SOLUTION: The length of time it takes to fill an order at a local sandwich shop is normally distributed with a mean of 4.1 and standard deviation of 1.3 minutes.
(a) what is the probabilit
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(a) what is the probabilit
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Question 678262: The length of time it takes to fill an order at a local sandwich shop is normally distributed with a mean of 4.1 and standard deviation of 1.3 minutes.
(a) what is the probability that the average waiting time for a random sample of ten customers is between 4.0 and 4.2 minutes?
(b) the probability is 95% that the average waiting time for a random sample of ten customers is greater than how long? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The length of time it takes to fill an order at a local sandwich shop is normally distributed with a mean of 4.1 and standard deviation of 1.3 minutes.
(a) what is the probability that the average waiting time for a random sample of ten customers is between 4.0 and 4.2 minutes?
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Note: std for all samples of size 10 = 1.3/sqrt(10)
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z(4) = (4-4.1)/[1.3/sqrt(10)] = -0.2436
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z(4.2) = (4.2-4.1)/(1.3/sqrt(10)) = +0.2436
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P(4.0<= x-bar <=4.2) = P(-0.2436<= z <=0.2436) = 0.1925
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(b) the probability is 95% that the average waiting time for a random sample of ten customers is greater than how long?
Find the z-value with a left-tail of 5%:
invNorm(0.05) = -1.645
Find the corresponding wait time using x = z*s+u
x = -1.645*(1.3/sqrt(10))+4.1 = 3.4237 minutes
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Cheers,
Stan H.
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