SOLUTION: The mean amount purchased by a typical customer at PQR Store is $21.60 with standard deviation of $8.80. Assume the distribution of amount purchase follows the normal probability.
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-> SOLUTION: The mean amount purchased by a typical customer at PQR Store is $21.60 with standard deviation of $8.80. Assume the distribution of amount purchase follows the normal probability.
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Question 1169033: The mean amount purchased by a typical customer at PQR Store is $21.60 with standard deviation of $8.80. Assume the distribution of amount purchase follows the normal probability. For a sample of 64 customers, what is the probability the sample mean is less than $20.00? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! population mean is 21.60
population standard deviation is 8.80
sample size is 64 customers.
probability sample mean is less than 20.00 is what?
standard error = standard deviation / square root of sample size.
standard error = 8.80/sqrt(64) = 8.80/8 = 1.10
z-score = (x-m)/s
x is the raw score
m is the mean
s is the standard error.
in this problem, .....
z-score = (20 - 21.60) / 1.10
solve for z-score to get:
z = -1.454545.....
round to -1.45
area to the left of a z-score of -1.45 is be equal to .0735.
that's the probability that the sample mean will be less than 20.00.
