Question

Majesty Video Production INC wants the mean length of its advertisements to be 34 seconds. Assume the distribution of the ad length follows the normal distrubution with a population standard deviation of 2 seconds. Supposed we select a sample of 13 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 to 2 decimal places) What percent of the sample means will be greater than 35.25 seconds? (Round your z values and the final answer to 2 decimal places) What persent of the sample means will be greater than 32.75 seconds? (Round your z values and the final answer to 2 decimal places) What percent of the sample means will be greater than 32.75 but less than 35.25 seconds? (Round your z values and the final answer to 2 decimal places)

Answer 1

Answer)

As the given data is normally distributed we can use standard normal z table to estimate the answers.

Z = (x-mean)/(s.d/√n)

Given mean = 34

S.d = 2

N = 13

-)

Shape of the distribution is normal

-)

Standard error = s.d/√n = 2/√7 = 0.75592894601 = 0.76

-)

P(x>35.25)

Z = (35.25 - 34)/(2/√13)

Z = 2.25

From z table, p(z>2.25) = 0.0122

= 1.22%

-)

P(x>32.75)

Z = (32.75-34)/(2/√13)

Z = -2.25

From z table, p(z>-2.25) = 0.9878

= 98.78%

-)

P(32.75<x<35.25) = p(x<35.25) - p(x<32.75)

P(x<35.25)

Z = 2.25

From z table p(z<2.25) = 0.9878

P(x<32.75)

Z = -2.25

From z table, p(z<-2.25) = 0.0122

Required answer is

0.9878 - 0.0122

= 0.9756

= 97.56%