CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD to be 128 seconds (2
ID: 3435163 • Letter: C
Question
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
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
and I end up with -3.8188. Its an introductory statistics class; the z-score chart I'm referencing doesn't reach -3. Am I missing something?
Explanation / Answer
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)