Question

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 $375.28 on Black Friday in 2010 with a standard deviation of $222.27.

1. 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?
   • It is not reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed.
   • It is reasonable to assume the amount of money spent by Black Friday shoppers is normally distributed
2. Completely describe the sampling distribution of the sample mean Black Friday expenditure when samples of size 81 are selected.
   • Mean: μ^-_y = _________
   • Standard deviation: σ^-_y = _________ (round to 4 decimal places)
   • Shape:the distribution of ^-_y is ____________ Select an answer: not normally distributed or normally distributed  because _______ Select an answer: the population of expenditures is normally distributed- the sample size is not large - the sample size is large - the population of expenditures is not normally distributed
3. Using the distribution described in part b, what is the probability of observing a sample mean of $441.538 or more?
   • z =   (round to 2 decimal places)
   • probability =   (include 4 decimal places)

Answer 1

Mean: μ^-_y = 375.28

std error=σy=σ/√n=222.27/sqrt(81)=

24.6967

the distribution of ^-_y is normally distributed   because the sample size is large

z =(441.538-375.28)/24.6967=2.68

probability = P(Z>2.68)=0.0037