Question

The incomes in a certain large population of college teachers have a normal distribution with mean $75,000 and standard deviation $10,000. Four teachers are selected at random from this population to serve on a salary review committee. The probability that their average salary is less than $65,000 is closest to ___________.

Answer 1

Answer:

The probability that their average salary is less than $65,000 is closest to 2.28%.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 75000, \sigma = 10000, n = 4, s = \frac{10000}{\sqrt{4}} = 5000$

The probability that their average salary is less than $65,000 is closest to

This is the pvalue of Z when X = 65000. So

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

By the Central Limit Theorem

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

$Z = \frac{65000 - 75000}{5000}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

The probability that their average salary is less than $65,000 is closest to 2.28%.