Question

Suppose the monthly earnings for all people who possess a bachelor’s degree is known to be normally distributed with a standard deviation of ? = $500. A person who earns more than $2300 per month earns more than 69% of earners. What is the mean ? of the distribution?

Answer 1

Solution :

x = 2300

standard deviation = = 500

Using standard normal table,

P(Z > z) = 69%

1 - P(Z < z) = 0.69

P(Z < z) = 1 - 0.69 = 0.31

P(Z < 2.326) = 0.31

z = -0.50

Using z-score formula,

= z * + x

= - 0.50 * 500 + 2300

= 2050

the mean    = 2050