Question

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?

Answer 1

Answer:

a = 33.32

Step-by-step explanation:

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

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

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

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:

$\mu = 38, \sigma = 6$

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

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

$Z = \frac{32 - 38}{6}$

$Z = -1$

$Z = -1$ 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.

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

$-0.78 = \frac{X - 38}{6}$

$X - 38 = -0.78*6$

$X = 33.32$

So a = 33.32