Answer to Question #145130 in Statistics and Probability for Julie

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.
a) What percent of people earn less than $40,000?
b) What percent of people earn between $45,000 and $65,000?
c) What percent of people earn more than $70,000?

Answer 1

a.

$\mu=50,000$

$\sigma=20000$

$P(x<40000)=P(z<\frac{40000-50000}{20000})$

$=P(z<-0.5)=0.3085$ from z tables.

Thus, 30.85% of people earn below $40,000

b.

P

P(45000<x<65000)

$P(45000<x<65000)=P(\frac{45000-50000}{20000}<z<\frac{65000-50000}{20000})$

$=P(-0.25<z<0.75)$

$=P(z<0.75)-P(z<-0.25)$

$=0.7734-0.4013$ from z tables

$=0.3721$

37.21% of people earn between $45000 and $65000

c.

P(x>70000)

$P(x>70000)=1-P(z<\frac{70000-50000}{20000}$

$=1-P(z<1)$

=1-0.8413 from z tables

=0.1587

15.87% of the people earn more than $70,000.