Question

We must show our work process

2. Majesty Video Production Inc. wants the mean length of its advertisements to be 30 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 25 ads produced by Majesty.

a. What can we say about the shape of the distribution of the sample mean time?

b. What is the standard error of the mean time?

   c. What percent of the sample means will be greater than 30.8 seconds?

   d. What percent of the sample means will be greater than 28.6 seconds?

   e. What percent of the sample means will be greater than 28.6 but less than 30.8 seconds?

Answer 1

a) The shape of the sample mean is approximately normally distributed.

b) SE = = 2/ = 0.4

c) P( > 30.8)

= P(( - )/( )) > (30.8 - )/( ))

= P(Z > (30.8 - 30)/0.4)

= P(Z > 2)

= 1 - P(Z < 2)

= 1 - 0.9772

= 0.0228 = 2.28%

d) P( > 28.6)

= P(( - )/( ) > (28.6 - )/( ))

= P(Z > (28.6 - 30)/0.4)

= P(Z > -3.5)

= 1 - P(Z < -3.5)

= 1 - 0 = 1 = 100%

e) P(28.6 < < 30.8)

= P((28.6 - )/( ) < ( - )/( ) < (30.8 - )/( ))

= P((28.6 - 30)/0.4 < Z < (30.8 - 30)/0.4)

= P(-3.5 < Z < 2)

= P(Z < 2) - P(Z < -3.5)

= 0.9772 - 0

= 0.9772 = 97.72%