Answer the followingThe image on the left shows the standard normaldistribution curve (with u = 0 and o = 1).Shade the portion of the graph that represents anarea of approximately 68.2%.68%95%99.7%mean-3s mean-2 mean-ismean mean+ls mean+2 mean+The image on the left shows the standard normaldistribution curve (with u = 0 and o = 1).Shade the portion of the graph that represents anarea of approximately 95.4%.The image on the left shows the standard normaldistribution curve (with µ = 0 and o = 1).Shade the portion of the graph that represents anarea of approximately 99.7%.---2

Answered: Image /qna-images/answer/60257ef0-993b-43de-8b9b-572bc37df7b0.jpg

Answered: Answer the following The image on the…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.053 Click to view page 1 of the table. Click to view page 2 of the table. The cumulative area corresponds to the z-score of. (Round to three decimal places as needed.) en 000 MacBook Air 2 8 DII DD B F10 F11 Next F12
Use the table below to find the percentage of data items in a normal distribution that lie between z= - 1 and z = 1. Click the icon to view the table. - X Data Table The percentage of data items in a normal distribution that lie between z = - 1 and z= 1 is %. z-SCORES AND PERCENTILES -Score Percentile z-Score Percentile z-Score Percentile -Score Percentile -4.0 0.003 -1.0 15.87 0.0 50.00 1.1 86.43 -3.5 0.02 -0.95 17.11 0.05 51.99 1.2 88.49 -3.0 0.13 -0.90 18.41 0.10 53.98 1.3 90.32 -2.9 0.19 -0.85 19.77 0.15 55.96 1.4 91.92 -2.8 0.26 -0.80 21.19 0.20 57.93 1.5 93.32 -2.7 0.35 -0.75 22.66 0.25 59.87 1.6 94.52 -2.6 0.47 -0.70 24.20 0.30 61.79 1.7 95.54 -2.5 0.62 -0.65 25.78 0.35 63.68 1.8 96.41 -2.4 0.82 -0.60 27.43 0.40 65.54 1.9 97.13 -2.3 1.07 -0.55 29.12 0.45 67.36 2.0 97.72 -2.2 1.39 -0.50 30.85 0.50 69.15 2.1 98.21 -2.1 1.79 -0.45 32.64 0.55 70.88 2.2 98.61 -2.0 2.28 -0.40 34.46 0.60 72.57 2.3 98.93 -1.9 2.87 -0.35 36.32 0.65 74.22 2.4 99.18 -1.8 3.59 -0.30 38.21 0.70 75.80 2.5…
One graph in the figure represents a normal distribution with mean µ = 6 and standard deviation o=3. The other graph represents a normal distribution with mean µ = 16 and standard deviation o = 3. Determine which graph is which and explain how you know. Choose the correct answer below. O A. Graph A has a mean of µ = 6 and graph B has a mean of μ = 16 because a larger mean shifts the graph to the right. B. Graph A has a mean of μ = 6 and graph B has a mean of µ = 16 because a larger mean shifts the graph to the left. C. Graph A has a mean of µ = 16 and graph B has a mean of μ = 6 because a larger mean shifts the graph to the left. D. Graph A has a mean of μ = 16 and graph B has a mean of µ = 6 because a larger mean shifts the graph to the right. A B s 6 16
Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the Z-score. 0.054 Click to view page 1 of the table. Click to view page 2 of the table. The cumulative area corresponds to the z-score of (Round to three decimal places as needed.)
the mean is 146, and standard deviation is 35. A score of 181 is how many z-scores above the mean?
One graph in the figure represents a normal distribution with mean u = 7 and standard deviation o = 1. The other graph represents a normal distribution with mean u = 12 and standard deviation o = 1. Determine which graph is which and explain how you know. 12 Choose the correct answer below. O A. Graph A has a mean of u = 12 and graph B has a mean of u =7 because larger mean shifts the graph to the left. O B. Graph A has a mean of u = 12 and graph B has a mean of u =7 because a larger mean shifts the graph to the right. O C. Graph A has a mean of u = 7 and graph B has a mean of u = 12 because a larger mean shifts the graph to the left. O D. Graph A has a mean of u = 7 and graph B has a mean of u = 12 because a larger mean shifts the graph to the right.
Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.051 Click to view page 1 of the table. Click to view page 2 of the table. The cumulative area corresponds to the z-score of (Round to three decimal places as needed.)
Use a calculator or computer software to generate a normal quantile plot for the data in the accompanying table. Then determine whether the data come from a normally distributed population. Click the icon to view the data set. Generate a normal quantile plot for the data. Choose the correct graph below. OA. 3.00 2.00- 1.00- 0.00- -1.00- -2.00- -3.00- 0 100 200 300 400 500 X Value Data table Space Flight Durations (hours) 0 350 100 400 300 200 50 150 200 OB. 3.00- 2.00- Print 1.00- 0.00- -1.00- -2.00- -3.00+ 0 BEELD 100 200 300 400 500 X Value 300 300 200 100 450 200 50 0 0 400 100 450 G Determine whether the data come from a normally distributed population. Choose the correct answer below. OA. The distribution is not normal. The points are not reasonably close to a straight line. OB. The distribution is normal. The points show a systematic pattern that is not a straight-line pattern. OC. The distribution is normal. The points are not reasonably close to a straight line and do not show…
Determine whether the graph shown could represent a variable with a normal distribution. Explain your reasoning. If the graph appears to represent a normal distribution, estimate the mean and standard deviation. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. OA. The graph could not represent a variable with a normal distribution because the graph skewed to the right. O B. The graph could represent a variable with a normal distribution because the curve symmetric and bell-shaped. Its mean is approximately, and its standard deviation is approximately. (Round o one decimal place as needed.) O C. The graph could not represent a variable with a normal distribution because the curve has two modes. O D. The graph could not represent a variable with a normal distribution because the graph is skewed to the left. O E. The graph could not represent a variable with a normmal distribution because the curve is constant. O F. The graph could not…
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the bone density test scores that can be used as cutoff values separating the lowest 10% and highest 10%, indicating levels that are too low or too high, respectively. Sketch the region containing the lowest 10% and highest 10%. Choose the correct graph below. OA. -Za Za Q The bone density scores are ☐ OB. -Za Za (Use a comma to separate answers as needed. Round to two decimal places as needed.) ○ C. Q Za -Za D. Q a G Za -Za
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Determine whether the graph shown could represent a variable with a normal distribution. Explain your reasoning. If the graph appears to represent a normal distribution, estimate the mean and standard deviation. 32 35 38 41 44 47 50 53 56 E. The graph could represent a variable with a normal distribution because the curve is symmetric and bell-shaped. Its mean is approximately 44, and its standard deviation is approximately 32. (Round to one decimal place as needed.) Q right. O C. The graph could not represent a variable with a normal distribution because the curve has two modes. O D. The graph could not represent a variable with a normal distribution because the curve crosses the x-axis. O F. The graph could not represent a variable with a normal distribution because the graph is skewed to the left. 4

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