For the manufacturing plant discussed in Exercise 8.10, the union president and the human resources director jointly select a simple random sample of 36 employees to engage in a discussion with regard to the company’s work rules and overtime policies. What is the probability that the average number of overtime hours last year for members of this sample would have been less than 65.0 hours? Between 55.0 and 65.0 hours?

(8.10) Employees in a large manufacturing plant worked and average of 62.0 hours of overtime last year, with a standard deviation of 15.0 hours. For a simple random sample of n=36 employees and x=the number of overtime hours worked last year, determine the z-score corresponding to each of the following sample means:
a. ȭ= 55.0 hours b. ȭ=60.0 hours
c. ȭ=65.0 hours d. ȭ= 47.0 hours

To find the probability that the average number of overtime hours for the sample is less than 65.0 hours, we need to convert this sample mean into a z-score using the formula:

z = (x - μ) / (σ / √n)

where:
x = sample mean (65.0 hours)
μ = population mean (62.0 hours)
σ = population standard deviation (15.0 hours)
n = sample size (36 employees)

Substituting the values into the formula, we get:

z = (65.0 - 62.0) / (15.0 / √36)
z = 3 / (15 / 6)
z = 3 / 2.5
z = 1.2

Now, we need to find the probability that the z-score is less than 1.2 using a standard normal distribution table or calculator. The cumulative probability up to 1.2 is approximately 0.8849.

Therefore, the probability that the average number of overtime hours for the sample is less than 65.0 hours is 0.8849 (or 88.49%).

To find the probability that the average number of overtime hours for the sample is between 55.0 and 65.0 hours, we first need to calculate the z-scores for both values.

For the sample mean of 55.0 hours:

z = (55.0 - 62.0) / (15.0 / √36)
z = -7 / (15 / 6)
z = -7 / 2.5
z = -2.8

For the sample mean of 65.0 hours, we have already calculated the z-score as 1.2.

Now, we need to find the probability that the z-score is between -2.8 and 1.2. Again, we can use a standard normal distribution table or calculator to find this cumulative probability.

The cumulative probability up to 1.2 is approximately 0.8849, and the cumulative probability up to -2.8 is approximately 0.0026. Subtracting the two probabilities, we get:

P(-2.8 < z < 1.2) = P(z < 1.2) - P(z < -2.8)
= 0.8849 - 0.0026
= 0.8823

Therefore, the probability that the average number of overtime hours for the sample is between 55.0 and 65.0 hours is 0.8823 (or 88.23%).