Question

The ages of personnel in the accounting department of a large company are normally distributed with a standard deviation of 7 years. There is a 0.02275 probability that the age of any randomly chosen person in the department is less than 22 and 0.15866 probabilty that the age of any randomly chosen person is greater than 43. What is the mean of this distribution?

Answer 1

Answer:

The mean of this distribution is 36 years.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\sigma = 7$

There is a 0.02275 probability that the age of any randomly chosen person in the department is less than 22

This means that when X = 22, Z has a pvalue of 0.02275. So when X = 22, Z = -2. So

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

$-2 = \frac{22 - \mu}{7}$

$22 - \mu = -2*7$

$\mu = 22 + 2*7$

$\mu = 36$

The mean of this distribution is 36 years.