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Normal distribution properties

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52521911
The graph shows two normal density curves with the same mean. Which curve has the larger standard deviation? If the standard deviation of curve \(a\) is \(5\), choose the most reasonable standard deviation for curve \(b\): \(2\), \(5\), or \(10\).
Figure for problem 525219

Hints

- Compare the horizontal spread of the two curves in the graph. - Which candidate value would make a normal curve more spread out than one with standard deviation \(5\)? - Check that your choice agrees with the relative shapes shown, not just with the size of the number.

Solution

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 \(10\).

Answer

Curve \(b\); its standard deviation is \(10\).
55557611
A normal model has mean \(72\) and standard deviation \(6\). What is the z-score of the value \(81\), and what does that z-score mean?

Hints

- Find the signed distance from the value to the mean. - Express that distance in units of the given standard deviation.

Solution

1. The value is \(81-72=9\) units above the mean. 2. In standard-deviation units, \(z=\frac{81-72}{6}=1.5\). 3. A z-score of \(1.5\) means the value is \(1.5\) standard deviations above the mean.

Answer

\(z=1.5\); the value \(81\) is \(1.5\) standard deviations above the mean.
52515811
The marked points on the normal density curve are its inflection points. Use the graph to find the mean \(\mu\) and standard deviation \(\sigma\).
Figure for problem 525158

Hints

- Read the x-coordinates of the two marked points before doing any calculation. - What does the symmetry of a normal curve tell you about a pair of corresponding marked points? - After locating the center, consider what special information inflection points carry about the curve's horizontal scale.

Solution

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 \(\sigma=2\).

Answer

\(\mu=2\) and \(\sigma=2\)
52519511
The fill weight \(X\), in kilograms, of a flour package is normally distributed with mean \(1.02\,\text{kg}\) and standard deviation \(0.02\,\text{kg}\). Find \(P(X<1.00)\) and interpret the result in context.

Hints

- Convert \(1.00\,\text{kg}\) to a z-score using the given mean and standard deviation. - Decide whether the probability is a lower-tail or upper-tail probability. - Translate the decimal probability into a percentage of packages.

Solution

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.

Answer

\(P(X<1.00)\approx0.1587\), so about \(15.87\%\) of packages are expected to be below the labeled weight.
52522011
Let \(X\) be normally distributed with mean \(10\) and standard deviation \(2\), and let \(Y\) be normally distributed with mean \(20\) and standard deviation \(8\). a) Give the interval within one standard deviation of the mean for each distribution. b) Approximately what percentage of each distribution lies in its interval from part a)? c) Which distribution has the flatter density curve? Explain using the standard deviations.

Hints

- For each distribution, identify the center and then move one standard deviation in both directions. - Which empirical-rule band corresponds to the interval you found in part a)? - Compare how much horizontal spread each standard deviation represents.

Solution

1. For \(X\), the interval is \([10-2,10+2]=[8,12]\). For \(Y\), the interval is \([20-8,20+8]=[12,28]\). 2. 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. The density 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.

Answer

a) \(X\): \([8,12]\); \(Y\): \([12,28]\) b) Approximately \(68\%\) for each distribution. c) The density curve for \(Y\) is flatter because \(8>2\).
52522111
A normal random variable \(X\) has mean \(\mu=50\). The cumulative distribution function satisfies \(\Phi_{50,\sigma}(45)\approx 0.3085\). Use symmetry to find \(\Phi_{50,\sigma}(55)\), and justify your reasoning.

Hints

- 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 \(55\).

Solution

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, \(\Phi_{50,\sigma}(55)\approx1-0.3085=0.6915\).

Answer

\(\Phi_{50,\sigma}(55)\approx0.6915\)
52527711
A pharmaceutical company produces tablets whose active-ingredient content \(X\), in milligrams, is normally distributed with mean \(250\) and standard deviation \(3\). a) Explain what the mean means in this context. b) Use the empirical rule to give the intervals containing approximately \(68\%\) and \(95\%\) of the tablets. c) In a batch of \(10{,}000\) tablets, approximately how many tablets would be expected to fall outside the \(95\%\) interval?

Hints

- Interpret the mean as a feature of the distribution in this tablet context. - For each requested central percentage, decide how many standard deviations from the mean the empirical rule uses. - For part c), use the complement of the central percentage before converting it to a batch count.

Solution

1. The mean \(250\,\text{mg}\) is the center of the distribution and the long-run average active-ingredient content. 2. 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. About \(5\%\) lie outside the \(95\%\) interval. The expected number is \(0.05\cdot10{,}000=500\) tablets.

Answer

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.
52535311
A normal distribution has mean \(30\) and standard deviation \(6\). a) Compare \(P(X<24)\) and \(P(X>36)\). b) If \(P(X<24)\approx0.1587\), find \(P(24\le X\le36)\). c) Explain how symmetry supports both answers.

Hints

- Compare each cutoff's signed distance from the mean. - Once the two tail probabilities are related, account for the total probability of the full distribution. - Use symmetry to explain why the same probability appears on opposite sides of the mean.

Solution

1. The cutoffs \(24=30-6\) and \(36=30+6\) are symmetric about the mean. Therefore, \(P(X<24)=P(X>36)\). 2. Each tail has probability approximately \(0.1587\). The central probability is \(1-2(0.1587)=0.6826\). 3. 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.

Answer

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.
52687311
A normal random variable \(X\) has mean \(4\) and standard deviation \(3\). a) Express the interval \([1,7]\) in the form \([\mu-\sigma,\mu+\sigma]\). b) Use the empirical rule to estimate \(P(1\le X\le7)\). c) Estimate the probability that \(X\) lies outside this interval.

Hints

- Compare how far each endpoint is from the mean. - Identify which empirical-rule band matches that distance from the mean. - The interval and its complement together have total probability \(1\).

Solution

1. Since \(4-3=1\) and \(4+3=7\), the interval is \([\mu-\sigma,\mu+\sigma]\). 2. 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. The probability outside the interval is approximately \(1-0.68=0.32\).

Answer

a) \([1,7]=[\mu-\sigma,\mu+\sigma]\) b) Approximately \(0.68\), or \(68\%\) c) Approximately \(0.32\), or \(32\%\)
53275711
The graph shows normal density curves \(f\) and \(g\). The marked points are the inflection points of the curves. Without calculating either mean or standard deviation numerically, compare the probability lying between the two marked x-values for \(f\) with the probability lying between the two marked x-values for \(g\). State the approximate probability for each and explain.
Figure for problem 532757

Hints

- Describe each pair of marked points relative to the center of its own curve. - Ask whether changing the numerical center or spread changes the standardized interval from one standard deviation below to one standard deviation above the mean. - Connect that standardized interval with the empirical rule.

Solution

1. For any normal density, the two inflection points occur at \(\mu-\sigma\) and \(\mu+\sigma\). Therefore, each marked interval is exactly one standard deviation on either side of its own mean. 2. By the empirical rule, the probability within one standard deviation of the mean is approximately \(0.68\), regardless of the numerical values of \(\mu\) and \(\sigma\). Thus the two probabilities are equal.

Answer

The probabilities are equal. Each is approximately \(0.68\), or \(68\%\).
53276811
Curve \(f\) is a normal density with mean \(10\) and standard deviation \(2\). Curve \(g\) in the graph has the same mean but twice the standard deviation. Explain why doubling the standard deviation makes curve \(g\) wider and lower than curve \(f\), and give the interval that lies within one standard deviation of the mean for curve \(g\).
Figure for problem 532768

Hints

- Determine the new horizontal scale from the stated change in standard deviation. - Both density curves must still have total area \(1\); consider what spreading that area farther horizontally does to the peak. - Build the requested interval symmetrically around the unchanged mean.

Solution

1. Doubling the standard deviation changes it from \(2\) to \(4\) while the mean stays \(10\). 2. The larger standard deviation spreads the same total area over a wider horizontal range, so curve \(g\) is wider and has a lower maximum than curve \(f\). 3. One standard deviation from the mean for curve \(g\) runs from \(10-4=6\) to \(10+4=14\), so the interval is \([6,14]\).

Answer

Curve \(g\) is wider and lower than curve \(f\), and its one-standard-deviation interval is \([6,14]\).
53478511
Three normal distributions have these parameters: (I) \(\mu=0\), \(\sigma=1\); (II) \(\mu=0\), \(\sigma=2\); (III) \(\mu=3\), \(\sigma=1\). Match graphs \(a\), \(b\), and \(c\) to the parameter pairs. Justify each match using the centers and spreads of the curves.
Figure for problem 534785

Hints

- 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.

Solution

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).

Answer

(I) Graph \(a\); (II) Graph \(b\); (III) Graph \(c\)
53478811
Consider a normal density function with \(\mu=4\) and \(\sigma=1.5\). a) State the intervals on which the function is strictly increasing and strictly decreasing. b) Find the y-coordinate of the maximum. Round to three decimal places.

Hints

- Relate the direction of change of a symmetric bell-shaped density to its center. - The maximum occurs where the exponent is largest. - At that x-value, simplify the density formula before evaluating numerically.

Solution

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

Answer

a) Strictly increasing on \(( -\infty,4]\); strictly decreasing on \([4,\infty)\) b) Approximately \(0.266\)
53479011
The graph shows two normal density curves, \(f\) and \(g\). Without finding exact parameter values, identify which curve has the larger standard deviation and explain why that same curve has the lower maximum.
Figure for problem 534790

Hints

- Compare the horizontal widths of the two curves rather than their locations. - Both curves enclose the same total probability. - Consider what must happen to peak height when the same total area is spread more widely.

Solution

1. Curve \(g\) is spread over a wider horizontal range around its center than curve \(f\), so \(g\) has the larger standard deviation. 2. Every density curve has total area \(1\). Spreading that area over a wider range lowers the peak. Equivalently, a normal density has maximum height \(\frac{1}{\sigma\sqrt{2\pi}}\), which decreases as \(\sigma\) increases.

Answer

Curve \(g\) has the larger standard deviation, and it has the lower maximum because the same total area is spread over a wider range.
53479111
The graph shows a normal density curve \(h\) with two marked points \(P\) and \(Q\) at the same height. Use the graph to find the mean \(\mu\). Explain how the equal heights determine the center.
Figure for problem 534791

Hints

- Read the two marked x-values from the graph. - Think about what equal heights on opposite sides of a normal curve imply about horizontal distance from the center. - The center lies halfway between symmetric x-values.

Solution

1. From the graph, the marked x-values are \(-2.5\) and \(0.5\). 2. A normal density is symmetric about its mean, and two distinct points at the same height lie the same horizontal distance from that center. 3. Therefore, the mean is their midpoint: \(\mu=\frac{-2.5+0.5}{2}=-1\).

Answer

\(\mu=-1\)
53479611
The histogram shows \(100\) measurements from a process. a) Describe the distribution's shape, center, and symmetry. b) Is a normal model reasonable for these data? Explain. c) About \(68\%\) of the measurements fall between \(3\) and \(5\). Use this fact to estimate the mean and standard deviation of a normal model.
Figure for problem 534796

Hints

- 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 \(\mu\pm\sigma\).

Solution

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 \([3,5]=[4-1,4+1]\), \(\sigma\approx1\).

Answer

a) Approximately symmetric and bell-shaped, centered near \(4\). b) Yes, a normal model is reasonable as an approximation. c) \(\mu\approx4\) and \(\sigma\approx1\).
53480211
A normal random variable \(X\) has mean \(4\) and standard deviation \(1.5\). Explain why \(P(X=4)=0\) but \(P(3.5<X<4.5)>0\).

Hints

- Think of probability for a continuous variable as area under a density curve. - Compare the horizontal width of one exact value with the width of an interval. - A positive-width interval around the mean lies where the density is positive.

Solution

1. A normal random variable is continuous, so probability is represented by area under its density curve. 2. A single exact value has zero horizontal width and therefore zero area, so \(P(X=4)=0\). 3. The interval \((3.5,4.5)\) has positive width and lies where the normal density is positive, so it has positive area and therefore positive probability.

Answer

\(P(X=4)=0\) because a single point has zero area, while \(P(3.5<X<4.5)>0\) because the interval has positive width and positive area under the density curve.
53480311
Two normally distributed random variables \(X_1\) and \(X_2\) have density functions \(f_1\) and \(f_2\), as shown. a) Compare the means \(\mu_1\) and \(\mu_2\). b) Compare the standard deviations \(\sigma_1\) and \(\sigma_2\). c) For which random variable is \(P(1.5\le X\le2.5)\) greater? Justify your answer using area under the curves.
Figure for problem 534803

Hints

- 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.

Solution

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 \(P(1.5\le X_1\le2.5)>P(1.5\le X_2\le2.5)\).

Answer

a) \(\mu_2>\mu_1\) b) \(\sigma_2>\sigma_1\) c) The probability is greater for \(X_1\).
53480511
The graph shows the density functions of two normally distributed random variables, \(X\) and \(Y\). The parameters of \(X\) are \(\mu_X=20\) and \(\sigma_X=5\). The marked points \(P\) and \(Q\) on curve \(Y\) are its inflection points. a) Determine \(\mu_Y\) and \(\sigma_Y\) from the graph. b) Describe how changing the mean affects a normal density graph. c) Describe how changing the standard deviation affects the width and maximum height.
Figure for problem 534805

Hints

- Compare the horizontal locations of the two peaks. - What do the marked inflection points tell you about the distance from the center of curve \(Y\)? - Both density curves must have total area \(1\); think about what that implies when one curve spreads out more.

Solution

1. Curve \(Y\) reaches its maximum at \(x=30\), so \(\mu_Y=30\). Its marked inflection points are at \(x=20\) and \(x=40\), each \(10\) units from the mean. Therefore, \(\sigma_Y=10\). 2. Changing the mean shifts the entire graph horizontally without changing its shape. 3. Increasing the standard deviation makes the curve wider and lowers its maximum height because the total area remains \(1\).

Answer

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.
53481911
The graph shows the cumulative distribution function \(F\) of a normally distributed random variable \(X\). Estimate each probability. a) \(P(X\le4)\) b) \(P(X\le5)\) c) \(P(3\le X\le5)\) d) \(P(X>6)\)
Figure for problem 534819

Hints

- 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.

Solution

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) \(P(X>6)=1-F(6)\approx1-0.98=0.02\).

Answer

a) \(0.5\) b) Approximately \(0.84\) c) Approximately \(0.68\) d) Approximately \(0.02\)
53482511
The graph shows two normal density functions, \(f_1\) and \(f_2\). a) Determine the means \(\mu_1\) and \(\mu_2\). b) Describe the transformation that maps \(f_1\) onto \(f_2\). What does this imply about the standard deviations?
Figure for problem 534825

Hints

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

Solution

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 \(\sigma_1=\sigma_2\).

Answer

a) \(\mu_1=3\), \(\mu_2=7\) b) Shift \(f_1\) four units right; \(\sigma_1=\sigma_2\).
53482611
The graph shows the normal density functions \(h_1\) and \(h_2\), which have the same mean \(\mu=5\). a) Which function has the larger standard deviation? Justify your answer from the graph. b) Use the fact that total probability is \(1\) to explain why the curve with the larger standard deviation must have a lower maximum.
Figure for problem 534826

Hints

- 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.

Solution

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.

Answer

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.
55015011
A test-score model is normal with mean \(70\) and standard deviation \(5\). Consider the interval from \(65\) to \(75\). a) How many standard deviations from the mean are the two endpoints? b) Use the empirical rule to estimate the percentage of scores in this interval. c) Find the z-score of a score of \(75\).

Hints

- Compare each endpoint with the mean using the stated standard deviation. - Which empirical-rule band matches the interval you identified? - A z-score describes signed distance from the mean in standard-deviation units.

Solution

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 \(z=\frac{75-70}{5}=1\).

Answer

a) Each endpoint is \(1\) standard deviation from the mean. b) Approximately \(68\%\). c) \(z=1\).
55557711
On Exam A, a score of \(84\) is compared with a mean of \(78\) and standard deviation \(3\). On Exam B, a score of \(92\) is compared with a mean of \(80\) and standard deviation \(8\). Which score is higher relative to its exam's distribution? Support your answer with z-scores.

Hints

- Raw scores from different distributions are not directly comparable. - Express each score's distance above its own mean in standard-deviation units. - Compare the two standardized values after calculating them.

Solution

1. For Exam A, \(z_A=\frac{84-78}{3}=2\). 2. For Exam B, \(z_B=\frac{92-80}{8}=1.5\). 3. Because \(2>1.5\), the score of \(84\) on Exam A is higher relative to its distribution.

Answer

Exam A. Its z-score is \(2\), compared with \(1.5\) for Exam B.
55557811
The two histograms show two different data sets collected with the same measurement scale. Which data set is more reasonably modeled by a normal distribution? Give one visual reason for your choice and one visual reason the other data set is less suitable.
Figure for problem 555578

Hints

- A normal model should have one main center rather than two separated peaks. - Compare the balance of the left and right sides of each histogram. - Focus on overall shape, not on one individual bar.

Solution

1. Data set a) is roughly symmetric, unimodal, and mound-shaped, which is consistent with a normal model. 2. Data set b) has two separated high regions with a low center, so it is strongly bimodal rather than bell-shaped. 3. Therefore, data set a) is more reasonably modeled by a normal distribution.

Answer

Data set a). It is roughly symmetric and unimodal; data set b) is strongly bimodal.
55558811
The histogram shows response times from one sample. Would a normal model be reasonable for these data? Give two visible features of the distribution that support your decision.
Figure for problem 555588

Hints

- Compare the overall shape with a symmetric bell-shaped distribution. - Look separately at the lengths of the left and right tails. - Check whether any isolated value or gap disrupts a smooth unimodal pattern.

Solution

1. A normal model would not be reasonable. 2. The main cluster has a long right tail rather than approximate symmetry. 3. There is also an isolated high response time after a visible gap, which is inconsistent with the smooth bell-shaped pattern expected from a normal distribution.

Answer

No. The distribution is strongly right-skewed and includes an isolated high value after a gap, so a normal model is not a reasonable description.
52517711
Curve \(a\) is the standard normal density. Curve \(b\) is the density \(f(x)=\frac{1}{2.5\sqrt{2\pi}}e^{-\frac12\left(\frac{x+4}{2.5}\right)^2}\). a) State the mean and standard deviation of \(f\). b) Describe the transformations that map curve \(a\) onto curve \(b\). c) Explain why curve \(b\) is wider and lower than curve \(a\).
Figure for problem 525177

Hints

- 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.

Solution

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.

Answer

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.
52522211
For the cumulative distribution function \(\Phi_{\mu,\sigma}\) of a normal random variable \(X\), \(\Phi_{\mu,\sigma}(\mu-a)=1-\Phi_{\mu,\sigma}(\mu+a)\) for \(a\ge 0\). Use this identity to derive a formula for \(P(\mu-a\le X\le\mu+a)\) that contains only \(\Phi_{\mu,\sigma}(\mu+a)\).

Hints

- 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.

Solution

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: \(P=\Phi_{\mu,\sigma}(\mu+a)-[1-\Phi_{\mu,\sigma}(\mu+a)]=2\Phi_{\mu,\sigma}(\mu+a)-1\).

Answer

\(P(\mu-a\le X\le\mu+a)=2\Phi_{\mu,\sigma}(\mu+a)-1\)
52527811
Daily smartphone use \(X\), in minutes, for a certain age group is normally distributed with \(\mu=180\) and \(\sigma=40\). 1) Explain why the theoretical probability that a randomly selected person uses a smartphone for exactly \(180\) minutes in one day is \(0\). 2) What percentage of people use a smartphone for more than \(3\) hours per day? 3) Describe how the density curve changes if the mean stays the same and the standard deviation decreases to \(20\).

Hints

- 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.

Solution

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 \(180\).

Answer

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.
52533311
Compare the standard normal distribution with a normal distribution \(X\) having mean \(4\) and standard deviation \(0.5\). a) Describe how the density curve of \(X\) differs from the standard normal curve in location and shape. b) Give the interval within one standard deviation of the mean for \(X\), and state the approximate percentage of values in that interval. c) Find the z-score of \(x=5\) and interpret it.

Hints

- Compare the mean and standard deviation of \(X\) with those of a standard normal distribution. - For part b), locate values one standard deviation on either side of the center. - For part c), express the signed distance from the mean in standard-deviation units.

Solution

1. The distribution of \(X\) is centered at \(4\), so its density is shifted \(4\) units to the right of the standard normal density. Because \(0.5<1\), it is narrower and taller than the standard normal density. 2. 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. The z-score is \(z=\frac{5-4}{0.5}=2\). Thus, \(5\) is two standard deviations above the mean.

Answer

a) The density of \(X\) 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.
52537911
The function \(f\) is defined for all real \(x\) by \(f(x)=\frac{1}{5\sqrt{2\pi}}e^{-0.02(x-12)^2}\). Explain why \(f\) is a normal probability density function, and identify its mean \(\mu\) and standard deviation \(\sigma\).

Hints

- Identify which part of the exponent controls the center of the bell-shaped density. - Determine what horizontal scale would make the exponent have the standard normal-density structure. - Use the coefficient in front as an independent check of the scale you identified.

Solution

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.

Answer

\(f\) is a normal density with \(\mu=12\) and \(\sigma=5\).
52538011
A normal random variable \(X\) has mean \(5\) and standard deviation \(4\). Define \(Z=\frac{2X-10}{8}\). a) Simplify the expression for \(Z\). b) Explain why \(Z\) has the standard normal distribution. c) Find the value of \(X\) that corresponds to \(Z=1.5\).

Hints

- 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 \(X\).

Solution

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 \(X=11\).

Answer

a) \(Z=\frac{X-5}{4}\) b) It standardizes \(X\), so \(Z\sim N(0,1)\). c) \(X=11\)
53274211
The graph shows normal density curves labeled \(f\) and \(g\). Determine the mean \(\mu\) and standard deviation \(\sigma\) for each curve. Round each standard deviation to the nearest tenth if needed.
Figure for problem 532742

Hints

- Start with a feature of each graph that identifies its center. - Then inspect how the curvature changes on each side of that center. - Use the horizontal scale of those shape changes to infer the spread parameter.

Solution

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 \(\sigma_g=1.0\).

Answer

Curve \(f\): \(\mu_f=-1\), \(\sigma_f=0.5\) Curve \(g\): \(\mu_g=2\), \(\sigma_g=1.0\)
53274511
The graph shows three normal density curves. The marked points on each curve are its inflection points. a) Determine the mean \(\mu\) and standard deviation \(\sigma\) for each distribution. Explain how you read each parameter. b) Describe the relationship between the maximum height of a normal density curve and its standard deviation. Justify your answer using the density formula.
Figure for problem 532745

Hints

- For each curve, first locate its center from the graph's symmetry and peak. - Use the marked points to compare horizontal distances from that center. - In part b), evaluate which parts of the density formula change when \(x\) is at the center of the curve.

Solution

1. 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. 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.

Answer

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.
53274611
The graph shows three functions, \(f\), \(g\), and \(h\). Exactly one is the probability density function of a normal distribution. a) For each of the other two graphs, explain why it cannot represent a normal density. b) For the remaining graph, estimate the mean \(\mu\) and standard deviation \(\sigma\).
Figure for problem 532746

Hints

- Check the defining visual properties of a normal density before doing any parameter calculation. - For a symmetric candidate, consider whether its height and horizontal spread are compatible with total area \(1\). - Once you identify the valid curve, use its center and changes in curvature to estimate its parameters.

Solution

1. 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. 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 \(\sigma=1\).

Answer

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 \(\sigma=1\).
53275011
The graph shows the cumulative distribution function \(F\) of a normal random variable \(X\). a) Estimate the mean \(\mu\) and standard deviation \(\sigma\) from the graph. Explain how you read them. b) Estimate \(P(2\le X\le6)\) by reading appropriate cumulative probabilities from the graph.
Figure for problem 532750

Hints

- First locate a cumulative probability that identifies the center of a symmetric normal distribution. - To estimate the scale, connect a familiar standardized normal location with its cumulative probability and then read the corresponding x-value from this graph. - For an interval probability, compare cumulative probabilities at its two endpoints.

Solution

1. 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. From the graph, \(F(6)\approx0.84\) and \(F(2)\approx0.16\). Therefore, \(P(2\le X\le6)=F(6)-F(2)\approx0.84-0.16=0.68\).

Answer

a) \(\mu\approx4\) and \(\sigma\approx2\) b) \(P(2\le X\le6)\approx0.68\)
53478911
A normal density has mean \(\mu=-2\) and standard deviation \(\sigma=1\). Its inflection points are at \(x=\mu-\sigma\) and \(x=\mu+\sigma\). a) Find the exact coordinates of the two inflection points. b) Show that the height at either inflection point is \(e^{-1/2}\) times the maximum height, and give this ratio as a decimal.

Hints

- Start with the given formulas for the inflection-point x-coordinates. - Compare the exponent at an inflection point with the exponent at the maximum. - Form a ratio of the two heights so common factors cancel.

Solution

1. 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. The maximum height is \(\frac{1}{\sqrt{2\pi}}\). Dividing the inflection height by the maximum gives \(e^{-1/2}\approx0.607\).

Answer

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.
53479211
The graph shows the cumulative distribution function \(F\) of a normally distributed random variable \(X\). a) Determine the mean \(\mu\) from the graph. b) Use the graph to estimate \(P(X\le4)\) and \(P(2\le X\le6)\). c) Describe the corresponding probability density function, including the location of its maximum and its line of symmetry.
Figure for problem 534792

Hints

- 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.

Solution

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 \(x=4\).

Answer

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 \(x=4\).
53479311
The graph shows the cumulative distribution functions \(F_1\) and \(F_2\) of two normally distributed random variables \(X_1\) and \(X_2\). a) Compare the means \(\mu_1\) and \(\mu_2\). b) Which random variable has the larger standard deviation? Explain from the shapes of the cumulative curves. c) From \(F_2\), estimate \(P(0.5\le X_2\le3.5)\). What does this interval suggest about \(\sigma_2\)?
Figure for problem 534793

Hints

- 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.

Solution

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 \(\sigma_2\approx1.5\).

Answer

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 \(\sigma_2\approx1.5\).
53479711
A function \(g\) is defined by \(g(x)=\frac{3}{\sqrt{2\pi}}e^{-4.5x^2+9x-4.5}\). Explain why \(g\) is a normal density function. Then find its mean, standard deviation, and the x-coordinate of its maximum.

Hints

- Rewrite the exponent so its center and horizontal scale are easier to see. - Compare both the exponent structure and the leading coefficient with the general form of a normal density. - Once the center is identified, relate it to the location of the density's maximum.

Solution

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 \(x=\mu=1\).

Answer

\(g\) is a normal density with \(\mu=1\), \(\sigma=\frac13\), and maximum at \(x=1\).
53480711
A flour-filling machine produces package weights \(X\), in grams, that are normally distributed with \(\mu=500\) and \(\sigma=10\). a) Determine \(P(X\le500)\). b) Use the empirical rule to estimate \(F(510)=P(X\le510)\). c) Describe the corresponding cumulative distribution function \(F\), including its shape, end behavior, and the point associated with the mean. d) Describe how the density graph changes if the machine becomes more precise while the mean remains \(500\).

Hints

- 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 \(1\).

Solution

1. A normal distribution is symmetric about its mean, so \(P(X\le500)=0.5\). 2. 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. 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. Greater precision means a smaller standard deviation. The density becomes narrower and taller while remaining centered at \(500\).

Answer

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.
53483211
Two machines produce metal pins whose diameters are approximately normally distributed. The graphs \(F_1\) and \(F_2\) show the corresponding cumulative distribution functions. a) Which machine produces pins with the larger mean diameter? Justify your answer from the graph. b) Greater precision corresponds to a smaller standard deviation. Which machine is more precise? Explain using the graphs. c) Estimate the standard deviation \(\sigma_1\) for Machine 1. Use the fact that about \(84.1\%\) of values in a normal distribution are at or below \(\mu+\sigma\).
Figure for problem 534832

Hints

- 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 \(0.841\).

Solution

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 \(\mu_1+\sigma_1\), \(\sigma_1\approx10.4-10.0=0.4\,\text{mm}\).

Answer

a) Machine 2 b) Machine 2 c) \(\sigma_1\approx0.4\,\text{mm}\)
54919911
Two production lines record the diameter of the same type of washer. The displays show samples from the two lines. Both samples have mean \(50.0\, \text{mm}\) and standard deviation about \(5.0\, \text{mm}\). a) Which line is more reasonably modeled by a normal distribution? Explain using the shape of its display. b) For that line, use a normal model to estimate the percentage of washers with diameters from \(40.0\, \text{mm}\) to \(60.0\, \text{mm}\).
Figure for problem 549199

Hints

- 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.

Solution

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.

Answer

a) Line A; its distribution is approximately symmetric and bell-shaped. b) Approximately \(95.4\%\).
54920011
A ceramics kiln produces tiles whose breaking strengths are approximately normal. A recent calibration gives mean \(72.4\, \text{MPa}\) and standard deviation \(3.1\, \text{MPa}\). The kiln will produce \(1200\) tiles tomorrow. Use technology or a standard normal table to estimate how many tiles will have breaking strengths outside the acceptable interval from \(68\, \text{MPa}\) through \(77\, \text{MPa}\). Round the expected count to the nearest whole tile.

Hints

- 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.

Solution

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 \(176\).

Answer

Approximately \(176\) tiles.
54920111
Completion times for a robotics assembly task are approximately normal with mean \(84\, \text{s}\) and standard deviation \(12\, \text{s}\). A coach will assign extra practice to the slowest \(8\%\) of students. Use an inverse-normal command or a standard normal table to find the completion-time cutoff. Round to the nearest tenth of a second.

Hints

- 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.

Solution

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: \(x=84+1.405\cdot 12\approx 100.9\).

Answer

Use a cutoff of approximately \(100.9\, \text{s}\); students taking longer than this are in the slowest \(8\%\).
54920211
The mass of a packaged product is normally distributed with mean \(310\,\text{g}\) and standard deviation \(18\,\text{g}\). Zofia writes: “About \(90\%\) of packages have masses between \(\mu-1.645\sigma\) and \(\mu+1.645\sigma\), so about \(10\%\) are heavier than the upper endpoint.” a) Identify and correct Zofia's error. b) Find the two endpoints to the nearest tenth of a gram. c) In a shipment of \(4000\) packages, estimate how many are heavier than the upper endpoint.

Hints

- 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.

Solution

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 \(310-1.645\cdot18\approx280.4\) and \(310+1.645\cdot18\approx339.6\), in grams. 3. The expected upper-tail count is \(0.05\cdot4000=200\).

Answer

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.
54920311
A sensor reading \(X\) is normally distributed with mean \(18\) and standard deviation \(2.5\). A converted reading is defined by \(Y=1.8X+32\). a) Find the mean and standard deviation of \(Y\). b) Use technology or a standard normal table to find \(P(60\le Y\le 70)\). Round the probability to four decimal places.

Hints

- 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.

Solution

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 \(\Phi(1.244)-\Phi(-0.978)\approx 0.7292\).

Answer

a) \(\mu_Y=64.4\) and \(\sigma_Y=4.5\). b) \(P(60\le Y\le 70)\approx 0.7292\).
54920611
Three large data sets have been standardized so each has mean \(0\) and standard deviation \(1\). Their observed percentages are shown. <table><tr><th>Data set</th><th>Between \(-1\) and \(1\)</th><th>Between \(-2\) and \(2\)</th><th>Between \(-3\) and \(3\)</th></tr><tr><td>A</td><td>\(68.4\%\)</td><td>\(89.0\%\)</td><td>\(99.6\%\)</td></tr><tr><td>B</td><td>\(61.0\%\)</td><td>\(95.2\%\)</td><td>\(99.7\%\)</td></tr><tr><td>C</td><td>\(68.1\%\)</td><td>\(95.5\%\)</td><td>\(99.7\%\)</td></tr></table> Which data set is most consistent with a normal distribution? For each other data set, identify one part of its three-interval pattern that differs meaningfully from a normal pattern.

Hints

- Recall the empirical-rule benchmarks for cumulative coverage within one, two, and three standard deviations. - Compare each data set across all three columns; no single column is enough to identify the best match. - For each mismatch, state both the interval and whether the observed percentage is too high or too low.

Solution

1. A normal distribution has approximately \(68.3\%\), \(95.4\%\), and \(99.7\%\) within one, two, and three standard deviations of the mean. 2. Data set C is close to all three benchmarks, so it is most consistent with a normal distribution. 3. Data set A is close at one and three standard deviations but has only \(89.0\%\) within two standard deviations, well below the normal benchmark of about \(95.4\%\). 4. Data set B is close at two and three standard deviations but has only \(61.0\%\) within one standard deviation, below the normal benchmark of about \(68.3\%\).

Answer

Data set C is most consistent with a normal distribution. Data set A has too little coverage within two standard deviations, and data set B has too little coverage within one standard deviation.
52515511
Find the mean \(\mu\) and standard deviation \(\sigma\) of the normal density function \(f(x)=\frac{1}{\sqrt{8\pi}}e^{-\frac{1}{8}x^2+\frac{1}{2}x-\frac{1}{2}}\).

Hints

- What x-value would make the quadratic part of the exponent largest? - Can you rewrite the exponent so that it is a constant multiple of one squared distance from that x-value? - After identifying the center and horizontal scale, check whether the coefficient in front of the exponential is consistent.

Solution

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 \(-\frac{1}{8}(x-2)^2=-\frac{(x-2)^2}{2\cdot4}\), \(\mu=2\) and \(\sigma^2=4\). 3. Therefore, \(\sigma=2\). The coefficient checks because \(\frac{1}{\sigma\sqrt{2\pi}}=\frac{1}{2\sqrt{2\pi}}=\frac{1}{\sqrt{8\pi}}\).

Answer

\(\mu=2\) and \(\sigma=2\)
52517811
A normal density graph is obtained from the standard normal density by these transformations: - Stretch horizontally by a factor \(k>0\). - Scale vertically so the total area remains \(1\). - Shift \(5\) units to the right. The transformed curve has maximum point \(\left(5,\frac{1}{4\sqrt{2\pi}}\right)\). Find \(\mu\), \(\sigma\), and the vertical scale factor.

Hints

- Which listed transformation determines the x-coordinate of the curve's center? - The same total area must be spread across the transformed curve. How does the given peak height constrain the horizontal stretch? - Once the horizontal stretch is known, think about how the vertical scale must compensate to preserve area \(1\).

Solution

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, \(\frac14=0.25\).

Answer

\(\mu=5\), \(\sigma=4\), and the vertical scale factor is \(\frac14=0.25\).
52533411
Consider the normal density function \(\varphi_{\mu,\sigma}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\). a) Show algebraically that its graph is symmetric about the line \(x=\mu\). b) A density function \(\varphi_{0,\sigma}\) has a maximum value of approximately \(0.133\). Estimate \(\sigma\). c) Explain how the maximum value changes when \(\sigma\) is multiplied by \(4\).

Hints

- For part a), think about what algebraic relationship function values on opposite sides of a vertical symmetry line must satisfy. - For part b), where on a normal curve must the exponent make the exponential factor as large as possible? - For part c), separate the part of the density's peak formula that changes with \(\sigma\) from the constant factors.

Solution

1. 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. 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. 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.

Answer

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 \(4\).
52687411
A normal random variable \(X\) has mean \(\mu=120\). Without using a calculator, determine all real values of \(k\) for which \(P(X\le120+k)\le P(X\ge150)\). Justify your reasoning using symmetry of the normal distribution.

Hints

- Use the mean as the mirror line for the two tails of the normal distribution. - Try to express both probabilities using cutoffs on the same side of the distribution. - Once both are cumulative probabilities of the same continuous distribution, what property lets you compare their cutoffs?

Solution

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, \(k\le-30\).

Answer

\(k\le-30\)
54920411
A normally distributed measurement has \(80\%\) of its values between \(41\) and \(59\). The interval is centered at the mean. Use an inverse-normal command or a standard normal table to find the mean and standard deviation. Round the standard deviation to two decimal places.

Hints

- 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.

Solution

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 \(59=50+z\sigma\), \(\sigma=\frac{9}{1.282}\approx 7.02\).

Answer

\(\mu=50\) and \(\sigma\approx 7.02\).
54920511
For a normal random variable \(X\), the 10th percentile is \(62\) and the 90th percentile is \(86\). a) Find \(\mu\) and \(\sigma\). Round \(\sigma\) to two decimal places. b) Use an inverse-normal command or a standard normal table to estimate the 75th percentile to the nearest tenth.

Hints

- 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.

Solution

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 \(x=74+0.674\cdot 9.36\approx 80.3\).

Answer

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

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