Question

Majesty Video Production Inc. wants the mean length of its advertisements to be 24 seconds. Assume the distribution of ad length follows the normal distribution with a population standard deviation of 2 seconds. Suppose we select a sample of 15 ads produced by Majesty. What can we say about the shape of the distribution of the sample mean time? What is the standard error of the mean time? (Round your answer to 2 decimal places.) What percent of the sample means will be greater than 25.50 seconds? (Round your z values and final answers to 2 decimal places.) What percent of the sample means will be greater than 22.50 seconds? (Round your z values and final answers to 2 decimal places.) What percent of the sample means will be greater than 22.50 but less than 25.50 seconds? (Round your z values and final answers to 2 decimal places.)

Answer 1

Given that mean =24 second and standard deviation=2 seconds and n=15 hence

a) Standard Error is calculated as

b) To find probability that mean is greater than 25.50 we need to find Z score at 25.50 sample mean as

HenceP(Z>2.91) is calculated by Z table shown below as 0.0018

c) Now at mean=22.50

Hence P(Z>-2.91)

=1-0.0018

=0.9982

d) Since Movie above Z=2.91 is 0.0 018 and above -2.91 is 0.9982 now for P(-2.91<Z<2.91)

=0.9982-0.0018

=0.9964

The Z table used for calculation is