Explanation:
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.
