SOLUTION: CRA CDs, Inc. wants the mean lengths of the �cuts� on a CD to be 135 seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have plenty of time for commercials wit

Question 414467: CRA CDs, Inc. wants the mean lengths of the �cuts� on a CD to be 135 seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs, Inc. What can we say about the shape of the distribution of the sample mean?
A. What can we say about the shape of the distribution of the sample mean?
B. What is the standard error of the mean?
C. What percent of the samples will be greater than 128 minutes?
D. What percent of the sample means will be greater than 128 but less than 140 minutes.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
CRA CDs, Inc. wants the mean lengths of the �cuts� on a CD to be 135 seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs, Inc. What can we say about the shape of the distribution of the sample mean?
A. What can we say about the shape of the distribution of the sample mean?
mean of the sample means = 135 seconds
std of the sample means = 8/sqrt(16) = 2
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B. What is the standard error of the mean?
2 seconds
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C. What percent of the samples will be greater than 128 seconds?
50%
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D. What percent of the sample means will be greater than 128 but less than 140 minutes.
z(128) = 0
z(140) = (140-128)/2 = 6
----
P(128 <= x <= 140) = P(0<= z <6) = 0.5000
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Cheers,
Stan H.
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