=\frac{117}{128}\approx0.9141\), or about \(91.41\%\) c) \(E(X)=0.75\,\text{mm}\); \(\sigma(X)\approx0.4873\,\text{mm}\)
52678412
Let \(X\) be a binomial random variable with parameters \(n\) and \(p\). a) For \(n=50\) and \(p=0.01\), find the numerical values of the variance \(V\) and standard deviation \(\sigma\), and compare them. b) Show that when \(n=2\), the numerical value of the standard deviation is greater than the numerical value of the variance for every \(p\in(0, 1)\).

Hints

- Use the binomial variance and standard-deviation formulas. - Determine the range of \(2p(1-p)\) for \(0<p<1\). - Find the maximum possible value of \(p(1-p)\). - Recall how square roots compare with numbers between \(0\) and \(1\).

Solution

1. For part a), \(V=np(1-p)=50\cdot0.01\cdot0.99=0.495\). 2. The standard deviation is \(\sigma=\sqrt{0.495}\approx0.7036\), so numerically \(\sigma>V\). 3. For part b), when \(n=2\), \(V=2p(1-p)\). 4. The expression \(p(1-p)\) has a maximum of \(\frac14\) at \(p=\frac12\). Therefore, \(0<V\le\frac12<1\) for every \(p\in(0, 1)\). 5. Since \(0<V<1\), \(\sqrt{V}>V\). Thus, the numerical value of \(\sigma\) is greater than that of \(V\).

Answer

a) \(V=0.495\) and \(\sigma\approx0.7036\), so numerically \(\sigma>V\). b) For \(n=2\), \(0<V\le\frac12\). Therefore, \(\sigma=\sqrt{V}>V\) for every \(p\in(0, 1)\).
52686612
At a carnival game, a player first rolls a fair four-sided die labeled \(1\) through \(4\). Only a player who rolls a \(1\) may spin a prize wheel with three sections: Grand Prize \((\$60)\), Consolation Prize \((\$15)\), and No Prize \((\$0)\). The No Prize section has a central angle of \(120^\circ\). The Grand Prize and Consolation Prize sections are designed so that the expected prize value per game entry is \(\$5.25\). Find the central angles of the Grand Prize and Consolation Prize sections.

Hints

- Find the probability of reaching the prize wheel. - Convert the \(120^\circ\) section into a probability. - Write one equation for the two unknown wheel probabilities and one for the expected prize. - Check that all three central angles add to \(360^\circ\).

Solution

1. The probability of qualifying to spin is \(\frac{1}{4}\). 2. The probability of No Prize on the wheel is \(\frac{120}{360}=\frac{1}{3}\). Therefore, the probabilities \(p_G\) and \(p_C\) of the other two sections satisfy \(p_G+p_C=\frac{2}{3}\). 3. The overall expected prize value gives \(\frac{1}{4}(60p_G+15p_C)=5.25\), so \(60p_G+15p_C=21\). 4. Substitute \(p_C=\frac{2}{3}-p_G\): \(60p_G+15\left(\frac{2}{3}-p_G\right)=21\). Thus, \(45p_G=11\), so \(p_G=\frac{11}{45}\). 5. Then \(p_C=\frac{2}{3}-\frac{11}{45}=\frac{19}{45}\). 6. The central angles are \(\frac{11}{45}\cdot360^\circ=88^\circ\) and \(\frac{19}{45}\cdot360^\circ=152^\circ\).

Answer

The Grand Prize section has a central angle of \(88^\circ\), and the Consolation Prize section has a central angle of \(152^\circ\).
52702412
In a manufacturing process, \(5\%\) of components are defective. Assume components are independently classified with this constant defect probability. A sample of \(80\) components is selected. Let \(Y\) be the number of defective components in the sample. Find the probabilities of these events: 1. More than \(6\) components are defective. 2. The number of defective components is within one standard deviation of the mean. Include exactly the integer values that lie in the interval.

Hints

- Decide whether the boundary value is included in “more than.” - Find the binomial mean and standard deviation. - Use a complement for the upper-tail probability. - Write the decimal endpoints of the interval before selecting the integer values.

Solution

1. The random variable is binomial with \(n=80\) and \(p=0.05\). 2. For the first event, \(P(Y>6)=1-P(Y\le6)\approx1-0.8947=0.1053\). 3. The mean is \(\mu=np=80\cdot0.05=4\), and the standard deviation is \(\sigma=\sqrt{80\cdot0.05\cdot0.95}=\sqrt{3.8}\approx1.949\). 4. The interval is approximately \([4-1.949, 4+1.949]=[2.051, 5.949]\). 5. The integer values in this interval are \(3, 4, 5\). 6. Therefore, \(P(3\le Y\le5)=P(Y\le5)-P(Y\le2)\approx0.7892-0.2306=0.5586\).

Answer

1. \(P(Y>6)\approx0.1053\) 2. \(P(3\le Y\le5)\approx0.5586\)
52713212
In an electronics manufacturing process, \(5\%\) of components are defective. Assume components are independently classified with this constant defect probability. A random sample of \(400\) components is selected. Let \(X\) be the number of defective components. 1. Find the interval \([\mu-2\sigma, \mu+2\sigma]\). 2. Find the probability that \(X\) lies outside this interval. 3. Interpret the result in context, referring to variability around the mean.

Hints

- Find \(\mu\) and \(\sigma\) first. - The interval endpoints need not be integers, but \(X\) can take only integer values. - “Outside” means below the lower endpoint or above the upper endpoint. - Add the two binomial tail probabilities.

Solution

1. The random variable is binomial with \(n=400\) and \(p=0.05\). Its mean is \(\mu=400\cdot0.05=20\), and its standard deviation is \(\sigma=\sqrt{400\cdot0.05\cdot0.95}=\sqrt{19}\approx4.36\). 2. Therefore, \([\mu-2\sigma, \mu+2\sigma]\approx[20-2\cdot4.36, 20+2\cdot4.36]=[11.28, 28.72]\). 3. Since \(X\) is an integer, lying outside the interval means \(X\le11\) or \(X\ge29\). 4. Using unrounded binomial probabilities, \(P(X\le11)\approx0.019046\) and \(P(X\ge29)\approx0.030705\). 5. Thus, \(P(|X-\mu|>2\sigma)\approx0.019046+0.030705=0.049751\approx0.0498\). 6. About \(5.0\%\) of samples have a defective-component count more than two standard deviations from the mean, while about \(95.0\%\) lie within the interval.

Answer

1. \([11.28, 28.72]\) 2. \(P(|X-\mu|>2\sigma)\approx0.0498\) 3. About \(5.0\%\) of samples fall outside the two-standard-deviation interval, and about \(95.0\%\) fall inside it.
52713412
For a binomial random variable \(X\) with parameters \(n\) and \(p\), where \(0<p<1\), let \(\mu\) be the mean and \(\sigma\) the standard deviation. a) Show that \(\sigma^2=\mu(1-p)\). b) Use the result from part a) to show that \(\sigma<\sqrt{\mu}\) for every \(0<p<1\). c) Find \(p\) in terms of \(n\) if the standard deviation is exactly one third of the mean.

Hints

- Substitute the standard formulas for the binomial mean and variance. - Consider what happens when a positive number is multiplied by a factor between \(0\) and \(1\). - Translate the condition in part c) into an equation. - Isolate \(p\) after simplifying.

Solution

1. Since \(\mu=np\) and \(\sigma^2=np(1-p)\), replacing \(np\) with \(\mu\) gives \(\sigma^2=\mu(1-p)\). 2. For \(0<p<1\), \(0<1-p<1\). Therefore, \(\sigma^2=\mu(1-p)<\mu\). 3. Both sides are positive, so taking square roots gives \(\sigma<\sqrt{\mu}\). 4. For part c), set \(\sigma=\frac13\mu\). Squaring gives \(\sigma^2=\frac19\mu^2\). 5. Equating the two expressions for \(\sigma^2\) gives \(\mu(1-p)=\frac19\mu^2\). 6. Since \(\mu=np>0\), divide by \(\mu\): \(1-p=\frac19np\). 7. Solving, \(1=p\left(1+\frac{n}{9}\right)\), so \(p=\frac{9}{n+9}\).

Answer

a) \(\sigma^2=\mu(1-p)\) b) Since \(0<1-p<1\), \(\sigma^2<\mu\), so \(\sigma<\sqrt{\mu}\). c) \(p=\frac{9}{n+9}\)
52714612
For a fixed number of trials \(n\), the variability of a binomial random variable depends on the success probability \(p\). a) Show algebraically that \(V(p)=np(1-p)\) is symmetric about \(p=0.5\) by verifying that \(V(0.5-h)=V(0.5+h)\) for any \(h\in[0, 0.5]\). b) Find the value of \(p\) that maximizes the standard deviation. Express the maximum standard deviation in terms of \(n\).

Hints

- Use the definition of symmetry about a vertical line. - Apply the difference-of-squares identity in part a). - The square-root function is increasing, so maximize the expression under the radical. - Use the vertex of the quadratic \(p(1-p)\).

Solution

1. Substitute \(0.5-h\): \(V(0.5-h)=n(0.5-h)[1-(0.5-h)]=n(0.5-h)(0.5+h)\). 2. Using the difference of squares, \(V(0.5-h)=n(0.25-h^2)\). 3. Similarly, \(V(0.5+h)=n(0.5+h)(0.5-h)=n(0.25-h^2)\). 4. Therefore, \(V(0.5-h)=V(0.5+h)\), proving symmetry about \(p=0.5\). 5. The standard deviation is greatest when the variance is greatest. The function \(p(1-p)\) reaches its maximum at \(p=0.5\). 6. Thus, \(\sigma_{\max}=\sqrt{n\cdot0.5\cdot0.5}=\frac{\sqrt{n}}{2}\).

Answer

a) \(V(0.5-h)=n(0.25-h^2)=V(0.5+h)\) b) The maximum occurs at \(p=0.5\), and \(\sigma_{\max}=\frac{\sqrt{n}}{2}\).
52714712
A binomial random variable \(X\) has parameters \(n\in\mathbb{N}\), \(n\ge1\), and \(0<p<1\). Find all values of \(p\) for which the standard deviation is less than one half of the mean. Express the result in terms of \(n\).

Hints

- Substitute the binomial formulas for the mean and standard deviation. - Square both sides after confirming that both are positive. - Divide only by quantities known to be positive. - Isolate \(p\) and then apply its domain restriction.

Solution

1. Write the required inequality: \(\sqrt{np(1-p)}<\frac12np\). 2. Both sides are positive, so square both sides: \(np(1-p)<\frac14n^2p^2\). 3. Divide by the positive quantity \(np\): \(1-p<\frac14np\). 4. Rearrange: \(1<p\left(1+\frac{n}{4}\right)\). 5. Therefore, \(p>\frac{1}{1+n/4}=\frac{4}{n+4}\). 6. Combining this with \(0<p<1\), the solution is \(p\in\left(\frac{4}{n+4}, 1\right)\).

Answer

\(p>\frac{4}{n+4}\), or \(p\in\left(\frac{4}{n+4}, 1\right)\)
52715612
The coefficient of variation of a random variable is the ratio of its standard deviation to its mean. Let \(Z_n\) be the sum of the measurements from \(n\) independent trials, where \(E(Z_n)=8n\) and \(\operatorname{Var}(Z_n)=2n\). Find \(n\) if the coefficient of variation of \(Z_n\) is exactly \(1.25\%\).

Hints

- Find the standard deviation from the variance. - Substitute the given expressions into the coefficient-of-variation formula. - Convert the percentage to a decimal or fraction. - Simplify \(\frac{\sqrt n}{n}\) before solving.

Solution

1. The standard deviation is \(\sigma(Z_n)=\sqrt{2n}\). 2. The coefficient of variation is \(\frac{\sigma(Z_n)}{E(Z_n)}=\frac{\sqrt{2n}}{8n}\). 3. Convert the target percentage: \(1.25\%=0.0125=\frac{1}{80}\). 4. Set up the equation \(\frac{\sqrt{2n}}{8n}=\frac{1}{80}\). 5. Since \(\frac{\sqrt{2n}}{8n}=\frac{\sqrt2}{8\sqrt n}\), the equation becomes \(\frac{\sqrt2}{8\sqrt n}=\frac{1}{80}\). 6. Solving gives \(8\sqrt n=80\sqrt2\), so \(\sqrt n=10\sqrt2\). 7. Squaring gives \(n=200\).

Answer

\(n=200\)
53097512
Two equally matched basketball teams play a best-of-seven playoff series. The first team to win \(4\) games wins the series, and each team has probability \(0.5\) of winning any game. The random variable \(X\) is the number of games played before the series ends. a) Determine the probability distribution of \(X\). b) Calculate the expected number of games, \(E(X)\).

Hints

- Identify all possible series lengths. - For the series to end in a particular game, the winner must win that last game and have exactly \(3\) earlier wins. - Account for either team being the series winner. - Compute an expected value by weighting each possible length by its probability.

Solution

1. The series can end after \(4\), \(5\), \(6\), or \(7\) games. 2. For a series to end in game \(m\), the eventual winner must have exactly \(3\) wins in the first \(m-1\) games and then win game \(m\). Either team can be the winner. Thus, \(P(X=m)=2\binom{m-1}{3}(0.5)^m\). 3. This gives \(P(X=4)=0.125\), \(P(X=5)=0.25\), \(P(X=6)=0.3125\), and \(P(X=7)=0.3125\). 4. The expected value is \(E(X)=4(0.125)+5(0.25)+6(0.3125)+7(0.3125)=5.8125\).

Answer

a) \(P(X=4)=0.125\), \(P(X=5)=0.25\), \(P(X=6)=0.3125\), and \(P(X=7)=0.3125\). b) \(E(X)=5.8125\) games.
53114812
A nursery studies how long newly planted seedlings survive during their first \(6\) weeks. By the end of week \(6\), none are alive. The table shows the proportion still alive after \(w\) weeks. <table> <tr> <td>Week \(w\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> <td>\(5\)</td> <td>\(6\)</td> </tr> <tr> <td>Proportion alive</td> <td>\(1.00\)</td> <td>\(0.92\)</td> <td>\(0.80\)</td> <td>\(0.55\)</td> <td>\(0.25\)</td> <td>\(0.05\)</td> <td>\(0.00\)</td> </tr> </table> Because the exact time of death within each week is unknown, the mean survival time can only be bounded. Find the lower and upper bounds for the mean survival time in weeks.

Hints

- Subtract consecutive survival proportions to find the proportion dying in each week. - For the lower bound, use the earliest possible time in each weekly interval. - For the upper bound, use the latest possible time in each interval. - Compare the two sets of representative times.

Solution

1. The proportions dying during weeks \(1\) through \(6\) are \(0.08,0.12,0.25,0.30,0.20,0.05\). 2. For the lower bound, place each death at the beginning of its week: \(E_{\min}=0\cdot0.08+1\cdot0.12+2\cdot0.25+3\cdot0.30+4\cdot0.20+5\cdot0.05=2.57\). 3. For the upper bound, place each death at the end of its week: \(E_{\max}=1\cdot0.08+2\cdot0.12+3\cdot0.25+4\cdot0.30+5\cdot0.20+6\cdot0.05=3.57\). 4. The upper bound is exactly one week greater because every representative time is shifted by \(1\) week.

Answer

The lower bound is \(2.57\) weeks, and the upper bound is \(3.57\) weeks.
53216012
A spinner has \(6\) equal sections: one red section labeled \(0\), three green sections labeled \(2\), \(4\), and \(6\), and two yellow sections labeled \(8\) and \(10\). Before playing, a player chooses to spin exactly \(n\) times. If any spin lands on \(0\), the game ends immediately and the payout is \(\$0\). If none of the \(n\) spins lands on \(0\), the payout is the sum of all numbers spun, in dollars. The expected payout has the form \(E(n)=anb^n\). a) Find \(a\) and \(b\). b) Find \(E(3)\) and \(E(4)\), rounded to the nearest cent. c) Show that two consecutive values of \(n\) give exactly the same expected payout, and identify them.
Figure for problem 532160

Hints

- Determine when the game pays anything at all. - Find the conditional mean of one nonzero spin. - Find the probability that all \(n\) spins avoid \(0\). - Substitute \(n=3\) and \(n=4\) into the expected-value formula. - For part c, solve \(E(n)=E(n+1)\).

Solution

1. The probability of avoiding \(0\) on one spin is \(b=\frac{5}{6}\). 2. Conditional on avoiding \(0\), the mean value of one spin is \(a=\frac{2+4+6+8+10}{5}=6\). 3. If all \(n\) spins are nonzero, their expected sum is \(6n\), and this event has probability \(\left(\frac{5}{6}\right)^n\). Therefore, \(E(n)=6n\left(\frac{5}{6}\right)^n\). 4. \(E(3)=18\left(\frac{5}{6}\right)^3=\frac{125}{12}\approx\$10.42\). 5. \(E(4)=24\left(\frac{5}{6}\right)^4=\frac{625}{54}\approx\$11.57\). 6. Set \(E(n)=E(n+1)\): \(6n\left(\frac{5}{6}\right)^n=6(n+1)\left(\frac{5}{6}\right)^{n+1}\). 7. Dividing by \(6\left(\frac{5}{6}\right)^n\) gives \(n=\frac{5}{6}(n+1)\), so \(n=5\). 8. Thus, \(E(5)=E(6)=\frac{15625}{1296}\approx\$12.06\).

Answer

a) \(a=6\) and \(b=\frac{5}{6}\) b) \(E(3)\approx\$10.42\) and \(E(4)\approx\$11.57\) c) \(n=5\) and \(n=6\)
53274712
At a coffee roastery, \(X\) is the amount, in grams, by which a package's fill weight exceeds the required minimum. The continuous random variable \(X\) has values in \([0, 4]\) and density function \(f(x)=\begin{cases}kx(4-x)^2 & \text{for }0\le x\le 4\\0 & \text{otherwise}\end{cases}\). The graph of \(f\) is shown. a) Find \(k\) so that \(f\) is a valid probability density function. b) Find \(P(1\le X\le 3)\). c) Estimate the mode from the graph, and then find it exactly by determining the maximum of \(f\) on \([0, 4]\). d) The mean is \(E(X)=1.6\,\text{g}\), and the standard deviation is \(0.8\,\text{g}\). Find the probability that \(X\) is within one standard deviation of the mean.
Figure for problem 532747

Hints

- What must the integral of a density over its full support equal? - Find an interval probability by integrating the density over that interval. - Use the first derivative to locate the maximum of \(f\). - First calculate the endpoints of the interval within one standard deviation of the mean.

Solution

1. a) A density must have total area \(1\): \(\int_0^4 kx(4-x)^2\,dx=1\). Since \(\int_0^4 x(4-x)^2\,dx=\frac{64}{3}\), \(k=\frac{3}{64}\). 2. b) An antiderivative is \(F(x)=\frac{3}{64}\left(8x^2-\frac{8}{3}x^3+\frac{1}{4}x^4\right)\). Thus, \(P(1\le X\le 3)=F(3)-F(1)=\frac{243}{256}-\frac{67}{256}=\frac{11}{16}=0.6875\). 3. c) The graph suggests a mode near \(1.3\). With \(f(x)=\frac{3}{64}(16x-8x^2+x^3)\), solve \(f'(x)=\frac{3}{64}(16-16x+3x^2)=0\). The critical values are \(x=4\) and \(x=\frac{4}{3}\). Comparing values at the critical points and endpoints shows the maximum occurs at \(x=\frac{4}{3}\approx 1.33\). 4. d) The interval within one standard deviation is \([1.6-0.8, 1.6+0.8]=[0.8, 2.4]\). Then \(P(0.8\le X\le 2.4)=F(2.4)-F(0.8)=0.8208-0.1808=0.64\).

Answer

a) \(k=\frac{3}{64}\) b) \(P(1\le X\le 3)=\frac{11}{16}=0.6875\), or \(68.75\%\) c) Estimate: about \(1.3\,\text{g}\); exact mode: \(\frac{4}{3}\,\text{g}\approx 1.33\,\text{g}\) d) \(P(0.8\le X\le 2.4)=0.64\), or \(64\%\)

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