Free math worksheets

- Compare the horizontal spread of the two curves. - A larger standard deviation makes a normal curve wider and lower. - Choose a value greater than the standard deviation of curve \(a\).

1. A larger standard deviation spreads a normal distribution over a wider range. 2. Curve \(b\) is wider, so it has the larger standard deviation. 3. Of the choices, \(10\) is larger than \(5\) and matches the wider curve. Therefore, curve \(b\) has standard deviation \(10\).

Curve \(b\); its standard deviation is \(10\).

- Read the x-coordinates of the two marked points. - The mean lies halfway between the inflection points of a normal curve. - The standard deviation is the horizontal distance from the mean to either inflection point.

1. A normal curve is symmetric about its mean. The inflection points are at \(x=0\) and \(x=4\), so their midpoint is \(\mu=\frac{0+4}{2}=2\). 2. Each inflection point is one standard deviation from the mean. The distance from \(2\) to either \(0\) or \(4\) is \(2\), so \(\sigma=2\).

\(\mu=2\) and \(\sigma=2\)

- Convert \(1.00\,\text{kg}\) to a z-score using the given mean and standard deviation. - The shaded region is to the left of the cutoff, so use a lower-tail probability. - Translate the decimal probability into a percentage of packages.

1. Standardize the labeled weight: \(z=\frac{1.00-1.02}{0.02}=-1\). 2. From the standard normal distribution, \(P(Z<-1)\approx0.1587\). Therefore, \(P(X<1.00)\approx0.1587\). 3. About \(15.87\%\) of packages are expected to contain less than \(1.00\,\text{kg}\), assuming the normal model is appropriate.

\(P(X<1.00)\approx0.1587\), so about \(15.87\%\) of packages are expected to be below the labeled weight.

- An interval within one standard deviation has endpoints \(\mu-\sigma\) and \(\mu+\sigma\). - Recall the first percentage in the empirical rule. - Compare the two standard deviations to determine which curve is more spread out.

1. a) For \(X\), the interval is \([10-2,10+2]=[8,12]\). For \(Y\), the interval is \([20-8,20+8]=[12,28]\). 2. b) By the empirical rule, approximately \(68\%\) of a normal distribution lies within one standard deviation of its mean. This percentage is the same for both distributions. 3. c) The curve for \(Y\) is flatter because \(\sigma_Y=8\) is greater than \(\sigma_X=2\). A larger standard deviation spreads the same total area over a wider range.

a) \(X\): \([8,12]\); \(Y\): \([12,28]\) b) Approximately \(68\%\) for each distribution. c) The curve for \(Y\) is flatter because \(8>2\).

- Compare the distances of \(45\) and \(55\) from the mean. - Reflect the lower cutoff across the mean. - Convert the right-tail area beyond \(55\) into the cumulative area to the left of \(55\).

1. The values \(45=50-5\) and \(55=50+5\) are equally far from the mean. 2. For a normal distribution, the area to the left of \(45\) equals the area to the right of \(55\). 3. Therefore, \(\Phi_{50,\sigma}(55)\approx1-0.3085=0.6915\).

\(\Phi_{50,\sigma}(55)\approx0.6915\)

- Measure each cutoff's distance from the mean. - Use the given left-tail probability for the matching right tail. - Subtract both tail probabilities from the total area \(1\).

1. a) The cutoffs \(24=30-6\) and \(36=30+6\) are symmetric about the mean. Therefore, \(P(X<24)=P(X>36)\). 2. b) Each tail has probability approximately \(0.1587\). The central probability is \(1-2(0.1587)=0.6826\). 3. c) Reflection across the mean maps the left tail onto the right tail, so the tail areas are equal. Removing both equal tails from total area \(1\) leaves the central area.

a) \(P(X<24)=P(X>36)\) b) \(P(24\le X\le36)\approx0.6826\) c) The cutoffs are equally far from the mean, so the two tail areas are equal.

- Compare each endpoint with the mean \(4\). - Recall the empirical-rule percentage within one standard deviation. - The shaded interval and its outside region together have total probability \(1\).

1. a) Since \(4-3=1\) and \(4+3=7\), the interval is \([\mu-\sigma,\mu+\sigma]\). 2. b) The empirical rule states that approximately \(68\%\) of a normal distribution lies within one standard deviation of the mean. Thus, \(P(1\le X\le7)\approx0.68\). 3. c) The probability outside the interval is approximately \(1-0.68=0.32\).

a) \([1,7]=[\mu-\sigma,\mu+\sigma]\) b) Approximately \(0.68\), or \(68\%\) c) Approximately \(0.32\), or \(32\%\)

- Read the mean from the x-coordinate of each curve's maximum. - Locate where each curve changes concavity on either side of its mean. - The horizontal distance from the mean to an inflection point equals the standard deviation.

1. For curve \(f\), the maximum occurs at \(x=-1\), so \(\mu_f=-1\). The inflection points are about \(-1.5\) and \(-0.5\), each \(0.5\) unit from the mean, so \(\sigma_f=0.5\). 2. For curve \(g\), the maximum occurs at \(x=2\), so \(\mu_g=2\). The inflection points are about \(1\) and \(3\), each \(1\) unit from the mean, so \(\sigma_g=1.0\).

Curve \(f\): \(\mu_f=-1\), \(\sigma_f=0.5\) Curve \(g\): \(\mu_g=2\), \(\sigma_g=1.0\)

- Match each curve's center to the value of \(\mu\). - For curves with the same center, the wider curve has the larger standard deviation. - Curves with equal standard deviations have the same shape even when their centers differ.

1. Graphs \(a\) and \(b\) are centered at \(0\), so they correspond to (I) and (II). Graph \(c\) is centered at \(3\), so it corresponds to (III). 2. Graph \(a\) is narrower than graph \(b\), so graph \(a\) has the smaller standard deviation. Therefore, graph \(a\) corresponds to (I), and graph \(b\) corresponds to (II). 3. Graph \(c\) has the same spread as graph \(a\) but is shifted \(3\) units to the right, confirming that it corresponds to (III).

(I) Graph \(a\); (II) Graph \(b\); (III) Graph \(c\)

- The maximum of a normal density occurs at its mean. - Compare function values as x moves toward and then away from the mean. - At the mean, the exponential factor in the density formula equals \(1\).

1. a) A normal density reaches its maximum at \(x=\mu=4\). It is strictly increasing on \(( -\infty,4]\) and strictly decreasing on \([4,\infty)\). 2. b) The maximum value is \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{1.5\sqrt{2\pi}}\approx0.266\).

a) Strictly increasing on \(( -\infty,4]\); strictly decreasing on \([4,\infty)\) b) Approximately \(0.266\)

- Read the mean from the maximum and measure the horizontal distance to either marked point. - The marked interval is within one standard deviation of the mean. - Use symmetry about the vertical line through the mean.

1. a) The maximum occurs at \(x=-1\), so \(\mu=-1\). The marked inflection points are at \(-2.5\) and \(0.5\), each \(1.5\) units from the mean, so \(\sigma=1.5\). 2. b) The marked values are \(\mu-\sigma\) and \(\mu+\sigma\). By the empirical rule, the probability between them is approximately \(0.68\). 3. c) The points are equally far from the mean, and a normal density is symmetric about its mean. Therefore, their heights are equal.

a) \(\mu=-1\), \(\sigma=1.5\) b) Approximately \(0.68\), or \(68\%\) c) The points are reflections across the mean, so they have equal heights.

- Look for one peak, approximate symmetry, and gradually decreasing frequencies away from the center. - A model may be reasonable without matching every bar perfectly. - Connect the central \(68\%\) interval with \(\mu\pm\sigma\).

1. a) The histogram is unimodal, approximately bell-shaped, and symmetric about \(4\). 2. b) A normal model is reasonable because the distribution is roughly symmetric, has one central peak, and its frequencies decrease gradually away from the center. The fit is approximate because the data are grouped into bins. 3. c) The center is about \(4\), so \(\mu\approx4\). In a normal distribution, about \(68\%\) of values lie within one standard deviation of the mean. Since \([3,5]=[4-1,4+1]\), \(\sigma\approx1\).

a) Approximately symmetric and bell-shaped, centered near \(4\). b) Yes, a normal model is reasonable as an approximation. c) \(\mu\approx4\) and \(\sigma\approx1\).

- Read each mean from the x-coordinate of the corresponding maximum. - Compare the horizontal spreads of the two curves. - Compare the areas above the same interval, not only the curve heights at one point.

1. a) The means are the x-coordinates of the maxima. Curve \(f_1\) is centered at \(2\), and curve \(f_2\) is centered at \(4\), so \(\mu_2>\mu_1\). 2. b) Curve \(f_2\) is wider and lower than curve \(f_1\), so \(\sigma_2>\sigma_1\). 3. c) Probability is area under a density curve. On \([1.5,2.5]\), curve \(f_1\) lies much higher than curve \(f_2\), so \(P(1.5\le X_1\le2.5)>P(1.5\le X_2\le2.5)\).

a) \(\mu_2>\mu_1\) b) \(\sigma_2>\sigma_1\) c) The probability is greater for \(X_1\).

- Read the mean from the x-coordinate of each maximum. - Compare the horizontal scales of the two curves. - Both density curves have total area \(1\).

1. a) Curve \(Y\) reaches its maximum at \(x=30\), so \(\mu_Y=30\). It is about twice as wide and half as high as curve \(X\). Therefore, its standard deviation is twice as large: \(\sigma_Y=10\). 2. b) Changing the mean shifts the entire graph horizontally without changing its shape. 3. c) Increasing the standard deviation makes the curve wider and lowers its maximum height because the total area remains \(1\).

a) \(\mu_Y=30\), \(\sigma_Y=10\) b) The mean shifts the graph horizontally. c) A larger standard deviation makes the curve wider and lower.

- A CDF value \(F(k)\) equals \(P(X\le k)\). - Find an interval probability by subtracting cumulative probabilities. - Use the complement for a greater-than probability.

1. a) \(P(X\le4)=F(4)=0.5\). 2. b) \(P(X\le5)=F(5)\approx0.84\). 3. c) \(P(3\le X\le5)=F(5)-F(3)\approx0.84-0.16=0.68\). 4. d) \(P(X>6)=1-F(6)\approx1-0.98=0.02\).

a) \(0.5\) b) Approximately \(0.84\) c) Approximately \(0.68\) d) Approximately \(0.02\)

- Read each mean from the corresponding maximum. - Compare the shapes before identifying the translation. - A horizontal shift changes location but not spread.

1. a) The maxima occur at \(x=3\) and \(x=7\), so \(\mu_1=3\) and \(\mu_2=7\). 2. b) Curve \(f_2\) is curve \(f_1\) shifted \(4\) units to the right. The curves have the same height and width, so \(\sigma_1=\sigma_2\).

a) \(\mu_1=3\), \(\mu_2=7\) b) Shift \(f_1\) four units right; \(\sigma_1=\sigma_2\).

- What does standard deviation indicate about the spread of a normal curve? - How are width and height related when the total area must remain constant? - Recall the defining total-area condition for a density function.

1. a) The graph of \(h_2\) is wider and flatter than the graph of \(h_1\). Greater spread around the mean means a larger standard deviation, so \(\sigma_2>\sigma_1\). 2. b) The total area under every density curve must equal \(1\). When a normal curve spreads over a wider range, its height must decrease so that the area remains \(1\). Therefore, the curve with the larger standard deviation has the lower maximum.

a) \(h_2\) has the larger standard deviation because its curve is wider. b) Both curves must have total area \(1\), so a wider curve must be lower to preserve that area.

- Compare each endpoint with the mean using the stated standard deviation. - Recall the empirical-rule percentage within one standard deviation. - A z-score measures signed distance from the mean in standard-deviation units.

1. The endpoints are \(70-5=65\) and \(70+5=75\), so each endpoint is one standard deviation from the mean. 2. The empirical rule says that approximately \(68\%\) of values in a normal distribution lie within one standard deviation of the mean. 3. The z-score is \(z=\frac{75-70}{5}=1\).

a) Each endpoint is \(1\) standard deviation from the mean. b) Approximately \(68\%\). c) \(z=1\).

- In the normal density formula, \(\mu\) controls the center and \(\sigma\) controls the horizontal scale. - Compare \(x+4\) with the form \(x-\mu\). - A horizontal stretch of a density requires a reciprocal vertical scale factor to preserve area.

1. a) Comparing with the normal density formula gives \(\mu=-4\) and \(\sigma=2.5\). 2. b) Starting with the standard normal curve, stretch horizontally by a factor of \(2.5\), compress vertically by a factor of \(\frac{1}{2.5}=0.4\), and shift \(4\) units left. 3. c) The horizontal stretch increases the spread. The matching vertical compression keeps the total area under the density equal to \(1\), so the wider curve must have a lower peak.

a) \(\mu=-4\) and \(\sigma=2.5\) b) Stretch horizontally by \(2.5\), compress vertically by \(0.4\), and shift \(4\) units left. c) A density must keep total area \(1\), so spreading the area over a wider interval lowers the peak.

- Express the probability of an interval as a difference of cumulative probabilities. - Replace the term containing \(\mu-a\) with the given symmetry identity. - Distribute the subtraction sign before combining like terms.

1. Write the interval probability as \(P(\mu-a\le X\le\mu+a)=\Phi_{\mu,\sigma}(\mu+a)-\Phi_{\mu,\sigma}(\mu-a)\). 2. Substitute \(\Phi_{\mu,\sigma}(\mu-a)=1-\Phi_{\mu,\sigma}(\mu+a)\). 3. Simplify: \(P=\Phi_{\mu,\sigma}(\mu+a)-[1-\Phi_{\mu,\sigma}(\mu+a)]=2\Phi_{\mu,\sigma}(\mu+a)-1\).

\(P(\mu-a\le X\le\mu+a)=2\Phi_{\mu,\sigma}(\mu+a)-1\)

- The mean describes the center of a normal distribution. - The empirical rule uses one standard deviation for about \(68\%\) and two for about \(95\%\). - Find the percentage outside the central \(95\%\) before applying it to the batch size.

1. a) The mean \(250\,\text{mg}\) is the center of the distribution and the long-run average active-ingredient content. 2. b) Approximately \(68\%\) of values lie within one standard deviation: \([250-3,250+3]=[247,253]\) milligrams. Approximately \(95\%\) lie within two standard deviations: \([250-6,250+6]=[244,256]\) milligrams. 3. c) About \(5\%\) lie outside the \(95\%\) interval. The expected number is \(0.05\cdot10{,}000=500\) tablets.

a) The mean is the long-run average and center, \(250\,\text{mg}\). b) Approximately \(68\%\): \([247,253]\,\text{mg}\); approximately \(95\%\): \([244,256]\,\text{mg}\) c) Approximately \(500\) tablets.

- Think of probability as area under a continuous density curve. - Convert \(3\) hours to minutes and locate that value relative to the mean. - A smaller standard deviation changes the spread but not the center.

1. A normal distribution is continuous. For a continuous random variable, a single point has no area under the density curve, so the probability of one exact value is \(0\). 2. Three hours is \(180\) minutes, which is the mean. By symmetry, half of the distribution lies above the mean, so \(P(X>180)=0.5\). 3. Reducing the standard deviation concentrates values more closely around the mean. The curve becomes narrower and taller while remaining centered at \(180\).

1) A single exact value has probability \(0\) in a continuous distribution. 2) \(50\%\) 3) The curve becomes narrower and taller, with the same center.

- The mean sets the center, and the standard deviation sets the horizontal spread. - An interval within one standard deviation is \([\mu-\sigma,\mu+\sigma]\). - A z-score measures signed distance from the mean in standard-deviation units.

1. a) Curve \(b\) is centered at \(4\), so it is shifted \(4\) units to the right. Because \(0.5<1\), it is narrower and taller than the standard normal curve. 2. b) The interval is \([4-0.5,4+0.5]=[3.5,4.5]\). By the empirical rule, approximately \(68\%\) of values lie in this interval. 3. c) The z-score is \(z=\frac{5-4}{0.5}=2\). Thus, \(5\) is two standard deviations above the mean.

a) Curve \(b\) is shifted \(4\) units right and is narrower and taller. b) \([3.5,4.5]\), containing approximately \(68\%\) of values. c) \(z=2\); the value \(5\) is two standard deviations above the mean.

- Compare the expression with the general normal density formula. - Rewrite \(0.02\) as a fraction and match the exponent to \(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\). - Confirm that the coefficient agrees with the standard deviation you identify.

1. Compare the function with \(\varphi_{\mu,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\). 2. Rewrite the exponent: \(-0.02(x-12)^2=-\frac{1}{50}(x-12)^2=-\frac{1}{2}\left(\frac{x-12}{5}\right)^2\). 3. This identifies \(\mu=12\) and \(\sigma=5\). The coefficient also matches because \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{5\sqrt{2\pi}}\). Therefore, \(f\) is a normal density.

\(f\) is a normal density with \(\mu=12\) and \(\sigma=5\).

- Factor \(2\) from the numerator before reducing the fraction. - Compare the simplified expression with \(\frac{X-\mu}{\sigma}\). - For part c), substitute the given z-score and solve backward for \(X\).

1. a) Factor the numerator: \(Z=\frac{2(X-5)}{8}=\frac{X-5}{4}\). 2. b) The expression \(\frac{X-5}{4}\) subtracts the mean of \(X\) and divides by its standard deviation. Standardizing a normal random variable produces a standard normal variable with mean \(0\) and standard deviation \(1\). 3. c) Solve \(1.5=\frac{X-5}{4}\). Then \(6=X-5\), so \(X=11\).

a) \(Z=\frac{X-5}{4}\) b) It standardizes \(X\), so \(Z\sim N(0,1)\). c) \(X=11\)

- Read each mean from the x-coordinate of the corresponding maximum. - Use the marked inflection points to measure one standard deviation from each mean. - Substitute \(x=\mu\) into the density formula to study the maximum height.

1. a) The mean is the x-coordinate of the maximum. Thus, \(\mu_1=-2\), \(\mu_2=3\), and \(\mu_3=0\). The marked inflection points are one standard deviation from each mean. Their horizontal distances from the means are \(1\), \(2\), and \(4\), so \(\sigma_1=1\), \(\sigma_2=2\), and \(\sigma_3=4\). 2. b) At \(x=\mu\), the exponential factor in \(f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12((x-\mu)/\sigma)^2}\) equals \(1\). Therefore, the maximum is \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}\), which is inversely proportional to \(\sigma\). A larger standard deviation produces a lower, wider curve.

a) Curve \(g_1\): \(\mu_1=-2\), \(\sigma_1=1\); curve \(g_2\): \(\mu_2=3\), \(\sigma_2=2\); curve \(g_3\): \(\mu_3=0\), \(\sigma_3=4\) b) The maximum is \(f(\mu)=\frac{1}{\sigma\sqrt{2\pi}}\), so it decreases as \(\sigma\) increases.

- Check symmetry, bell shape, and nonnegative values first. - Compare the apparent spread of a curve with the maximum height required by \(\frac{1}{\sigma\sqrt{2\pi}}\). - For the valid curve, read the mean at the maximum and the standard deviation from the inflection points.

1. a) Curve \(g\) is symmetric about \(x=2\) and has the same horizontal scale as a normal curve with \(\sigma=1\), but its maximum height is \(0.6\). A normal density with \(\sigma=1\) has maximum height \(\frac{1}{\sqrt{2\pi}}\approx0.40\), so \(g\) is not normalized. 2. Curve \(h\) is not symmetric about its maximum, while every normal density is symmetric about its mean. 3. b) Curve \(f\) has its maximum and axis of symmetry at \(x=2\), so \(\mu=2\). Its inflection points are approximately \(1\) and \(3\), one unit from the mean, so \(\sigma=1\).

a) Curve \(g\) has the wrong height for its spread, and curve \(h\) is asymmetric. b) Curve \(f\) is the normal density, with \(\mu=2\) and \(\sigma=1\).

- Locate the x-value where the cumulative probability is \(0.5\). - One standard deviation above the mean has cumulative probability about \(0.84\). - Subtract the cumulative probability at the lower endpoint from the value at the upper endpoint.

1. a) The mean is the value where \(F(x)=0.5\), which occurs at \(x=4\). Thus, \(\mu=4\). For a normal distribution, \(F(\mu+\sigma)\approx0.8413\). The graph reaches about \(0.84\) at \(x=6\), so \(\sigma\approx6-4=2\). 2. b) From the graph, \(F(6)\approx0.84\) and \(F(2)\approx0.16\). Therefore, \(P(2\le X\le6)=F(6)-F(2)\approx0.84-0.16=0.68\).

a) \(\mu\approx4\) and \(\sigma\approx2\) b) \(P(2\le X\le6)\approx0.68\)

- Read each mean at the corresponding maximum and use the marked inflection points for the standard deviation. - Symmetry divides the area at the mean into two equal halves. - Compare the endpoints \(3\) and \(7\) with \(\mu_2\pm\sigma_2\).

1. a) For \(f\), the maximum is at \(2\) and the marked inflection points are at \(1\) and \(3\), so \(\mu_1=2\) and \(\sigma_1=1\). For \(g\), the maximum is at \(5\) and the marked inflection points are at \(3\) and \(7\), so \(\mu_2=5\) and \(\sigma_2=2\). 2. b) Half of a normal distribution lies on each side of its mean. Therefore, \(P(X_1\le2)=0.5\) and \(P(X_2\ge5)=0.5\). 3. c) The interval \([3,7]\) is \([\mu_2-\sigma_2,\mu_2+\sigma_2]\). By the empirical rule, \(P(3\le X_2\le7)\approx0.68\).

a) Curve \(f\): \(\mu_1=2\), \(\sigma_1=1\); curve \(g\): \(\mu_2=5\), \(\sigma_2=2\) b) \(P(X_1\le2)=0.5\); \(P(X_2\ge5)=0.5\) c) \(P(3\le X_2\le7)\approx0.68\)

- Use the maximum for the mean and the marked inflection points for the standard deviation. - Compare the endpoints \(6\) and \(14\) with \(\mu\pm2\sigma\). - Doubling the standard deviation doubles every horizontal distance measured from the mean.

1. a) The maximum occurs at \(x=10\), so \(\mu=10\). The marked inflection points are at \(8\) and \(12\), each \(2\) units from the mean, so \(\sigma=2\). 2. b) The interval \([6,14]\) is \([\mu-2\sigma,\mu+2\sigma]\). By the empirical rule, \(P(6\le X\le14)\approx0.95\). 3. c) The new standard deviation is \(4\). The curve remains centered at \(10\) but becomes wider and lower. Its interval within one standard deviation is \([10-4,10+4]=[6,14]\).

a) \(\mu=10\), \(\sigma=2\) b) Approximately \(0.95\); the interval is within two standard deviations of the mean. c) The curve becomes wider and lower, and its one-standard-deviation interval is \([6,14]\).

- Substitute \(\mu\) and \(\sigma\) into \(\mu\pm\sigma\). - At an inflection point, the standardized squared distance equals \(1\). - Form a ratio so the common coefficient cancels.

1. a) The x-coordinates are \(-2-1=-3\) and \(-2+1=-1\). At either value, \(\left(\frac{x-\mu}{\sigma}\right)^2=1\), so the common height is \(\frac{1}{\sqrt{2\pi}}e^{-1/2}=\frac{1}{\sqrt{2\pi e}}\). Thus, the points are \(\left(-3,\frac{1}{\sqrt{2\pi e}}\right)\) and \(\left(-1,\frac{1}{\sqrt{2\pi e}}\right)\). 2. b) The maximum height is \(\frac{1}{\sqrt{2\pi}}\). Dividing the inflection height by the maximum gives \(e^{-1/2}\approx0.607\).

a) \(\left(-3,\frac{1}{\sqrt{2\pi e}}\right)\) and \(\left(-1,\frac{1}{\sqrt{2\pi e}}\right)\) b) The inflection-point height is \(e^{-1/2}\approx0.607\) times the maximum height.

- Use each maximum for the mean and the marked inflection points for the standard deviation. - The interval within one standard deviation is \([\mu-\sigma,\mu+\sigma]\). - Both curves enclose total area \(1\).

1. a) Curve \(f\) has maximum at \(0\) and inflection points at \(-1\) and \(1\), so \(\mu_f=0\) and \(\sigma_f=1\). Curve \(g\) has maximum at \(3\) and inflection points at \(1\) and \(5\), so \(\mu_g=3\) and \(\sigma_g=2\). 2. b) For \(f\), the interval is \([-1,1]\). For \(g\), it is \([1,5]\). By the empirical rule, each interval contains approximately \(68\%\) of its distribution. 3. c) Curve \(g\) has the larger standard deviation, so its total area of \(1\) is spread over a wider range. Its maximum must therefore be lower.

a) \(f\): \(\mu=0\), \(\sigma=1\); \(g\): \(\mu=3\), \(\sigma=2\) b) \(f\): \([-1,1]\); \(g\): \([1,5]\); approximately \(68\%\) in each. c) The wider curve spreads the same total area over a larger range, so its peak is lower.

- Find the x-value where the cumulative probability is \(0.5\). - Use a difference of cumulative values for the interval probability. - The mean determines both the maximum and the symmetry line of a normal density.

1. a) For a normal distribution, \(F(\mu)=0.5\). The graph passes through \((4,0.5)\), so \(\mu=4\). 2. b) From the graph, \(P(X\le4)=F(4)=0.5\). Also, \(P(2\le X\le6)=F(6)-F(2)\approx0.909-0.091=0.818\), or about \(0.82\). 3. c) The density is bell-shaped with its maximum at \(x=4\). It is symmetric about the line \(x=4\).

a) \(\mu=4\) b) \(P(X\le4)=0.5\); \(P(2\le X\le6)\approx0.82\) c) The density is bell-shaped, has its maximum at \(x=4\), and is symmetric about \(x=4\).

- A normal CDF equals \(0.5\) at the mean. - Greater spread makes a cumulative curve less steep near its center. - Subtract the CDF values at the endpoints, then compare the central probability with the empirical rule.

1. a) Both cumulative distribution functions equal \(0.5\) at \(x=2\). Therefore, \(\mu_1=\mu_2=2\). 2. b) Curve \(F_2\) rises more gradually around the mean. A more gradual cumulative curve corresponds to a more spread-out distribution, so \(X_2\) has the larger standard deviation. 3. c) From the graph, \(F_2(0.5)\approx0.159\) and \(F_2(3.5)\approx0.841\). Thus, \(P(0.5\le X_2\le3.5)\approx0.841-0.159=0.682\). The endpoints are \(1.5\) units below and above the mean, and the central probability is about \(68\%\), so \(\sigma_2\approx1.5\).

a) \(\mu_1=\mu_2=2\) b) \(X_2\) has the larger standard deviation because \(F_2\) rises more gradually. c) Approximately \(0.682\); this suggests \(\sigma_2\approx1.5\).

- Complete the square in the exponent. - Match the squared coefficient with \(\frac{1}{2\sigma^2}\). - Verify that the coefficient in front of the exponential equals \(\frac{1}{\sigma\sqrt{2\pi}}\).

1. Complete the square: \(-4.5x^2+9x-4.5=-4.5(x^2-2x+1)=-4.5(x-1)^2\). 2. Compare \(-4.5(x-1)^2\) with \(-\frac{(x-\mu)^2}{2\sigma^2}\). This gives \(\mu=1\) and \(\frac{1}{2\sigma^2}=4.5\), so \(\sigma^2=\frac19\) and \(\sigma=\frac13\). 3. The required leading factor is \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{3}{\sqrt{2\pi}}\), which matches the function. Therefore, \(g\) is a normal density, and its maximum occurs at \(x=\mu=1\).

\(g\) is a normal density with \(\mu=1\), \(\sigma=\frac13\), and maximum at \(x=1\).

- Read the mean where the cumulative probability is \(0.5\). - One standard deviation above the mean has cumulative probability about \(0.84\). - Subtract the cumulative probabilities at the two symmetric endpoints.

1. a) The mean is the x-value where \(F(x)=0.5\). From the graph, \(\mu=4\). 2. b) For a normal distribution, \(F(\mu+\sigma)\approx0.8413\). The graph reaches about \(0.84\) near \(x=5.5\), so \(\sigma\approx5.5-4=1.5\). 3. c) The endpoints are approximately \(\mu-\sigma=2.5\) and \(\mu+\sigma=5.5\). The graph gives cumulative probabilities near \(0.16\) and \(0.84\), so the interval probability is about \(0.84-0.16=0.68\), consistent with the empirical rule.

a) \(\mu=4\) b) \(\sigma\approx1.5\) c) \(P(2.5\le X\le5.5)\approx0.68\), the central interval within one standard deviation.

- Use the maximum and the marked inflection points to read the parameters. - Distinguish a single point from an interval with positive width. - Probability for a continuous variable is represented by area under the density curve.

1. a) The density curve reaches its maximum at \(x=4\), so \(\mu=4\). The marked inflection points are at \(2.5\) and \(5.5\), each \(1.5\) units from the mean, so \(\sigma=1.5\). 2. b) A normal random variable is continuous. One exact value corresponds to a point with zero width and therefore zero area under the density curve, so \(P(X=4)=0\). 3. c) No. The interval from \(3.5\) to \(4.5\) has positive width and lies where the density is positive, so it has positive area and positive probability.

a) \(\mu=4\), \(\sigma=1.5\) b) \(P(X=4)=0\) because a single point has zero area. c) No. \(P(3.5<X<4.5)>0\) because the interval has positive width.

- Symmetry places half of the area on each side of the mean. - Recognize \(510\) as \(\mu+\sigma\) and use half of the central \(68\%\). - A cumulative distribution records area to the left and always rises from near \(0\) to near \(1\).

1. a) A normal distribution is symmetric about its mean, so \(P(X\le500)=0.5\). 2. b) The value \(510\) is one standard deviation above the mean. Approximately \(68\%\) lies between \(490\) and \(510\), so about \(34\%\) lies between \(500\) and \(510\). Therefore, \(F(510)\approx0.50+0.34=0.84\). 3. c) The cumulative distribution function is S-shaped, approaches \(0\) as \(x\) becomes small, approaches \(1\) as \(x\) becomes large, and passes through \((500,0.5)\). 4. d) Greater precision means a smaller standard deviation. The density becomes narrower and taller while remaining centered at \(500\).

a) \(0.5\) b) Approximately \(0.84\) c) An S-shaped curve approaching \(0\) and \(1\), passing through \((500,0.5)\) d) The density becomes narrower and taller, with the same center.

- Read each mean where the corresponding cumulative curve reaches \(0.5\). - A steeper cumulative curve indicates less spread. - Compare the x-values where \(F_1(x)\) is approximately \(0.5\) and \(0.841\).

1. a) A normal cumulative distribution function equals \(0.5\) at the mean. The graph shows \(\mu_1\approx10.0\,\text{mm}\) and \(\mu_2\approx10.5\,\text{mm}\), so Machine 2 has the larger mean diameter. 2. b) A smaller standard deviation gives a steeper cumulative distribution function near the mean. Since \(F_2\) is steeper, Machine 2 is more precise. 3. c) For Machine 1, \(F_1(x)\approx0.841\) at \(x\approx10.4\). Since this input is approximately \(\mu_1+\sigma_1\), \(\sigma_1\approx10.4-10.0=0.4\,\text{mm}\).

a) Machine 2 b) Machine 2 c) \(\sigma_1\approx0.4\,\text{mm}\)

- Compare each display with the overall shape expected from a normal model. - Relate each endpoint of the interval to the given center and spread. - Use the normal-distribution benchmark for a symmetric interval around the mean.

1. Line A is approximately symmetric, unimodal, and bell-shaped; Line B is strongly right-skewed. A normal model is reasonable for Line A but not for Line B. 2. The interval \([40, 60]\) is \(50\pm2\cdot5\), so it lies within two standard deviations of the mean. 3. A normal model places approximately \(95.4\%\) of values within two standard deviations of the mean.

a) Line A; its distribution is approximately symmetric and bell-shaped. b) Approximately \(95.4\%\).

- Treat the two unacceptable regions together by first finding the probability of the acceptable middle region. - Express each endpoint in terms of its distance from the mean measured in standard deviations. - Convert the resulting probability into an expected count for the full production run.

1. Standardize the endpoints: \(z_1=\frac{68-72.4}{3.1}\approx -1.419\) and \(z_2=\frac{77-72.4}{3.1}\approx 1.484\). 2. The normal-model probability inside the interval is \(P(68\le X\le 77)\approx 0.8532\). 3. The expected number outside is \(1200\cdot(1-0.8532)\approx 176.2\), which rounds to \(176\).

Approximately \(176\) tiles.

- Translate the upper-tail percentage into a cumulative percentage from the left. - Locate the corresponding standardized position before returning to the original units. - Check that the cutoff is above the mean because it separates unusually long times.

1. The slowest \(8\%\) begins at the 92nd percentile because \(1-0.08=0.92\). 2. The standard normal value with cumulative area \(0.92\) is \(z\approx 1.405\). 3. Convert back to seconds: \(x=84+1.405\cdot 12\approx 100.9\).

Use a cutoff of approximately \(100.9\, \text{s}\); students taking longer than this are in the slowest \(8\%\).

- Separate the percentage inside the interval from the total percentage outside it. - Use the symmetry of the distribution to divide the outside percentage between the two ends. - Translate the standardized endpoints into the original measurement scale.

1. The middle \(90\%\) leaves \(10\%\) outside the interval, split equally between the two tails by symmetry. Thus only \(5\%\) are above the upper endpoint. 2. The endpoints are \(310-1.645\cdot 18\approx 280.4\) and \(310+1.645\cdot 18\approx 339.6\). 3. The expected upper-tail count is \(0.05\cdot 4000=200\).

a) The \(10\%\) outside is split into two \(5\%\) tails, so about \(5\%\) are above the upper endpoint. b) Approximately \(280.4\, \text{g}\) and \(339.6\, \text{g}\). c) Approximately \(200\) packages.

- Think about how multiplying and then shifting every observation changes the center. - A shift changes location but not spread; a positive scale factor changes both in predictable ways. - Work entirely in the converted scale or translate the interval back to the original scale.

1. A positive linear transformation changes the mean to \(\mu_Y=1.8\cdot 18+32=64.4\) and the standard deviation to \(\sigma_Y=1.8\cdot 2.5=4.5\). 2. Standardize the endpoints: \(z_1=\frac{60-64.4}{4.5}\approx -0.978\) and \(z_2=\frac{70-64.4}{4.5}\approx 1.244\). 3. The probability is \(\Phi(1.244)-\Phi(-0.978)\approx 0.7292\).

a) \(\mu_Y=64.4\) and \(\sigma_Y=4.5\). b) \(P(60\le Y\le 70)\approx 0.7292\).

- Recall the approximate normal percentages for one, two, and three standard deviations from the mean. - Compare the full pattern across all three columns rather than choosing from one percentage alone. - For a mismatch, decide whether observations appear too concentrated near the center or too common in the tails.

1. A normal distribution has approximately \(68.3\%\), \(95.4\%\), and \(99.7\%\) within one, two, and three standard deviations. 2. Data set A is close to all three benchmarks, so it is most consistent with a normal distribution. 3. Data set B has too few observations in every central interval, indicating substantially heavier tails or more extreme values than a normal distribution. 4. Data set C has too many observations within one standard deviation but fewer than expected within three standard deviations, indicating unusually strong concentration near the center together with heavier extreme tails.

Data set A is most consistent with a normal distribution. Data set B has much less central coverage than expected. Data set C is too concentrated within one standard deviation while still having more extreme values beyond three standard deviations than a normal model predicts.

- Rewrite the quadratic exponent by completing the square. - Compare the rewritten exponent with \(-\frac{(x-\mu)^2}{2\sigma^2}\). - Check that the coefficient in front of the exponential agrees with the value of \(\sigma\).

1. Complete the square in the exponent: \(-\frac{1}{8}x^2+\frac{1}{2}x-\frac{1}{2}=-\frac{1}{8}(x^2-4x+4)=-\frac{1}{8}(x-2)^2\). 2. Compare with the normal exponent \(-\frac{(x-\mu)^2}{2\sigma^2}\). Since \(-\frac{1}{8}(x-2)^2=-\frac{(x-2)^2}{2\cdot 4}\), \(\mu=2\) and \(\sigma^2=4\). 3. Therefore, \(\sigma=2\). The coefficient checks because \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{2\sqrt{2\pi}}=\frac{1}{\sqrt{8\pi}}\).

\(\mu=2\) and \(\sigma=2\)

- The x-coordinate of a normal curve's maximum is its mean. - Compare the maximum height with \(\frac{1}{\sigma\sqrt{2\pi}}\). - A horizontal stretch by \(k\) requires a vertical scale factor of \(\frac{1}{k}\) to preserve area.

1. The horizontal shift places the center and maximum at \(x=5\), so \(\mu=5\). 2. The maximum height of a normal density is \(\frac{1}{\sigma\sqrt{2\pi}}\). Comparing this with \(\frac{1}{4\sqrt{2\pi}}\) gives \(\sigma=4\). 3. The horizontal stretch factor is \(k=4\). To preserve area, the vertical scale factor is the reciprocal, \(\frac14=0.25\).

\(\mu=5\), \(\sigma=4\), and the vertical scale factor is \(\frac14=0.25\).

- To prove symmetry about \(x=a\), compare the function values at \(a-h\) and \(a+h\). - Substitute the mean into the density to obtain its maximum value. - Examine how \(\sigma\) appears in the coefficient of the density function.

1. a) For any real \(h\), substitute \(\mu-h\) and \(\mu+h\). The exponents contain \(\left(\frac{-h}{\sigma}\right)^2\) and \(\left(\frac{h}{\sigma}\right)^2\), which are equal. Therefore, \(\varphi_{\mu,\sigma}(\mu-h)=\varphi_{\mu,\sigma}(\mu+h)\), proving symmetry about \(x=\mu\). 2. b) The maximum occurs at \(x=0\) and equals \(\frac{1}{\sigma\sqrt{2\pi}}\). Solving \(\frac{1}{\sigma\sqrt{2\pi}}\approx0.133\) gives \(\sigma\approx\frac{1}{0.133\sqrt{2\pi}}\approx3.00\). 3. c) The maximum value is inversely proportional to \(\sigma\). Replacing \(\sigma\) with \(4\sigma\) changes the maximum to \(\frac{1}{4\sigma\sqrt{2\pi}}\), one-fourth of its original value.

a) \(\varphi_{\mu,\sigma}(\mu-h)=\varphi_{\mu,\sigma}(\mu+h)\), so the graph is symmetric about \(x=\mu\). b) \(\sigma\approx3.00\) c) The maximum value is divided by \(4\).

- Reflect \(150\) across the mean \(120\). - Rewrite the right-tail probability as an equivalent left-tail probability. - Use the increasing nature of a cumulative distribution function to compare the two cutoffs.

1. Since \(150=120+30\), symmetry about \(\mu=120\) gives \(P(X\ge150)=P(X\le120-30)=P(X\le90)\). 2. The inequality becomes \(P(X\le120+k)\le P(X\le90)\). 3. A normal cumulative distribution function is strictly increasing, so \(120+k\le90\). Therefore, \(k\le-30\).

\(k\le-30\)

- Use the symmetry of the stated interval to locate its center. - Determine how the percentage outside the interval is divided between the two tails. - Relate the distance from the center to a standardized percentile position.

1. The midpoint of the symmetric interval is the mean: \(\mu=\frac{41+59}{2}=50\). 2. A central area of \(0.80\) leaves \(0.10\) in each tail, so the upper endpoint is the 90th percentile, with \(z\approx 1.282\). 3. Since \(59=50+z\sigma\), \(\sigma=\frac{9}{1.282}\approx 7.02\).

\(\mu=50\) and \(\sigma\approx 7.02\).

- Look for a symmetry relationship between the two given percentiles. - Use the same distance from the mean on opposite sides of a normal curve. - After finding the parameters, convert the requested percentile from standardized units.

1. The 10th and 90th percentiles are symmetric about the mean, so \(\mu=\frac{62+86}{2}=74\). 2. Their standardized values are approximately \(-1.282\) and \(1.282\). Thus \(\sigma=\frac{86-74}{1.282}\approx 9.36\). 3. The 75th-percentile standard score is \(z\approx 0.674\), so \(x=74+0.674\cdot 9.36\approx 80.3\).

a) \(\mu=74\) and \(\sigma\approx 9.36\). b) The 75th percentile is approximately \(80.3\).