Question

The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. A random sample of 36 intermediate level executives is selected. What is the probability that the mean annual salary of the sample is between $71000 and $73500?

Answer 1

11.51% probability that the mean annual salary of the sample is between $71000 and $73500

Explanation:

To solve this question, we need 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 = (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 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.

In this question, we have that:

`\mu = 74000, \sigma = 2500, n = 36, s = (2500)/(√(36)) = 416.67`

What is the probability that the mean annual salary of the sample is between $71000 and $73500?

This is the pvalue of Z when X = 73500 subtracted by the pvalue of Z when X = 71000. So

X = 73500

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

By the Central Limit Theorem

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

`Z = (73500 - 74000)/(416.67)`

`Z = -1.2`

`Z = -1.2` has a pvalue of 0.1151

X = 71000

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

`Z = (71000 - 74000)/(416.67)`

`Z = -7.2`

`Z = -7.2` has a pvalue of 0.

0.1151 - 0 = 0.1151

11.51% probability that the mean annual salary of the sample is between $71000 and $73500