Question

Majesty Video Production Inc. wants the mean length of its advertisements to be 26 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 14 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 27.25 seconds? (Round your z values and final answers to 2 decimal places.) What percent of the sample means will be greater than 24.50 seconds? (Round your z values and final answers to 2 decimal places.) What percent of the sample means will be greater than 24.50 but less than 27.25 seconds? (Round your z values and final answers to 2 decimal places.)

Answer 1

Solution :

Given that ,

mean = = 26

standard deviation = = 2

n = 14

= = 26  

= / n = 2 / 14 = 0.5345

1) P( >27.25 ) = 1 - P( < 27.25 )

= 1 - P[( - ) / < (27.25 - 26) / 0.5345 ]

= 1 - P(z < 2.34 )

= 1- 0.9904 = 0.0096

Probability = 0.0096 = 0.96%

Answer = 0.96%

2)

P( >24.50 ) = 1 - P( < 24.50 )

= 1 - P[( - ) / < (24.50 - 26) / 0.5345 ]

= 1 - P(z < -2.81 )

= 1- 0.0025 = 0.9975

Probability = 0.9975= 99.75%

Answer = 99.75%

3)

P(24.50< < 27.25 )  

= P[(24.50 - 26) 0.5345/ < ( - ) / < (27.25 - 26) /0.5345 )]

= P(-2.81 < Z <2.34 )

= P(Z <2.34 ) - P(Z < -2.81)

=0.9904 -0.0025 = 0.9879

probability = 0.9879=98.79%

Answer = 98.79%