Question

The annual salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $20,000. What percent of people earn more than $75,000

Answer 1

Let x denotes the annual salaries of employees in a large company

Mean µ = 50,000 and S.D σ = 20,000

Standard normal variate z = `(x - mu)/sigma`

P(people earn more than$75,000) = P(X > 70000)

When x = 75,000

z = `(75,000 - 50, 000)/(20,000)`

= `25000/20000`

= `5/4`

z = 1.25

P(X > 75,000) = P(X > 1.25)

= P(0 < z < `oo`) – P(0 < z < 1.25)

= 0.5 – 0.3944

= 0.1056

P(x > 750,000) in percent

= 01056 × 100

= 10.56