CRA CDs Inc wants the mean lengths of the cuts on a CD to be

CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD to be 128 seconds (2 minutes and 8 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 population standard deviation of 6 seconds. Suppose we select a sample of 21 cuts from various CDs sold by CRA CDs Inc.

--> What percent of the sample means will be greater than 123 seconds?

additional info:

I set the problem up as

Solution

Usually z score greater than -3 is that the probablity is closed to 1 that you may not be able to find on the standard normal table.

The probability is

P(xbar>123) = P((xbar-mean)/(s/vn) >(123-128)/(6/sqrt(21)))

=P(Z>-3.82) =0.9999 (from standard normal table)