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?

Question

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?

PsyDAG · Answer

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.