Question

g Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distributed with mean $39725 and standard deviation $7320. Suppose a school takes a sample of 125 such graduates and records the annual starting salary of each. The probability that the sample mean would be at least $39000 is about

Answer 1

Answer:

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.

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 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:

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

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

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

Mean $39725 and standard deviation $7320.

This means that $\mu = 39725, \sigma = 7320$

Sample of 125

This means that $n = 125, s = \frac{7320}{\sqrt{125}}$

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

$Z = \frac{39000 - 39725}{\frac{7320}{\sqrt{125}}}$

$Z = -1.11$

$Z = -1.11$ has a pvalue of 0.1335

1 - 0.1335 = 0.8665

The probability that the sample mean would be at least $39000 is of 0.8665 = 86.65%.