Question

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

Answer 1

Answer:

97.72% probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 40000, \sigma = 5000$

What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

This is 1 subtracted by the pvalue of Z when X = 30000. So

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

$Z = \frac{30000 - 40000}{5000}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

1 - 0.0228 = 0.9772

97.72% probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000