Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 53, \sigma = 2.8[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{58.6 - 53}{2.8}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] 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.
