Question

The average starting salary for this year's graduates at a large university (LU) is $30,000 with a standard deviation of $6,000. Furthermore, it is known that the starting salaries are normally distributed. Answer the following question - Individuals with starting salaries of less than $15,600 receive a low income tax break. What percentage of the graduates will receive the tax break

Answer 1

Answer:

0.82% of the graduates will receive the tax break

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 30000, \sigma = 6000$

What percentage of the graduates will receive the tax break

This is the pvalue of Z when X = 15600. So

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

$Z = \frac{15600 - 30000}{6000}$

$Z = -2.4$

$Z = -2.4$ has a pvalue of 0.0082

0.82% of the graduates will receive the tax break