SOLUTION: CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 135seconds (2 min 15 sec). This will allow the disk jockeys to have plenty of time for commercials within each 10-mi

Question 664958: CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 135seconds (2 min 15 sec). 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 population standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs Inc.
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 sample means will be greater than 140 seconds?
d) What percent of the sample means will be greater than 128 seconds?
e) What percent of the sample means will be greater than 128 but less than 140 seconds?
This is just one sample problem for my homework. I cannot seem to understand the process to answering these questions. Please show all formulas and calculations so that I can learn from this
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 135seconds (2 min 15 sec). 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 population standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs Inc.
a) What can we say about the shape of the distribution of the sample mean?
According to the Central Limit Theorem:
The mean of the sample means = the mean of the population = 135 sec
The std of the sample means = 8/sqrt(16) = 4
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b) What is the standard error of the mean?
standard error = std of the sample means = 4
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c) What percent of the sample means will be greater than 140 seconds?
z(140) = (140-135)/4 = 5/4
P(x-bar > 140) = P(z > 5/4) = normalcdf(5/4,100) = 0.1056
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d) What percent of the sample means will be greater than 128 seconds?
z(128) = (128-135)/4 = -7/4
P(x-bar > 128) = P(z > -7/4) = normalcdf(-7/4,100) = 0.9599
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e) What percent of the sample means will be greater than 128 but less than 140 seconds?
Ans: = P(-7/4 < z < 5/4) = normalcdf(-7/4,-5/4) = 0.8543
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Cheers,
Stan H.
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