Question

According to a recent survey, the salaries of entry-level positions at a large company have a mean of $40,756 and a standard deviation of $7500. Assuming that the salaries of these entry-level positions are normally distributed, find the proportion of

employees in entry-level positions at the company who earn at most $53,000. Round your answer to at least four decimal places.

Answer 1

We can use the standard normal distribution to solve this problem, by standardizing the salary value of $53,000 using the given mean and standard deviation:

z = (X - μ) / σ

where X is the salary value of $53,000, μ is the mean of $40,756, and σ is the standard deviation of $7500.

z = (53,000 - 40,756) / 7500 = 1.63547

Using a standard normal distribution table or calculator, we can find the proportion of employees who earn at most $53,000 by finding the area to the left of the standardized value of 1.63547:

P(Z ≤ 1.63547) = 0.9514

Therefore, approximately 95.14% of employees in entry-level positions at the company earn at most $53,000.