Question

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

Answer 1

Answer:

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

Step-by-step explanation:

Answer 2

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

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

• It 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, which is the percentile of X.

In this problem:

• The mean is of 80, hence $\mu = 80$.
• The standard deviation is of 7, hence $\sigma = 7$.

The minimum value is the 80th percentile, which means that it is $X_m$ 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 $P(X_m < X < 90)$, which is the p-value of Z when X = 90 subtracted by the p-value of Z when $X = X_m$, hence the p-value of Z when X = 90 subtracted by 0.8.

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

$Z = \frac{90 - 80}{7}$

$Z = 1.43$

$Z = 1.43$ 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 https://brainly.com/question/24663213