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- Combine the centers of the two stages directly. - Use the stated relationship between the stages when combining their spreads. - Convert the combined variance back to the original time unit.

1. The mean of the sum is \(\mu_T=4.2+3.5=7.7\). 2. Independence allows the variances to add: \(\sigma_T^2=0.6^2+0.8^2=1\). 3. Therefore, \(\sigma_T=1.0\) second. 4. Total scan times average \(7.7\) seconds and typically differ from that mean by about \(1.0\) second.

\(\mu_T=7.7\,\text{seconds}\) and \(\sigma_T=1.0\,\text{second}\). Total scan times average \(7.7\) seconds and typically differ from that mean by about \(1.0\) second.

- Express the total charge as a linear transformation of repair time. - Separate the role of the fixed charge from the hourly multiplier. - Interpret both parameters in dollars.

1. The mean is \(\mu_C=65+42(2.8)=182.6\). 2. The fixed charge does not affect spread, so \(\sigma_C=42(0.7)=29.4\). 3. Repair charges average \(\$182.60\), with a typical deviation of \(\$29.40\) from that average.

\(\mu_C=\$182.60\) and \(\sigma_C=\$29.40\). Repair charges average \(\$182.60\), with a typical deviation of \(\$29.40\) from that average.

- Apply the full transformation to the center. - Standard deviation responds to the magnitude of a multiplier, not its sign. - Compare two hypothetical original scores to see what the negative coefficient does to their order.

1. The mean is \(\mu_Y=100-1.5(40)=40\). 2. The standard deviation is \(\sigma_Y=|-1.5|(6)=9\). 3. Because the coefficient is negative, a larger value of \(X\) produces a smaller value of \(Y\), so the ordering is reversed.

a) \(\mu_Y=40\) and \(\sigma_Y=9\). b) The negative coefficient reverses the ranking: higher \(X\)-values correspond to lower \(Y\)-values.

- Handle the subtraction and division as separate transformation steps. - Identify which step changes the center and which changes the scale. - Interpret the new unit in relation to the original spread.

1. The mean is \(\mu_Z=\frac{65-65}{7}=0\). 2. Subtracting \(65\) does not change spread, and dividing by \(7\) gives \(\sigma_Z=\frac{7}{7}=1\). 3. The transformation recenters the variable at \(0\) and rescales one standard deviation of \(X\) to one unit of \(Z\).

\(\mu_Z=0\) and \(\sigma_Z=1\). The transformation expresses values relative to the original mean in units of the original standard deviation.

- Treat the total as a sum of identical independent variables. - Express the average as a scaled version of the total. - Keep track of how the scaling affects both parameters.

1. For the total, \(\mu_S=10(45)=450\) and \(\sigma_S=9\sqrt{10}\approx28.46\). 2. Since \(A=\frac{S}{10}\), \(\mu_A=45\) and \(\sigma_A=\frac{9}{\sqrt{10}}\approx2.85\). 3. The average keeps the component mean but has less spread than one daily value.

\(\mu_S=450\) units and \(\sigma_S\approx28.46\) units. \(\mu_A=45\) units and \(\sigma_A\approx2.85\) units.

- Express the average as a scaled sum. - Track variance through the sum before returning to standard deviation. - Compare how the number of independent measurements changes the spread.

1. The analyst divided the component standard deviation directly by the number of measurements. 2. For an average of \(25\) independent measurements, the standard deviation is divided by \(\sqrt{25}\), not by \(25\). 3. The correct value is \(\frac{10}{\sqrt{25}}=2\) units.

The analyst used \(25\) instead of \(\sqrt{25}\). The correct standard deviation is \(2\) units.

- Handle the center and spread of each scoring rule separately. - A fixed added amount affects only one of the two parameters. - Compare variability using nonnegative spread values.

1. For \(A=50+10X\), \(\mu_A=50+10(4)=90\) and \(\sigma_A=10(1.2)=12\). 2. For \(B=80-6X\), \(\mu_B=80-6(4)=56\) and \(\sigma_B=|-6|(1.2)=7.2\). 3. Since \(12>7.2\), score \(A\) has greater variability.

\(\mu_A=90\), \(\sigma_A=12\); \(\mu_B=56\), \(\sigma_B=7.2\). Score \(A\) has greater variability.

- Find the probability that one pollen grain lands in one specified cell. - Model the number of grains in that cell using repeated independent trials. - Multiply the one-cell probability by the number of cells to obtain the expected count.

1. For one selected cell, each pollen grain lands in that cell with probability \(p=\frac{1}{100}=0.01\). Let \(X\) be the number of pollen grains in the cell. Then \(X\) is binomial with \(n=120\) and \(p=0.01\). 2. \(P(X=0)=(0.99)^{120}\approx0.2994\). 3. \(P(X=1)=\binom{120}{1}(0.01)(0.99)^{119}\approx0.3629\). 4. \(P(X=2)=\binom{120}{2}(0.01)^2(0.99)^{118}\approx0.2181\). 5. By linearity of expectation, the expected number of cells containing exactly one pollen grain is \(100P(X=1)\approx36.29\).

a) \(P(X=0)\approx0.2994\) b) \(P(X=1)\approx0.3629\) c) \(P(X=2)\approx0.2181\) The expected number of cells containing exactly one pollen grain is about \(36.29\).

- List the \(25\) ordered pairs from two spins. - Group pairs that produce the same product. - Use the probability distribution to find the mean. - For independent spins, the expected product equals the product of the expected values. - In part c, find the new mean of one spin first.

1. There are \(5\cdot5=25\) equally likely ordered pairs. Counting pairs that produce each product gives the distribution in the answer. 2. From the distribution, \(E(X)=\frac{1\cdot1+2\cdot2+3\cdot2+4\cdot3+5\cdot2+6\cdot2+8\cdot2+9\cdot1+10\cdot2+12\cdot2+15\cdot2+16\cdot1+20\cdot2+25\cdot1}{25}=9\). 3. Alternatively, the spins are independent, so the expected product is the product of the single-spin means: \(E(X)=3\cdot3=9\). 4. After replacing \(5\) with \(k\), the mean of one spin is \(\frac{1+2+3+4+k}{5}=\frac{10+k}{5}\). 5. Require \(\left(\frac{10+k}{5}\right)^2=16\). Since the mean is positive, \(\frac{10+k}{5}=4\), giving \(k=10\).

a) <table> <tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td><td>\(6\)</td><td>\(8\)</td><td>\(9\)</td><td>\(10\)</td><td>\(12\)</td><td>\(15\)</td><td>\(16\)</td><td>\(20\)</td><td>\(25\)</td></tr> <tr><td>\(P(X=x)\)</td><td>\(\frac{1}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{3}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{1}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{1}{25}\)</td><td>\(\frac{2}{25}\)</td><td>\(\frac{1}{25}\)</td></tr> </table> b) \(E(X)=9\) c) \(k=10\)

- Treat the center and spread as separate calculations. - Consider what happens to a negative coefficient when it is squared. - Use the independence information when combining variability.

1. The mean is \(\mu_D=52-47=5\). 2. Independence allows the variances to add even for a difference: \(\sigma_D^2=4^2+(-1)^2(3^2)=25\). 3. Therefore, \(\sigma_D=5\). 4. Standard deviations are not subtracted because variability is combined through squared coefficients and variances.

a) \(\mu_D=5\) requests and \(\sigma_D=5\) requests. b) The sign in a difference changes the mean contribution but disappears when the coefficient is squared for variance, so the variances add rather than the standard deviations subtracting.

- Work backward from the parameter rules for a sum. - Treat the center and variability equations separately. - Remember to solve for a nonnegative standard deviation.

1. From the mean of the sum, \(\mu_Y=50-18=32\). 2. Independence gives \(6.5^2=2.5^2+\sigma_Y^2\). 3. Therefore, \(\sigma_Y^2=42.25-6.25=36\), so \(\sigma_Y=6\).

\(\mu_Y=32\) and \(\sigma_Y=6\).

- First determine the spread of each sample average. - Combine the two centers with the sign used in the definition of the difference. - Use the independence of the samples when combining variability.

1. The sample means have \(\mu_{\overline{X}_A}=100\), \(\sigma_{\overline{X}_A}=\frac{6}{\sqrt{4}}=3\), \(\mu_{\overline{X}_B}=98\), and \(\sigma_{\overline{X}_B}=\frac{9}{\sqrt{9}}=3\). 2. The mean difference is \(\mu_D=100-98=2\,\text{mm}\). 3. Independence gives \(\sigma_D=\sqrt{3^2+3^2}=\sqrt{18}\approx4.24\,\text{mm}\).

\(\mu_D=2\,\text{mm}\) and \(\sigma_D\approx4.24\,\text{mm}\).

- Use the target center to determine the unknown coefficient. - The added constant affects the center but not the spread. - After finding the coefficient, consider its magnitude when transforming standard deviation.

1. The mean condition gives \(8a+7=31\). 2. Solving gives \(a=3\). 3. The standard deviation is multiplied by \(|a|\), so \(\sigma_Y=3(2.5)=7.5\).

\(a=3\) and \(\sigma_Y=7.5\).

- Compute the two centers using their respective signs. - Compare how the coefficient of the second variable enters the variability calculation. - Focus on the coefficient after it is squared.

1. The means are \(\mu_S=20+14=34\) and \(\mu_D=20-14=6\). 2. Independence gives \(\sigma_S^2=5^2+2^2=29\). 3. For the difference, the coefficient of \(Y\) is \(-1\), whose square is \(1\), so \(\sigma_D^2=5^2+(-1)^2(2^2)=29\). 4. Thus, \(\sigma_S=\sigma_D=\sqrt{29}\approx5.39\).

\(\mu_S=34\), \(\sigma_S\approx5.39\); \(\mu_D=6\), \(\sigma_D\approx5.39\). The spreads are equal because changing the sign of \(Y\) does not change its squared contribution to variance.

- List every pair of possible component values and its resulting sum. - Use the relationship between the variables to obtain each pair probability. - Combine probabilities when different pairs produce the same sum.

1. Independence gives \(P(S=1)=P(X=0)P(Y=1)=0.70(0.40)=0.28\). 2. The value \(S=3\) can occur as \(0+3\) or \(2+1\), so \(P(S=3)=0.70(0.60)+0.30(0.40)=0.54\). 3. The value \(S=5\) occurs as \(2+3\), so \(P(S=5)=0.30(0.60)=0.18\). 4. Therefore, \(P(S\ge3)=0.54+0.18=0.72\).

<table> <tr><th>\(s\)</th><th>\(1\)</th><th>\(3\)</th><th>\(5\)</th></tr> <tr><th>\(P(S=s)\)</th><td>\(0.28\)</td><td>\(0.54\)</td><td>\(0.18\)</td></tr> </table> \(P(S\ge3)=0.72\)

- Treat production hours as a transformation of planning hours. - Simplify the sum before applying any variance rule. - Consider how one component must change when the other component changes.

1. The transformation \(Y=30-X\) gives \(\mu_Y=30-12=18\) and \(\sigma_Y=|-1|(3)=3\). 2. Since \(X+Y=X+(30-X)=30\), the sum is constant. 3. Therefore, \(\mu_{X+Y}=30\) and \(\sigma_{X+Y}=0\). 4. The variables are perfectly dependent: an increase in planning hours forces an equal decrease in production hours, so their variations cancel.

a) \(\mu_Y=18\) hours and \(\sigma_Y=3\) hours. b) \(\mu_{X+Y}=30\) hours and \(\sigma_{X+Y}=0\) hours. c) The variables are not independent; they always move in opposite directions by equal amounts.

- Match each coefficient to the corresponding component parameter. - Combine centers using the stated weights. - For spread, apply the coefficients before combining the independent contributions.

1. The mean is \(\mu_G=0.2(80)+0.3(74)+0.5(90)=83.2\). 2. Independence gives \(\sigma_G^2=(0.2)^2(6^2)+(0.3)^2(8^2)+(0.5)^2(4^2)=11.2\). 3. Therefore, \(\sigma_G=\sqrt{11.2}\approx3.35\).

\(\mu_G=83.2\) points and \(\sigma_G\approx3.35\) points.

- Relate the total variability to the number of independent components. - Work with variance before solving for the count. - Once the count is known, combine the component centers.

1. Independence gives \(4.8^2=n(1.2^2)\). 2. Therefore, \(n=\left(\frac{4.8}{1.2}\right)^2=16\). 3. The total mean is \(16(0.3)=4.8\) units.

a) \(n=16\) b) The mean total error is \(4.8\) units.

- Treat each plan as a sum of independent, identically distributed components. - Compare the centers and spreads separately. - Use standard deviation, not the number of containers, to decide which total is more variable.

1. Plan A has mean \(4(12)=48\,\text{kg}\) and standard deviation \(2\sqrt{4}=4\,\text{kg}\). 2. Plan B has mean \(2(24)=48\,\text{kg}\) and standard deviation \(3\sqrt{2}\approx4.24\,\text{kg}\). 3. The means are equal, but Plan B has the larger standard deviation, so its total weight is more variable.

Plan A: mean \(48\,\text{kg}\), standard deviation \(4\,\text{kg}\). Plan B: mean \(48\,\text{kg}\), standard deviation \(3\sqrt{2}\approx4.24\,\text{kg}\). Plan B has the more variable total weight.

- Include the constant only in the center calculation. - Track each coefficient, including its sign, through the two parameter calculations. - Use independence before combining the variability contributions.

1. The mean is \(\mu_R=2(30)-0.5(18)+6=57\). 2. Independence gives \(\sigma_R^2=2^2(4^2)+(-0.5)^2(6^2)=73\). 3. Therefore, \(\sigma_R=\sqrt{73}\approx8.54\).

\(\mu_R=57\) and \(\sigma_R\approx8.54\).

- Evaluate the difference for every joint outcome in the table. - Combine probabilities from cells that produce the same difference. - Select only positive values of the combined variable for the final event.

1. Map each joint outcome to \(X-Y\): the six cells give \(1,-1,-4,4,2,-1\). 2. The value \(-1\) occurs in two cells, so its probability is \(0.20+0.10=0.30\). 3. The full distribution is \(P(D=-4)=0.15\), \(P(D=-1)=0.30\), \(P(D=1)=0.10\), \(P(D=2)=0.20\), and \(P(D=4)=0.25\). 4. Therefore, \(P(D>0)=0.10+0.20+0.25=0.55\).

<table> <tr><th>\(d\)</th><th>\(-4\)</th><th>\(-1\)</th><th>\(1\)</th><th>\(2\)</th><th>\(4\)</th></tr> <tr><th>\(P(D=d)\)</th><td>\(0.15\)</td><td>\(0.30\)</td><td>\(0.10\)</td><td>\(0.20\)</td><td>\(0.25\)</td></tr> </table> \(P(D>0)=0.55\)

- Relate the center of an average to the center of its components. - Work backward from the way averaging changes spread. - Use the number of independent measurements in the scaling factor.

1. An average has the same mean as each identically distributed component, so \(\mu_X=12\). 2. For nine independent measurements, \(\sigma_{\overline{X}}=\frac{\sigma_X}{\sqrt{9}}\). 3. Therefore, \(1.5=\frac{\sigma_X}{3}\), so \(\sigma_X=4.5\).

One measurement has mean \(12\) and standard deviation \(4.5\).

- First transform the original variable to obtain the parameters of the second variable. - Substitute the definition of the second variable into the total and simplify. - Check whether the two variables can vary independently of one another.

1. For \(Y=2X+1\), \(\mu_Y=2(5)+1=11\) and \(\sigma_Y=2(2)=4\). 2. Since \(T=X+(2X+1)=3X+1\), \(\mu_T=3(5)+1=16\) and \(\sigma_T=3(2)=6\). 3. The variables \(X\) and \(Y\) are perfectly dependent because \(Y\) is determined by \(X\), so the independent-variance rule does not apply.

a) \(\mu_Y=11\) and \(\sigma_Y=4\). b) \(\mu_T=16\) and \(\sigma_T=6\). c) \(Y\) is determined by \(X\), so the two variables are not independent.

- Express the spread of an average in terms of the number of independent readings. - Translate “at most” into an inequality. - Check that the final value is an integer count and satisfies the requirement.

1. The standard deviation of the average is \(\frac{3.2}{\sqrt{n}}\). 2. Require \(\frac{3.2}{\sqrt{n}}\le0.8\), which gives \(\sqrt{n}\ge4\). 3. Therefore, \(n\ge16\), so the smallest integer is \(16\).

\(n=16\) readings.

- Combine the centers separately from the spreads. - Use the relationship between the two standardized scores. - Compare the sum of the coefficient squares with the sum of the coefficients.

1. The mean is \(\mu_C=0.6(0)+0.8(0)=0\). 2. Independence gives \(\sigma_C^2=(0.6)^2(1^2)+(0.8)^2(1^2)=0.36+0.64=1\). 3. Therefore, \(\sigma_C=1\). 4. Variability depends on the squares of the coefficients, and \((0.6)^2+(0.8)^2=1\); the ordinary coefficient sum is not the relevant condition.

\(\mu_C=0\) and \(\sigma_C=1\). The squared coefficients, not the coefficients themselves, determine the variance contribution.

- Use the target center to determine the unknown coefficient first. - Treat a negative coefficient carefully when finding spread. - Use the stated relationship between the component variables.

1. The mean condition is \(4a+7=-1\). 2. Solving gives \(a=-2\). 3. Independence gives \(\sigma_W=\sqrt{(-2)^2(1.5)^2+2^2}=\sqrt{13}\approx3.61\).

\(a=-2\) and \(\sigma_W=\sqrt{13}\approx3.61\).

- Write both converted temperatures before taking their difference. - Look for terms that cancel when the two expressions are subtracted. - Apply the independence information only when combining variability.

1. Subtracting the two conversions gives \(D=(1.8C_1+32)-(1.8C_2+32)=1.8(C_1-C_2)\). 2. The mean is \(\mu_D=1.8(18-12)=10.8^\circ\text{F}\). 3. Independence gives \(\sigma_D=1.8\sqrt{2^2+1.5^2}=1.8(2.5)=4.5^\circ\text{F}\).

The mean difference is \(10.8^\circ\text{F}\), and the standard deviation is \(4.5^\circ\text{F}\).

- Translate the spread of the combination into an equation. - Account for the coefficients before isolating the unknown spread. - The final standard deviation must be nonnegative.

1. Independence gives \(13^2=2^2(4^2)+3^2\sigma_Y^2\). 2. Thus, \(169=64+9\sigma_Y^2\), so \(\sigma_Y^2=\frac{105}{9}=\frac{35}{3}\). 3. Therefore, \(\sigma_Y=\sqrt{\frac{35}{3}}\approx3.42\).

\(\sigma_Y=\sqrt{\frac{35}{3}}\approx3.42\).

- Substitute the displayed-reading definitions into the difference. - Simplify before trying to combine any parameters. - Use independence only for variables that remain after simplification.

1. The common offset cancels: \(D=(X+C)-(Y+C)=X-Y\). 2. The mean is \(\mu_D=8-5=3\) volts. 3. Since \(X\) and \(Y\) are independent, \(\sigma_D=\sqrt{1.5^2+2^2}=2.5\) volts. 4. The random variable \(C\) does not appear in the simplified difference.

\(\mu_D=3\) volts and \(\sigma_D=2.5\) volts. The shared calibration offset cancels exactly.

- Model the number of landings on one specified section with a binomial distribution. - For more than \(2\) landings, use a complement. - For part b, first find the probability that one particular section is never reached. - Use indicator variables and linearity of expectation across all \(32\) sections.

1. Let \(X\) be the number of times the spinner lands on \(13\). Then \(X\) is binomial with \(n=100\) and \(p=\frac{1}{32}\). 2. \(P(X=0)=\left(\frac{31}{32}\right)^{100}\approx0.0418\). 3. \(P(X=1)=100\cdot\frac{1}{32}\left(\frac{31}{32}\right)^{99}\approx0.1348\). 4. \(P(X=2)=\binom{100}{2}\left(\frac{1}{32}\right)^2\left(\frac{31}{32}\right)^{98}\approx0.2153\). 5. \(P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\approx0.6081\). 6. For any particular section, the probability of being landed on at least once is \(1-\left(\frac{31}{32}\right)^{100}\). By linearity of expectation, the expected number of different sections reached is \(32\left[1-\left(\frac{31}{32}\right)^{100}\right]\approx30.66\).

a) \(P(X=0)\approx0.0418\), \(P(X=1)\approx0.1348\), \(P(X=2)\approx0.2153\), and \(P(X>2)\approx0.6081\) b) About \(30.66\) different sections

- For one error event, find the probability that it occurs on the specified server. - Model the number of errors on one server with a binomial distribution. - Use a complement for at least one error. - Apply linearity of expectation to the \(50\) server indicators.

1. For a specified server, the number of errors \(X\) is binomial with \(n=120\) and \(p=\frac{1}{50}=0.02\). 2. \(P(X=0)=0.98^{120}\approx0.0885\). 3. \(P(X=1)=120(0.02)(0.98)^{119}\approx0.2168\). 4. \(P(X=2)=\binom{120}{2}(0.02)^2(0.98)^{118}\approx0.2633\). 5. \(P(X>2)=1-P(X=0)-P(X=1)-P(X=2)\approx0.4313\). 6. A specified server has probability \(1-0.98^{120}\) of experiencing at least one error. By linearity of expectation, the expected number of affected servers is \(50(1-0.98^{120})\approx45.57\).

a) \(P(X=0)\approx0.0885\), \(P(X=1)\approx0.2168\), \(P(X=2)\approx0.2633\), and \(P(X>2)\approx0.4313\) b) About \(45.57\) servers

- Treat each defect as one trial. - For one defect, find the probability that it lands on the specified chip. - Multiply the probability that one chip has exactly \(2\) defects by \(1000\). - Identify the independence and equal-probability assumptions.

1. For a specified chip, the number of defects \(X\) is binomial with \(n=50\) and \(p=\frac{1}{1000}=0.001\). 2. \(P(X=0)=(0.999)^{50}\approx0.9512\). 3. \(P(X=2)=\binom{50}{2}(0.001)^2(0.999)^{48}\approx0.0011676\). 4. By linearity of expectation, the expected number of chips with exactly \(2\) defects is \(1000\cdot0.0011676\approx1.17\). 5. The model assumes independent defect placements and the same probability for every chip. A success means that one defect lands on the specified chip, and the \(50\) defects are the \(50\) trials.

a) About \(0.9512\), or \(95.12\%\) b) About \(1.17\) chips c) Defect placements are independent and equally likely across chips. A success is one defect landing on the specified chip, and \(n=50\) is the number of defects.

- Use a complement for “at least \(3\).” - Think of the \(80\) colonies as separate trials. - Find the probability that one specified area is empty. - Use linearity of expectation across all \(200\) areas.

1. For one specified area, the number of colonies \(X\) is binomial with \(n=80\) and \(p=\frac{1}{200}=0.005\). 2. Use the complement: \(P(X\ge3)=1-P(X=0)-P(X=1)-P(X=2)\). 3. The needed probabilities are \(P(X=0)=(0.995)^{80}\approx0.6696\), \(P(X=1)=80(0.005)(0.995)^{79}\approx0.2692\), and \(P(X=2)=\binom{80}{2}(0.005)^2(0.995)^{78}\approx0.0534\). 4. Therefore, \(P(X\ge3)\approx0.0077\). 5. The expected number of empty areas is \(200P(X=0)\approx200(0.6696)=133.93\), or about \(134\). 6. The model treats each colony placement as an independent trial with the same probability of landing in the specified area.

a) About \(0.0077\), or \(0.77\%\) b) About \(134\) sample areas c) Each colony placement is modeled as independent, with a constant probability of landing in the specified area.

- For one specified number, find the probability that it does not occur on one roll. - Raise that probability to the number of rolls. - Use linearity of expectation for the number of missing values. - A logarithm can solve an equation with \(n\) in the exponent.

1. The probability of not rolling \(20\) on one roll is \(\frac{19}{20}=0.95\). Thus, the probability of no \(20\) in \(15\) rolls is \((0.95)^{15}\approx0.4633\). 2. For any specified number, the probability that it is still missing after \(n\) rolls is \((0.95)^n\). 3. By linearity of expectation, the expected number of missing values is \(20(0.95)^n\). Set this equal to \(4\): \(20(0.95)^n=4\). 4. Then \((0.95)^n=0.2\), so \(n=\frac{\ln(0.2)}{\ln(0.95)}\approx31.38\). The estimate is about \(31\) rolls.

a) \((0.95)^{15}\approx0.4633\), or about \(46.3\%\) b) About \(31\) rolls

- For one raisin, find the probability that it is placed in a specified roll. - Use a binomial model for the number of raisins in that roll. - Multiply the probability of more than \(4\) raisins by \(150\) for part b). - Use logarithms to solve the inequality in part c).

1. For a specified roll, each raisin has probability \(p=\frac{1}{150}\) of being placed in that roll. 2. With \(450\) raisins, \(X\) is binomial with \(n=450\) and \(p=\frac{1}{150}\). Thus, \(P(X=3)=\binom{450}{3}\left(\frac{1}{150}\right)^3\left(\frac{149}{150}\right)^{447}\approx0.2248\). 3. \(P(X>4)=1-\sum_{k=0}^{4}\binom{450}{k}\left(\frac{1}{150}\right)^k\left(\frac{149}{150}\right)^{450-k}\approx0.1842\). 4. The expected number of rolls with more than \(4\) raisins is \(150\cdot0.1842\approx27.63\), or about \(28\) rolls. 5. If \(n\) raisins are used, require \(P(X=0)=\left(\frac{149}{150}\right)^n\le0.05\). 6. Solving gives \(n\ge\frac{\ln(0.05)}{\ln(149/150)}\approx447.86\), so the minimum whole number is \(448\).

a) About \(0.2248\), or \(22.48\%\) b) About \(28\) rolls c) At least \(448\) raisins

- Convert the average number of defects per monitor into a total number of defects. - “At least one defect” is the complement of no defects. - In part b, use the observed proportion of defect-free monitors as an estimate of \(P(X=0)\).

1. An average of \(1.5\) defects on each of \(200\) monitors corresponds to \(n=300\) defects. For a specified monitor, each defect has probability \(\frac{1}{200}\) of landing there. 2. The probability of no defects is \(P(X=0)=\left(\frac{199}{200}\right)^{300}\approx0.2223\). 3. The probability of at least one defect is about \(0.7777\), so the expected number of affected monitors is \(200\cdot0.7777\approx155.54\), or about \(156\). 4. In part b, use the observed proportion \(\frac{90}{200}=0.45\) to estimate \(P(X=0)\). Solve \(\left(\frac{199}{200}\right)^n=0.45\). 5. This gives \(n=\frac{\ln(0.45)}{\ln(199/200)}\approx159.30\), so the estimated total is about \(159\) defects.

a) The probability of no pixel defects is about \(0.2223\), or \(22.23\%\). About \(156\) monitors are expected to have at least one defect. b) About \(159\) pixel defects

- For one specified value, the success probability on each roll is \(\frac{1}{6}\). - Use a binomial model for its number of appearances. - For part 3, use an indicator for whether each die value is missing. - Apply linearity of expectation to all six values.

1. For the specified number, the count \(X\) is binomial with \(n=6\) and \(p=\frac{1}{6}\). Thus, \(P(X=0)=\left(\frac{5}{6}\right)^6\approx0.3349\). 2. \(P(X=1)=\binom{6}{1}\frac{1}{6}\left(\frac{5}{6}\right)^5=\left(\frac{5}{6}\right)^5\approx0.4019\). 3. Each of the six values has probability \(\left(\frac{5}{6}\right)^6\) of being missing. By linearity of expectation, the expected number missing is \(6\left(\frac{5}{6}\right)^6\approx2.0094\). 4. One-third of \(6\) is \(2\), and \(2.0094>2\), so the expected number missing is slightly more than one-third of the possible values.

1. \(P(X=0)\approx0.3349\) 2. \(P(X=1)\approx0.4019\) 3. About \(2.01\) values are expected to be missing, which is slightly more than one-third of \(6\).

- Begin with one specified day and model the number of birthdays on that day. - Use linearity of expectation to extend a one-day probability to \(365\) days. - “More than one” is the complement of zero or exactly one. - The model assumes independent, uniformly distributed birthdays.

1. For one specified day, the number of employee birthdays \(X\) is binomial with \(n=500\) and \(p=\frac{1}{365}\). 2. \(P(X=0)=\left(\frac{364}{365}\right)^{500}\approx0.2537\). 3. By linearity of expectation, the expected number of days with no birthdays is \(365P(X=0)\approx92.59\). 4. The probability of exactly one birthday on a specified day is \(P(X=1)=500\cdot\frac{1}{365}\left(\frac{364}{365}\right)^{499}\approx0.3484\). 5. Therefore, \(P(X>1)=1-P(X=0)-P(X=1)\approx0.3979\). 6. The expected number of days with more than one birthday is \(365P(X>1)\approx145.23\).

a) About \(0.2537\), or \(25.37\%\) b) About \(92.59\) days c) About \(145.23\) days

- Focus first on one fixed integer. - Its number of appearances follows a binomial distribution. - Use indicator variables for whether each integer has the requested frequency. - Apply linearity of expectation across the \(100\) possible integers.

1. For one specified integer, its number of appearances \(X\) is binomial with \(n=100\) and \(p=0.01\). 2. The probability that it does not appear is \(P(X=0)=(0.99)^{100}\approx0.3660\). 3. By linearity of expectation, the expected number of missing integers is \(100(0.99)^{100}\approx36.60\). 4. The probability that one specified integer appears exactly twice is \(P(X=2)=\binom{100}{2}(0.01)^2(0.99)^{98}\approx0.1849\). 5. Therefore, the expected number of integers that appear exactly twice is \(100P(X=2)\approx18.49\).

a) About \(36.60\) values b) About \(18.49\) values

- Count the total equally likely outcomes for each experiment separately. - List all ordered pairs producing the requested sums. - “At least \(7\)” includes every sum of \(7\) or more. - Use linearity of expectation for each sum.

1. Experiment A has \(16\) equally likely ordered outcomes. Four produce a sum of \(5\), so \(P(X=5)=\frac{4}{16}=0.25\). 2. Experiment B has \(12\) equally likely ordered outcomes. Two produce a sum of \(5\), so \(P(Y=5)=\frac{2}{12}=\frac{1}{6}\approx0.1667\). Experiment A has the greater probability. 3. In Experiment A, sums of at least \(7\) occur in \(3\) outcomes, so \(P(X\ge7)=\frac{3}{16}=0.1875\). In Experiment B, they occur in \(3\) outcomes, so \(P(Y\ge7)=\frac{3}{12}=0.25\). The probabilities are not equal. 4. By linearity of expectation, \(E(X)=2.5+2.5=5\) and \(E(Y)=1.5+3.5=5\).

a) \(P(X=5)=0.25\) and \(P(Y=5)=\frac{1}{6}\approx0.1667\); Experiment A is greater. b) No. \(P(X\ge7)=0.1875\) and \(P(Y\ge7)=0.25\). c) \(E(X)=E(Y)=5\).

- For one selection, find the probability of choosing the specified song. - Use the complement of zero or one appearance for part b). - For part c, consider an indicator for whether each song is never selected. - Apply linearity of expectation across all \(50\) songs.

1. The number of times a specified song is played is binomial with \(n=50\) and \(p=\frac{1}{50}=0.02\). 2. \(P(X=0)=(0.98)^{50}\approx0.3642\). 3. \(P(X=1)=50\cdot0.02\cdot(0.98)^{49}=(0.98)^{49}\approx0.3716\). 4. Therefore, \(P(X\ge2)=1-P(X=0)-P(X=1)\approx0.2642\). 5. By linearity of expectation, the expected number of songs never played is \(50(0.98)^{50}\approx18.21\).

a) About \(0.3642\), or \(36.4\%\) b) About \(0.2642\), or \(26.4\%\) c) About \(18.21\) songs

- Determine the probability of each label on one spin. - List all possible sums from two spins. - Add the probabilities of all ordered pairs that produce the same sum. - Compute an expected value by multiplying each possible value by its probability. - You can also use the expected value of one spin and linearity of expectation.

1. On one spin, \(P(0)=\frac{3}{6}=\frac{1}{2}\), \(P(2)=\frac{2}{6}=\frac{1}{3}\), and \(P(5)=\frac{1}{6}\). 2. Combining the two independent spins gives \(P(X=0)=\frac{1}{4}\), \(P(X=2)=\frac{1}{3}\), \(P(X=4)=\frac{1}{9}\), \(P(X=5)=\frac{1}{6}\), \(P(X=7)=\frac{1}{9}\), and \(P(X=10)=\frac{1}{36}\). 3. The expected value is \(E(X)=0\left(\frac{1}{4}\right)+2\left(\frac{1}{3}\right)+4\left(\frac{1}{9}\right)+5\left(\frac{1}{6}\right)+7\left(\frac{1}{9}\right)+10\left(\frac{1}{36}\right)=3\). 4. Equivalently, one spin has expected value \(0\left(\frac{1}{2}\right)+2\left(\frac{1}{3}\right)+5\left(\frac{1}{6}\right)=1.5\), so two spins have expected sum \(2\cdot1.5=3\).

a) <table> <thead> <tr><th>\(k\)</th><th>\(0\)</th><th>\(2\)</th><th>\(4\)</th><th>\(5\)</th><th>\(7\)</th><th>\(10\)</th></tr> </thead> <tbody> <tr><th>\(P(X=k)\)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{36}\)</td></tr> </tbody> </table> b) \(E(X)=3\)

- Express the event count as a sum of indicator variables. - Independence removes covariance; mutual exclusivity makes it negative. - Compare which moments depend only on marginal probabilities.

1. Always, \(E(N)=P(A)+P(B)=0.80\). 2. Under independence, the indicators have zero covariance, so \(\operatorname{Var}(N)=0.30\cdot0.70+0.50\cdot0.50=0.21+0.25=0.46\). 3. Under mutual exclusivity, \(E(I_AI_B)=0\), so \(\operatorname{Cov}(I_A,I_B)=0-0.30\cdot0.50=-0.15\). 4. Thus \(\operatorname{Var}(N)=0.46+2\cdot(-0.15)=0.16\). 5. Expectation adds without independence, but variance includes covariance and therefore depends on the relationship.

a) Independent: \(\operatorname{Var}(N)=0.46\). b) Mutually exclusive: \(\operatorname{Var}(N)=0.16\). c) The mean uses only the marginals; the variance also uses covariance.

- Use the target spread to determine the possible magnitudes of the multiplier. - Remember that two multiplier signs can produce the same standard deviation. - Use the target mean separately for each possible multiplier.

1. The standard deviation condition gives \(|a|(4)=6\), so \(a=1.5\) or \(a=-1.5\). 2. If \(a=1.5\), the mean condition \(1.5(10)+b=25\) gives \(b=10\). 3. If \(a=-1.5\), the mean condition \(-1.5(10)+b=25\) gives \(b=40\). 4. Both transformations have the required center and spread.

\((a,b)=(1.5,10)\) or \((a,b)=(-1.5,40)\).

- The center of a sum does not require independence. - Rewrite the sum algebraically in the cases where one variable is defined from the other. - Compare how much the two variables reinforce or cancel one another.

1. In every case, \(\mu_S=10+10=20\). 2. In a), independence gives \(\sigma_S=\sqrt{2^2+2^2}=\sqrt{8}\approx2.83\). 3. In b), \(S=2X\), so \(\sigma_S=2(2)=4\). 4. In c), \(S=X+(20-X)=20\), so \(\sigma_S=0\). 5. The variability ranking is c), a), b).

a) \(\mu_S=20\), \(\sigma_S=\sqrt{8}\approx2.83\). b) \(\mu_S=20\), \(\sigma_S=4\). c) \(\mu_S=20\), \(\sigma_S=0\). Least to greatest variability: c), a), b).

- Identify every pair of component values that can produce the stated combined value. - Express the known probability in terms of the unknown component probability. - After recovering the component distribution, group all pairs by their transformed totals.

1. The value \(T=4\) occurs from \((X,Y)=(0,4)\) or \((1,1)\). 2. Independence gives \(P(T=4)=0.60q+0.40(1-q)=0.40+0.20q\). 3. Solving \(0.40+0.20q=0.46\) gives \(q=0.30\). 4. Then \(P(T=1)=0.60(0.70)=0.42\), \(P(T=4)=0.46\), and \(P(T=7)=0.40(0.30)=0.12\).

a) \(q=0.30\). b) <table> <tr><th>\(t\)</th><th>\(1\)</th><th>\(4\)</th><th>\(7\)</th></tr> <tr><th>\(P(T=t)\)</th><td>\(0.42\)</td><td>\(0.46\)</td><td>\(0.12\)</td></tr> </table>

- Express the spread of the weighted score in terms of the unknown weight. - Use the fact that the two weights add to one. - Check every algebraic solution against the allowed interval.

1. Independence gives \(\sigma_W^2=a^2(10^2)+(1-a)^2(10^2)\). 2. The target variance condition becomes \(a^2+(1-a)^2=0.58\). 3. Simplifying gives \(2a^2-2a+0.42=0\). 4. The solutions are \(a=0.3\) and \(a=0.7\), and both lie in \([0,1]\).

\(a=0.3\) or \(a=0.7\).

- Write the total variability as a function of the allocation. - Rewrite the resulting quadratic so its minimum is visible. - Confirm that the minimizing value is an allowed whole-number allocation.

1. The variance is \(\sigma_T^2=16a^2+4(10-a)^2\). 2. Expanding gives \(\sigma_T^2=20a^2-80a+400=20(a-2)^2+320\). 3. The minimum occurs at the allowed integer \(a=2\). 4. The minimum standard deviation is \(\sqrt{320}=8\sqrt{5}\approx17.89\).

Choose \(a=2\) contracts tied to A and \(8\) tied to B. The minimum standard deviation is \(8\sqrt{5}\approx17.89\).

- Compare the sample sizes used by the half-sample and full-sample proportions. - Think about how averaging two independent quantities changes variability. - Check whether averaging equal-size group proportions reproduces the overall proportion.

1. Each half-sample proportion has standard deviation \(\sqrt{\frac{(0.36)(0.64)}{50}}\approx 0.06788\). 2. Because the half-sample proportions are independent, the standard deviation of their average is \(\frac{0.06788}{\sqrt{2}}=0.048\). 3. The full-sample proportion has standard deviation \(\sqrt{\frac{(0.36)(0.64)}{100}}=0.048\). 4. The average of the two equal-size half-sample proportions is the same estimator as the full-sample proportion, so their standard deviations match.

a) Approximately \(0.06788\) for each half-sample proportion. b) \(0.048\). c) The full-sample proportion also has standard deviation \(0.048\).