Detemine whether the graph shown could represent a variable with a nomal distribution. Explain your reasoning. If the graph appears to represent a normal distribution, estimate the meanand standard deviation.4245485154Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.O A. The graph could represent a variable with a normal distribution because the curve is symmetric and bell-shaped. Its mean is approximately, and its standard deviation is approximately(Round to one decimal place as needed.)O B. The graph could not represent a variable with a normal distribution because the graph is skewed to the left.O C. The graph could not represent a variable with a normal distribution because the curve crosses the x-axis.O D. The graph could not represent a variable with a normal distribution because the curve is constant.O E. The graph could not represent a variable with a normal distribution because the curve has two modes.O F. The graph could not represent a variable with a normal distribution because the graph is skewed to the right.

Answered: Image /qna-images/answer/beea28b8-a312-477c-85f0-d7a6d5b5d9ee.jpg

Answered: Detemine whether the graph shown could…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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. 4 ..... Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. O A. The graph could represent a variable with a normal distribution because the curve is symmetric and bell-shaped. Its mean is approximately and standard deviation is approximately (Round to one decimal place as needed.) O B. The graph could not represent a variable with a normal distribution because the curve crosses the x-axis. 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 right. O E. The graph could not represent a variable with a normal distribution because the graph is skewed to the left. O F. The…
Determine whether the following graph can represent a variable with a nomal distribution. Explain your reasoning. If the graph appears to represent a normal distribution, estimate the mean and standard deviation. 10 15 20 25 30 Could the graph represent a variable with a normal distribution? Explain your reasoning. Select the correct choice below and, if necessary, fill in the answer boxes within your choice. O A. No, because the graph crosses the x-axis. O B. No, because the graph is skewed nght. O C. No, because the graph is skewed left. O D. Yes, the graph fulfills the properties of the normal distribution. The mean is approximately and the standard deviation is about- (Type whole numbers.) Clear all Final chec Help me solve this View an example Get more help -
Suppose that you are using independent samples to compare two population proportions. Fill in the blanks. a. The mean of all possible differences between the two sample proportions equals the_______.b. For large samples, the possible differences between the two samplel proportions have approximately a_________distribution.
Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table. LOADING... Click to view page 2 of the table. LOADING... x 0.9 A graph with a bell-shaped curve, divided into 2 regions by a line from top to bottom, on the left side. The region right of the line is shaded and labeled 0.9. The x-axis below the line is labeled "x". Question content area bottom Part 1 The indicated IQ score, x, is enter your response here. (Round to one decimal place as needed.)
Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.05 to test for a difference between the weights of discarded paper (in pounds) and weights of discarded plastic (in pounds). E Click the icon to view the data. In this example, Hg is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis test? O A. Ho: Ha = 0 H,: Ha #0 O B. Ho: Ha #0 H1: Hd =0 O C. Ho: Ha #0 O D. Ho: Ha = 0 H1: Hd 0
The amount of caffeine in a sample of five-ounce servings of brewed coffee is shown in the histogram. Make a frequency distribution for the data. Then use the table to estimate the sample mean and the sample standard deviation of the data set. W Click the icon to view the histogram. .. Complete the table. Round values to the nearest tenth as needed. Graph/chart f Midpoint x xf 70.5 Ay 30- 25- 24 20- |14 15- 10- 17 5- X 48.5 70.5 92.5 114.5 136.5 158.5
Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P9, the 9th percentile. This is the bone density score separating the bottom 9% from the top 91%. Which graph represents P9? Choose the correct graph below.
The distribution of the number of hours that a random sample of people spend doing chores per week is shown in the pie chart. Use 32 as the midpoint for "30+ hours." Make a frequency distribution for the data. Then use the table to estimate the sample mean and the sample standard deviation of the data set. Click the icon to view the pie chart. First construct the frequency distribution. Weekly study hours Class Frequency, f 0-4 5-9 0-4 hours (3 people) 10-14 5-9 hours (14 people) 10-14 hours (19 people) 15-19 15-19 hours (21 people) 20-24 20-24 hours (14 people) O 25-29 hours (13 people) 25-29 30+ hours (4 people) 30+ Print Done
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table. Click to view page 2 of the table. 70 105 The area of the shaded region is (Round to four decimal places as needed.)
Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find P7, the 7th percentile. This is the bone density score separating the bottom 7% from the top 93%. Which graph represents P7? Choose the correct graph below. A. P7 The bone density score corresponding to P7 is (Round to two decimal places as needed.) B. A P7 2 C. ^ P7 Ly D. A P7
The distribution of the number of hours that a random sample of people spend doing chores per week is shown in the pie chart. Use 32 as the midpoint for "30+ hours." Make a frequency distribution for the data. Then use the table to estimate the sample mean and the sample standard deviation of the data set. W Click the icon to view the pie chart. First construct the frequency distribution. Weekly study hours Class Frequency, f 0-4 I 0-4 hours (4 people) 5-9 1 5-9 hours (12 people) O 10-14 hours (23 people) O 15-19 hours (21 people) O 20-24 hours (16 people) O 25-29 hours (14 people) O 30+ hours (5 people) 10-14 15-19 20-24 25-29 30+ Print Done OOOOD OD
The graph shows the distribution of the number of text Which statement describes the distribution? messages young adults send per day. The distribution is approximately Normal, with a Daily Text Messaging mean of 128 messages and a standard deviation of 98 messages. O The distribution is approximately Normal, with a mean of 128 messages and a standard deviation of 30 messages. O The distribution is approximately Normal, with a mean of 30 messages and a standard deviation of 128 messages. O The distribution is uniform, with a mean of 128 messages and a standard deviation of 30 messages. 8. 38 68 98 128 158 188 218 248 Number of Texts

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