The ages of adults in a certain community follow a normal distribution with mean 38 and standard deviation 6. The random variable x represents the age of a randomly-selected adult from this community. Given that P(32 < x < a) = .0590, what is the value of a?

a = 33.32

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

P(32 < x < a) = .0590

This is the pvalue of Z when X = a subtracted by the pvalue of Z when X = 32.

X = 32

has a pvalue of 0.1587

X = a

p - 0.1587 = 0.0590

p = 0.0590 + 0.1587

p = 0.2177

So when X = a, Z has a pvalue of 0.2177. So when X = a, Z = -0.78.

So a = 33.32

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23