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

Answer:

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

Step-by-step 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:

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

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 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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 74000, \sigma = 2500, n = 36, s = \frac{2500}{\sqrt{36}} = 416.67$

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

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

By the Central Limit Theorem

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

$Z = \frac{73500 - 74000}{416.67}$

$Z = -1.2$

$Z = -1.2$ has a pvalue of 0.1151

X = 71000

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

$Z = \frac{71000 - 74000}{416.67}$

$Z = -7.2$

$Z = -7.2$ 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