Question

CRA CDs Inc. wants the mean lengths of the “cuts” on a CD to be 148 seconds (2 minutes and 28 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows a normal distribution with a standard deviation of eight seconds. Suppose that we select a sample of 26 cuts from various CDs sold by CRA CDs Inc. Use Appendix B.1 for the z values. a. What can we say about the shape of the distribution of the sample mean? Shape of the distribution is (Click to select) b. What is the standard error of the mean? (Round the final answer to 2 decimal places.) Standard error of the mean seconds. c. What percentage of the sample means will be greater than 152 seconds? (Round the z values to 2 decimal places and the final answers to 2 decimal places.) Percentage % d. What percentage of the sample means will be greater than 144 seconds? (Round the z values to 2 decimal places and the final answers to 2 decimal places.) Percentage % e. What percentage of the sample means will be greater than 144 but less than 152 seconds? (Round the z values to 2 decimal places and the final answers to 2 decimal places.) Percentage %

Answer 1

a. The shape of the distribution of the sample mean will be approximately normal, according to the Central Limit Theorem.

b. The standard error of the mean is given by:

SE = σ / sqrt(n)

where σ is the population standard deviation (8 seconds), and n is the sample size (26). Substituting the given values, we get:

SE = 8 / sqrt(26) ≈ 1.57 seconds

Rounded to 2 decimal places, the standard error of the mean is 1.57 seconds.

c. To find the percentage of sample means that will be greater than 152 seconds, we need to calculate the z-score corresponding to a sample mean of 152 seconds:

z = (x - μ) / (σ / sqrt(n))

where x is the sample mean (152 seconds), μ is the population mean (148 seconds), σ is the population standard deviation (8 seconds), and n is the sample size (26).

Substituting the given values, we get:

z = (152 - 148) / (8 / sqrt(26)) ≈ 1.98

Using Appendix B.1, we find that the area to the right of a z-score of 1.98 is 0.0242, or 2.42%. Therefore, approximately 2.42% of the sample means will be greater than 152 seconds.

d. To find the percentage of sample means that will be greater than 144 seconds, we need to calculate the z-score corresponding to a sample mean of 144 seconds:

z = (x - μ) / (σ / sqrt(n))

where x is the sample mean (144 seconds), μ is the population mean (148 seconds), σ is the population standard deviation (8 seconds), and n is the sample size (26).

Substituting the given values, we get:

z = (144 - 148) / (8 / sqrt(26)) ≈ -1.98

Using Appendix B.1, we find that the area to the right of a z-score of -1.98 is also 0.0242, or 2.42%. Therefore, approximately 2.42% of the sample means will be less than 144 seconds.

e. To find the percentage of sample means that will be greater than 144 but less than 152 seconds, we need to find the area between the z-scores corresponding to sample means of 144 and 152 seconds.

The z-score corresponding to a sample mean of 144 seconds is:

z1 = (144 - 148) / (8 / sqrt(26)) ≈ -1.98

The z-score corresponding to a sample mean of 152 seconds is:

z2 = (152 - 148) / (8 / sqrt(26)) ≈ 1.98

Using Appendix B.1, we find that the area to the right of a z-score of -1.98 is 0.0242, and the area to the right of a z-score of 1.98 is 0.0242. Therefore, the area between these two z-scores is:

0.5 - 0.0242 - 0.0242 = 0.4516

Multiplying by 100, we get that approximately 45.16% of the sample means will be greater than 144 but less than 152 seconds.