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 between $45,000 and $65,000?

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 ear between $45,000 and $65,000)

P(45000 < x < 65000)

When x = 45,000

z = `(45, 000 - 50, 000)/(20, 000)`

= `(-5000)/(20, 000)`

= `(-1)/4`

z = – 0.25

When x = 65,000

z = `(5, 000 - 50, 000)/(20,000)`

= `(15000)/(20, 000)`

= `3/4`

z = 0.75

P(45000 < x < 65000) = P(– 0.25 < z < 0.75)

= P(– 0.25 < z < 0) + P(0 < z < 0.75)

= p(0 < z < 0.25) + P(0 < z < 0.75)

= P(0 < z < 0.25) + P(0 < z < 0.75)

= 0.0987 + 0.2734

= 0.3721

P(45000 < x < 65000) in percentage

= 0.3721 × 100

= 37.21