SOLUTION: A manager of a grocery store took a random sample of 100 customers. Their average checkout time is 3 minutes. The standard deviation of the population is 2 minutes. What is the pro

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Question 526332: A manager of a grocery store took a random sample of 100 customers. Their average checkout time is 3 minutes. The standard deviation of the population is 2 minutes. What is the probability that the average time is greater than 3.6 minutes? Less than 3.4 minutes?
I would like to identify a z score for each of the questions, but I don't have a population mean? Can I solve this without that number?

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A manager of a grocery store took a random sample of 100 customers. Their average checkout time is 3 minutes. The standard deviation of the population is 2 minutes. What is the probability that the average time is greater than 3.6 minutes? Less than 3.4 minutes?
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I would like to identify a z score for each of the questions, but I don't have a population mean? Can I solve this without that number?
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x-bar = 3 min
std for the mean of sample means = 2/sqrt(100) = 0.2
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z(3.6) = (3.6-3)/0.2 = 0.6/0.2 = 3
P(x-bar > 3.6) = P(z > 3) = 0.0013
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z(3.4) = (3.4-3)/0.2 = 0.4/0.2 = 2
P(x-bar < 3.4) = P(z < 2) = 0.9772
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Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

What makes you think you don't have a population mean? What do you think the statement "Their average checkout time is 3 minutes" means?

John

My calculator said it, I believe it, that settles it