Question 670561
Checkout times in a Wal-Mart store follow a normal distribution with a mean time of 8 minutes and standard deviation of 1.5 minutes.
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For each of these problems you have to convert the raw numbers to z-numbers,
then look up the probability on a z-chart or use a calculator.
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a. What is the probability that the checkout time for the next customer will be 6 minutes or less?
z(6) = (6-8)/1.5 = -2/1.5 = -4/3
P(x <=6) = P(z <= -4/3) = normalcdf(-100,-4/3) = 0.0912
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b. 80% of the customers need more than C minutes to checkout. Calculate C. 
A sample of 40 clients is selected randomly.
c. What is the probability distribution of the sample mean? Explain.
mean of sample means = 8
std of sample means = 1.5/sqrt(40)
Stated in The Central Limit Theorem
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d. What is the probability that the mean of the sample is between 6 and 7 minutes?
z(6) = (6-8)/[1.5/sqrt(40)) = -8.4327
z(7) = (7-8)/(1.5/sqrt(40)) = -4.2164
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P(6 <= x <= 7) = P(-8.4327  <= z <= -4.2164) = normalcdf(-8.4327,-4.2164)
= 0.00001242
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Cheers,
Stan H.