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?

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 and standard deviation , the z-score of a measure X is given by:

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

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

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.

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.