CRA CDs Inc. wants the mean lengths of the \"cuts\" on a CD to be 160 seconds (2
ID: 3021211 • Letter: C
Question
CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 160 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 the normal distribution with a population standard deviation of 8 seconds. Suppose we select a sample of 15 cuts from various CDs sold by CRA CDs Inc.
a.What can we say about the shape of the distribution of the sample mean? Sample mean (Click to select)NormalBinomialUniform
b. What is the standard error of the mean? (Round your z-value to 2 decimal places and final answer to 2 decimal places.) Standard error of the mean seconds.
c. What percent of the sample means will be greater than 162 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.) Percent %
d. What percent of the sample means will be greater than 153 seconds? (Round your answer to 2 decimal places.) Percent %
e. What percent of the sample means will be greater than 153 but less than 162 seconds? (Round your z-value to 2 decimal places and final answer to 2 decimal places.) Percent %
Explanation / Answer
a.) Normal
b.) Standard error = 8/15 = 2.07
c.) P( X > 162) = P( Z > (162 - 160)/2.07)
= P( Z > 0.97)
= 0.166
16.6% of the sample mean will be greater than 162 seconds.
d.) P( X > 153)
= P( Z > (153 - 160)/2.07)
= P( Z > -3.38)
= 0.9996
99.96% of the sample mean is greater than 153 seconds.
e.) P( 153 < X < 162)
= P( -3.38 < Z < 0.97)
= P( Z < 0.97) - P( Z < -3.38)
= 0.834 - 0.0004
= 0.8336
83.36% of the sample mean will be greater than 153 but less than 162 seconds.