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 ?.

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

Z = \frac{X - \mu}{\sigma}Z=σX−μ

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 = 80μ=80 .

The standard deviation is of 7, hence \sigma = 7σ=7 .

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

Z = \frac{X - \mu}{\sigma}Z=σX−μ

Z = \frac{90 - 80}{7}Z=790−80

Z = 1.43Z=1.43

Z = 1.43Z=1.43 has a p-value of 0.9236.

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

lhmarianateixeira lhmarianateixeira · Answer 2 · 2021-12-21T17:08:13+01:00

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:

Letter B

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

[tex]Z = \frac{X - \mu}{\sigma}\\Z=\sigma X - \mu[/tex]

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 [tex]\mu = 80[/tex].

The standard deviation is of 7, hence [tex]\sigma = 7[/tex].

The minimum value is the 80th percentile, which means that it is [tex]X_m[/tex] 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, which is the p-value of Z when X = 90 subtracted by the p-value of Z when X = X_mX=Xm , hence the p-value of Z when X = 90 subtracted by 0.8.

[tex]Z = \frac{X - \mu}{\sigma}\\Z=\sigma X - \mu\\Z = \frac{90 - 80}{7}\\Z=790 - 80\\Z = 1.43Z=1.43[/tex]

Has a p-value of 0.9236.

[tex]0.9236 - 0.8 = 0.1236[/tex]

Hence closest to 0.123.

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