Question

The incomes in a certain large population of college teachers have a normal distribution with mean $75,000 and standard deviation $8,000. Sixteen teachers are selected at random from this population to serve on a committee. What is the probability that their average salary is more than $77,500?

Answer 1

0.1056 = 10.56% probability that their average salary is more than $77,500.

Explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (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 p-value, 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 normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $75,000 and standard deviation $8,000.

This means that
`\mu = 75000, \sigma = 8000`

Sample of 16

This means that
`n = 16, s = (8000)/(√(16)) = 2000`

What is the probability that their average salary is more than $77,500?

This is 1 subtracted by the pvalue of Z when X = 77500. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (77500 - 75000)/(2000)`

`Z = 1.25`

`Z = 1.25` has a pvalue of 0.8944

1 - 0.8944 = 0.1056

0.1056 = 10.56% probability that their average salary is more than $77,500.