Question

Majesty Video Production Inc. wants the mean length of its advertisements to be 22 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 10 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 23.50 seconds? (Round your z values and final answers to 2 decimal places.)

What percent of the sample means will be greater than 20.50 seconds? (Round your z values and final answers to 2 decimal places.)

What percent of the sample means will be greater than 20.50 but less than 23.50 seconds? (Round your z values and final answers to 2 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 22

standard deviation = = 2

n = 10

= = 22 and

= / n = 2 / 10 = 0.63

The standard error of the mean time is 0.63

P( > 23.50) = 1 - P( < 23.50)

= 1 - P(( - ) / < (23.50 - 22) / 0.63)

= 1 - P(z < 2.38)

= 0.9913 Using standard normal table.

= 99.13%

99.13% of the sample means will be greater than 23.50 seconds.

P( > 20.50) = 1 - P( < 20.50)

= 1 - P(( - ) / < (20.50 - 22) / 0.63)

= 1 - P(z < -2.38)

= 0.0087 Using standard normal table.

= 0.87%

0.87% of the sample means will be greater than 20.50 seconds.

P(20.50 < < 23.50) = P((20.50 -22) / 0.63<( - ) / < (23.50 - 22) / 0.63))

= P(-2.38 < Z < 2.38)

= P(Z < 2.38) - P(Z < -2.38) Using z table,

= 0.9913 - 0.0087

= 0.9826

= 98.26%

98.26%  of the sample means will be greater than 20.50 but less than 23.50 seconds.