Black Friday - the annual shopping tradition the day after Thanksgiving - is often the day which puts retailers "in the black." According to a CNN Money report, consumers spent an average of $363.99 on Black Friday in 2010 with a standard deviation of $235.28. a. Draw and label a normal curve which would be used to describe the Black Friday expenditures. Based on the values calculated, would it be reasonable to assume the money spent is normally distributed? OIt is reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed It is not reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed. b. Completely describe the sampling distribution of the sample mean Black Friday expenditure when samples of size 55 are selected. o Mean: µj o Standard deviation: 0, = (round to 4 decimal places) 0 because o Shape:the distribution of j is Select an answer Select an answer c. Using the distribution described in part b, what is the probability of observing a sample mean of $429.456 or more? O z= (round to 2 decimal places) probability = (include 4 decimal places) d. Based on the probability found, what conclusion can be reached? o The probability would be classified as 8 . So, there O sufficient evidence to conclude the mean amount Select an answer Select an answer spent by customers on Black Friday is greater than 363.99.
Black Friday - the annual shopping tradition the day after Thanksgiving - is often the day which puts retailers "in the black." According to a CNN Money report, consumers spent an average of $363.99 on Black Friday in 2010 with a standard deviation of $235.28. a. Draw and label a normal curve which would be used to describe the Black Friday expenditures. Based on the values calculated, would it be reasonable to assume the money spent is normally distributed? OIt is reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed It is not reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed. b. Completely describe the sampling distribution of the sample mean Black Friday expenditure when samples of size 55 are selected. o Mean: µj o Standard deviation: 0, = (round to 4 decimal places) 0 because o Shape:the distribution of j is Select an answer Select an answer c. Using the distribution described in part b, what is the probability of observing a sample mean of $429.456 or more? O z= (round to 2 decimal places) probability = (include 4 decimal places) d. Based on the probability found, what conclusion can be reached? o The probability would be classified as 8 . So, there O sufficient evidence to conclude the mean amount Select an answer Select an answer spent by customers on Black Friday is greater than 363.99.
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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