The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and standard deviation 7. Volunteers in the top 20 percent of hours worked will
receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given that the
number of hours the volunteer worked is less than 90?
A 0.077
B 0.123
C 0 0.618
D 0.618
E 0.923

Using the normal distribution, it is found that the probability that the volunteer selected will receive a certificate of merit given that the  number of hours the volunteer worked is less than 90 is closest to:

B 0.123

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

In this problem:

The minimum value is the 80th percentile, which means that it is when Z has a p-value of 0.8.

The probability that the volunteer selected will receive a certificate of merit given that the number of hours the volunteer worked is less than 90 is , which is the p-value of Z when X = 90 subtracted by the p-value of Z when , hence the p-value of Z when X = 90 subtracted by 0.8.

has a p-value of 0.9236.

0.9236 - 0.8 = 0.1236, hence closest to 0.123, option B.

A similar problem is given at link

correct answer is 0.134? it's not on the list

Step-by-step explanation:

the answer is c just got it on ed

step-by-step explanation:

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632