Standard deviation and z-scores

Question

Assume that the salaries of elementary teachers in the United States are approximately normally distributed with a mean of $32,000 and a standard deviation of $6,000. What is the approximate percentage of teachers who would earn salaries below $30,000? What percentage would earn between $30,000 and $38,000?

 

In respect for the academic honesty policy, I am really just looking for the formulas to solve this question. The first part of the question asked what percent would make less than $30,000. I found the z-score of -0.33 so I came up with 37.07%. I am hoping that is correct? 

Answer 1

That is correct! The formula for the z-score is (x-μ)/σ. Here, x=30,000 μ=32,000 and σ=6,000. Plugging this in we get (30,000-32,000)/6,000 = -.33333. Rounding to two decimals gives -.33. To find the probability of teachers earning less than 30,000 we can use a table to find the solution.

 

Keep in mind if you have a TI-83/84 calculator you can use the normcdf(lower, upper, mean, standard deviation) to come to a similar answer. This one will be more accurate. (Above, we rounded the -0.33333... to two decimals of -0.33. Try the next solution and I will check your answer!