O HelpOne graph in the figure represents a normal distribution with mean u= 15 and standard deviation o =1. The other graph represents anomal distribution with mean u=10 and standard deviation o=1. Determine which graph is which and explain how you know.me10 15TestsGradebookScorectivityChoose the correct answer below.OA. Graph A has a mean of u= 10 and graph B has a mean of u = 15 because a larger mean shifts the graph to the right.%3DO B. Graph A has a mean of u= 10 and graph B has a mean of u = 15 because a larger mean shifts the graph to the left.%3DOC. Graph A has a mean of u= 15 and graph B has a mean of u = 10 because a larger mean shifts the graph to the right.94.05%mchO D. Graph A has a mean of µ= 15 and graph B has a mean of u = 10 because a larger mean shifts the graph to the left.Contentsor SuccessNextGradebookScoremedia LibraryAttemptsSample Tests and Quizzesse Options301 PM5/4/2022ly sunnyhpinsprt scfgf12deletehomeend40numbackspace

Here's first one mean is 15 and standard deviation is 1. Other graph has mean is 10 and standard…

Answered: O Help One graph in the figure…

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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What is the FSTAT and P-Value?
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
I do not understand and need help answering
The histogram has a longer tail/appears roughly the same/has a longer right tail so the distribution of the data is not skewed/skewed to the left/ skewed to the right
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 -
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. 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- 92.5 26 25- 114.5 136.5 20- 158.5 15- 12 Ef = Exf = 10- Find the mean of the data set. 5- X = (Round to the nearest tenth as needed.) 48.5 70.5 92.5 114.5 136.5 158.5 Complete the table. Round values to the nearest tenth as needed. Midpoint x (x-x)? (x-x)?f X-X 70.5 Print Done 92.5 114.5 136.5 158.5 (x-x)?r= ] of
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
The first picture is the question and the second is the ones that I need help with. Thanks
Please see attached.
Find the mean, the median and the mode of the given data set. (Round your answers to four decimal places. If there is more than one mode, enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) S = {45, 45, 60, 70, 75, 75, 85} mean median mode
I vacuum cleaners. The weights are classified according to vacuum cleaner type. At a=0.10, can you conclude that at least one mean vacuum cleaner weight is different The table available below shows the weights (in pounds) for a sample from the others? E Click the icon to view the vacuum cleaner weight data. Let Hgu. PBLU and PTC represent the mean weights for bagged upright, bagless upright, and top canister vacuums respectively. What are the hypotheses for this test? O A. Ho: Heu * PBLU HTC H PBU PBLU =HTC Weight of Vacuum Cleaners by Type O B. Ho: At least one of the means is different. H PBu = HBLUu =PTC &c. Họ: PBU = HBLU =HTC Bagged upright 22 19 Bagless upright Top canister 21 22 23 25 H: At least one of the means different. 17 20 15 18 18 19 O D. Ho: PBU = PBLU "HTC 23 20 26 25 22| 23 Ha: Peu * PBLU *HTC What is the test statistic? Print Done F= (Round to two decimal places as needed.)

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