A pop-music record firm wants the distribution of lengths of cuts on its record to have an average of 129 seconds and a standard deviation of 10 seconds so that disc jockeys will have plenty of time for commercials within each 5-minute period. The population times for cuts is approximately normally distributed with only a negligible skew to the right. You have just timed the cuts on a new release and have found that the 15 cuts average 149 seconds.

What percentage of the time will the average be 149 seconds or longer if the new release is randomly selected? (Give your answer correct to four decimal places.)


(b) If the music firm wants 15 cuts to average 149 seconds less than 5% of the time, what must the population mean be, given that the standard deviation remains at 10 seconds? (Give your answer correct to one decimal place.)

To solve this problem, we can use the concept of sampling distributions and the properties of the normal distribution.

(a) To find the percentage of the time that the average will be 149 seconds or longer, we need to calculate the z-score corresponding to that value and then find the area under the normal curve to the right of that z-score.

The formula to calculate the z-score is:

z = (x - μ) / (σ / √n)

where:
x = sample mean (149 seconds)
μ = population mean (unknown)
σ = standard deviation (10 seconds)
n = sample size (15 cuts)

We can rearrange the formula to solve for μ:

μ = x - (z * σ / √n)

Substituting the given values:

μ = 149 - (z * 10 / √15)

To find the z-score corresponding to 149 seconds or longer, we need to use the standard normal distribution table or a calculator. From the table, we find that the z-score is approximately 1.297.

Now we can substitute the z-score into the formula to calculate μ:

μ = 149 - (1.297 * 10 / √15)

Calculating this expression:

μ ≈ 147.358

To find the percentage of the time that the average will be 149 seconds or longer, we need to find the area under the normal curve to the right of the z-score (1.297). We can use a standard normal distribution table or a calculator to find this value.

Using a standard normal distribution table, the area to the right of 1.297 is approximately 0.0974.

Converting this to a percentage:

Percentage = 0.0974 * 100 ≈ 9.74%

Therefore, the percentage of the time the average will be 149 seconds or longer if the new release is randomly selected is approximately 9.74%.

(b) To find the population mean (μ) that ensures the 15 cuts average 149 seconds less than 5% of the time, we need to calculate the z-score corresponding to 5% and then use it to solve for μ.

The z-score corresponding to 5% can be found using a standard normal distribution table or a calculator. From the table, we find that the z-score is approximately -1.645.

Using the formula:

μ = x - (z * σ / √n)

Substituting the given values and rearranging:

μ = 149 - (-1.645 * 10 / √15)

Calculating this expression:

μ ≈ 154.926

Therefore, the population mean must be approximately 154.9 seconds to ensure that the 15 cuts average 149 seconds less than 5% of the time.

To answer these questions, we need to use the concept of the sampling distribution of the mean.

(a) To find the percentage of the time that the average will be 149 seconds or longer, we can calculate the z-score and use the standard normal distribution.

1. Calculate the z-score:
z = (x - μ) / (σ / √n)
Here, x = 149 (sample mean), μ = 129 (population mean), σ = 10 (standard deviation), and n = 15 (number of observations).

z = (149 - 129) / (10 / √15)
z = 20 / (10 / √15)
z ≈ 6.12

2. Use the standard normal distribution to find the percentage:
Find the cumulative probability P(Z ≥ 6.12) using a z-table or calculator.
The percentage will be 1 - P(Z < 6.12).

(b) To find the population mean that results in a 15-cut average of 149 seconds less than 5% of the time, we need to find the critical z-score corresponding to a cumulative probability of 0.05.

1. Find the critical z-score:
Find the z-score corresponding to a cumulative probability of 0.05 using a z-table or calculator.
Let's denote this critical z-score as z_critical.

2. Solve for the population mean:
Using the formula for z-score:
z_critical = (x - μ) / (σ / √n)

Rearrange the equation to solve for μ:
μ = x - (z_critical * (σ / √n))
Plug in the given values for x = 149, σ = 10, n = 15, and z_critical from step 1.

Once you have solved the equation, round the result to one decimal place.

Please note that the z-table or calculator is required to find the cumulative probabilities and critical z-scores.