Question

CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 150 seconds (2 minutes and 40 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 a normal distribution with a standard deviation of nine seconds. Suppose that we select a sample of 12 cuts from various CDs sold by CRA CDs Inc.

What percentage of the sample means will be greater than 145 but less than 157 seconds?

(Round the final answer to 2 decimal places.)

Answer 1

From the given information,
By using calculator,
The required correct answer is,

P(145<Xbar<157)= 0.97

(Note: There is some confusion. Given mean is 150 seconds which should be 2 minutes and 30 seconds but in the given information it is 2 minutes and 40 minutes so using this mean will be 160 seconds. So above answer is with mean=150.)

If mean=160 seconds then

P(145<Xbar<157)= 0.12

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