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 and a standard deviation of 5.5 hours. As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.

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

2. What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal places.)

3. What proportion of the samples will have a mean useful life of more than 38 hours? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

4. What proportion of the sample will have a mean useful life greater than 36.5 hours? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

5. What proportion of the sample will have a mean useful life between 36.5 and 38 hours? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

Answer 1

Solution:

Given:

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 and a standard deviation of 5.5 hours.

That is: X ~ N( µ= 37 , σ = 5.5 )

Sample size = n = 25

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

Since the mean life of these batteries follows the normal probability distribution , shape of the distribution of the sample mean is also a Normal distribution, that is symmetrical.

Part b) What is the standard error of the distribution of the sample mean?

The standard error of the distribution of the sample mean is given by:

Part c) What proportion of the samples will have a mean useful life of more than 38 hours?

Find z score:

Thus we get:

Look in z table for z = 0.9 and 0.01 and find corresponding area.

Thus from z table we get:

P( Z < 0.91) = 0.8186

Thus

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

Find z score:

Thus we get:

Look in z table for z = -0.4 and 0.05 and find corresponding area.

Thus from z table we get:

P( Z < -0.45 ) = 0.3264

Thus

Part e) What proportion of the sample will have a mean useful life between 36.5 and 38 hours?

We z values for 38 and 36.5 as 0.91 and -0.45 respectively.

From part c) we have P( Z < 0.91 ) = 0.8186 and P( Z < -0.45) = 0.3264