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.
a. What is the probability that the checkout time for the next customer will be 6 minutes or less?
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.
d. What is the probability that the mean of the sample is between 6 and 7 minutes?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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