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

To find the percentage of faculty members who work more than 58.6 hours a week, calculate the z-score and find the area to the right of the z-score using a standard normal distribution table or calculator. Approximately 1.92% of faculty members work more than 58.6 hours a week.

Step-by-step explanation:

To find the percentage of faculty members who work more than 58.6 hours a week, we can use the z-score formula:

z = (x - μ) / σ

where
x = 58.6 (target value),
μ = 53 (mean),
σ = 2.8 (standard deviation)

Calculate the z-score:

z = (58.6 - 53) / 2.8 = 2.07

Use a standard normal distribution table or calculator to find the area to the left of the z-score (2.07). The area to the right represents the percentage of faculty members who work more than 58.6 hours a week.

The area to the right is approximately 0.0192, which is equivalent to 1.92%. Therefore, approximately 1.92% of faculty members work more than 58.6 hours a week.

Answer 2

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 = (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 = (X - \mu)/(\sigma)`

`Z = (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.