Question

Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 37 hours. hours and a standard deviation of 5.5 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 9 batteries. a. What can you say about the shape of the distribution of the sample mean? b. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.) c. What proportion of the samples will have a mean useful life of more than 38.5 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) d. What proportion of the sample will have a mean useful life greater than 36.5 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.) e. What proportion of the sample will have a mean useful life between 36.5 and 38.5 hours? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Given the mean life of these batteries follows the normal probability distribution with a mean of = 37 hours. hours and a standard deviation of = 5.5 hours.

As a part of its quality assurance program, Power +, Inc. tests samples of n = 9 batteries.

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

Since the given population is normally distributed hence using the central limit theorem we can assume that the sample distribution is also normally distributed.

b. What is the standard error of the distribution of the sample mean?

The standard error of the distribution of the sample mean is calculated as:

c. What proportion of the samples will have a mean useful life of more than 38.5 hours?

To calculate the proportion we need to find the Z score at mean = 38.5 as:

now the proportion P( > 38.5) is P( Z > 0.86) is computed using the excel formula for normal distribution which is =1-NORM.S.DIST(0.86, TRUE), thus the proportion is computed as 0.1949.

d. What proportion of the sample will have a mean useful life greater than 36.5 hours?

again the Z score is calculated as:

now the proportion P( > 36.5) is P( Z > -0.29) is computed using the excel formula for normal distribution which is =1-NORM.S.DIST(-0.29, TRUE), thus the proportion is computed as 0.6141.

e) What proportion of the sample will have a mean useful life between 36.5 and 38.5 hours?

As the scores are calculated as:

and

This proportion is P(36.5< < 38.5) = P( -0.29< Z< 0.86) is computed using the excel fomula for normal distribution which is =NORM.S.DIST(0.86, TRUE) - NORM.S.DIST(-0.29, TRUE), thus the proportion is computed as 0.4192.