8.11) 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 answer this question, we need to calculate the z-scores corresponding to each of the given sample means, and then use the z-score table to find the probabilities.

The formula for calculating the z-score is:

z = (x - μ) / (σ / √n)

Where:
- x is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size

In Exercise 8.10, we are given that the population mean (μ) is 62.0 hours and the population standard deviation (σ) is 15.0 hours.

a) For ȭ = 55.0 hours:
z = (55.0 - 62.0) / (15.0 / √36) = -1.40

b) For ȭ = 60.0 hours:
z = (60.0 - 62.0) / (15.0 / √36) = -0.40

c) For ȭ = 65.0 hours:
z = (65.0 - 62.0) / (15.0 / √36) = 0.80

d) For ȭ = 47.0 hours:
z = (47.0 - 62.0) / (15.0 / √36) = -2.00

To find the probabilities, we need to look up the z-scores in the standard normal distribution (z-score) table. The table provides probabilities of obtaining a z-score below a given value.

Using the table, we can find the probabilities from the z-scores:

a) The probability of obtaining a z-score less than -1.4 is approximately 0.0808 or 8.08%.
b) The probability of obtaining a z-score less than -0.4 is approximately 0.3446 or 34.46%.
c) The probability of obtaining a z-score less than 0.8 is approximately 0.7881 or 78.81%.
d) The probability of obtaining a z-score less than -2.0 is approximately 0.0228 or 2.28%.

To find the probability between 55.0 and 65.0 hours, we need to find the difference between the probabilities of the two z-scores.

The probability between 55.0 and 65.0 hours is:
P(55.0 < ȭ < 65.0) = P(ȭ < 65.0) - P(ȭ < 55.0)

Using the z-scores calculated earlier:
P(ȭ < 65.0) = 0.78
P(ȭ < 55.0) = 0.08

P(55.0 < ȭ < 65.0) = 0.78 - 0.08 = 0.70 or 70%

Therefore:
- The probability that the average number of overtime hours last year for members of this sample would have been less than 65.0 hours is approximately 78.81%.
- The probability that the average number of overtime hours last year for members of this sample would have been between 55.0 and 65.0 hours is approximately 70%.

47hours