Question

The average full-time faculty member in a college works an average of 53 hours per week. a) If we assume the standard deviation is 2.8 hours, what percentage of faculty members work more than 58.6 hours a week?

Answer 1

Answer:

2.28% of faculty members work more than 58.6 hours a week.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The average full-time faculty member in a college works an average of 53 hours per week. Standard deviation of 2.8 hours.

This means that $\mu = 53, \sigma = 2.8$

What percentage of faculty members work more than 58.6 hours a week?

The proportion is 1 subtracted by the p-value of Z when X = 58.6. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{58.6 - 53}{2.8}$

$Z = 2$

$Z = 2$ has a p-value of 0.9772

1 - 0.9772 = 0.0228

0.0228*100% = 2.28%

2.28% of faculty members work more than 58.6 hours a week.