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?

Need mean (μ) and standard deviation (SD). Calculate Z scores for values and find probabilities in table in the back of statistics text labeled something like "areas under normal distribution."

Z = (x - μ)/SE

SE = Standard Error of the mean = SD/sqrt n

I hope this helps. Thanks for asking.

To answer this question, we need to have information about the population average and standard deviation of overtime hours. Unfortunately, that information is not provided in the question. Without that information, we cannot accurately calculate the probabilities.

However, I can explain the general steps you would need to take if you had the population parameters.

To calculate the probability that the average number of overtime hours for this sample is less than 65.0 hours, you would need to use the Central Limit Theorem. This theorem states that regardless of the shape of the population distribution, the distribution of sample means will approach a normal distribution as the sample size increases.

Here are the general steps to calculate the probability:

1. Determine the population mean (μ) and population standard deviation (σ) of overtime hours.
2. Calculate the standard error of the mean (SE), which is equal to the population standard deviation divided by the square root of the sample size. SE = σ / √n.
3. Convert the value of 65.0 hours to a z-score using the formula:
z = (x - μ) / SE
where x is the value you are interested in (in this case, 65.0 hours), μ is the population mean, and SE is the standard error of the mean.
4. Look up the z-score in the standard normal distribution table (or use a calculator or software) to find the corresponding probability.
5. Repeat the steps above to calculate the probability for the range between 55.0 and 65.0 hours. In this case, you would calculate two z-scores: one for 55.0 hours and one for 65.0 hours. Then, find the area between these two z-scores in the standard normal distribution.

Again, without the population parameters, we cannot provide specific probabilities. However, you can refer to the steps outlined above if you have the necessary information.