Question

According to the Bureau of Labor Statistics (BLS), the mean weekly earnings of salary workers in the United States in 2018 was

µ =$870. Assume that the weekly earnings are approximately normally distributed with a standard deviation of σ = $90.

1. If a salary worker is selected at random, what is the probability that his/her weekly earnings exceed $700?
2. What are the weekly earnings for which 95% of salary workers’ earnings are in this range?
3. What is the weekly earning of the top 10% of salary workers in the United States?
4. What is weekly earning of the bottom 10% of salary workers in the United States?

Answer 1

Let X denote the weekly earnings of salary workers in the United States in 2018

We are given :

a.

Probability that his/her weekly earnings exceed $700 = P(X>700)

We use Excel function "NORMSDIST()" to find the above probability as :

Probability that his/her weekly earnings exceed $700 = 0.971

b.

According to empirical rule,

95% of the data lies within 2 standard deviations

Hence,

95% of salary workers’ earnings are in between $690 and $1050

c.

Let P₉₀ denote the weekly earning of the top 10% of salary workers in the United States

We have,

P(X<P₉₀) = 0.90

We use Excel function "NORMSINV()" :

So,

Weekly earning of the top 10% of salary workers in the United States = $ 985.34

d.

Let P₁₀ denote the weekly earning of the bottom 10% of salary workers in the United States

We have,

P(X<P₁₀) = 0.10

We use Excel function "NORMSINV()" :

So,

Weekly earning of the bottom 10% of salary workers in the United States = $ 754.66