Answer to Question #152039 in Statistics and Probability for JG

Question

Based on a study conducted on residents in the United States, we found their salaries follow the normally distribution with a mean of $48600, and a standard deviation of $14000.
a) If we randomly select 50 residents from the United States, approximately how many of them will have salaries above $70000?
b) If we randomly select 10 residents from the United States, what is the probability that their average salary is between $45000 to $50000?
c) Knowing Jackie’s salary is 17% below the third quartile, what is her actual salary?

Answer 1

a)

$P(X > 70000) = P(Z > \frac{70000-48600}{14000}) \\ = P(Z> 1.53) \\ = 0.063$

Therefore Expected number of people with salaries above $70000:

$= 0.063 \times 50 = 3.15$

b)

$P(45000< \bar{X} < 50000) = P(\frac{45000-48600}{\frac{14000}{\sqrt{10}}} <Z< \frac{50000-48600}{\frac{14000}{\sqrt{10}}}) \\ = P(-0.81 < Z < 0.32) \\ = P(Z<0.32)-P(Z <-0.81) \\ = 0.6255 -0.2090 \\ = 0.4165$

c) From standard normal tables:

$P(Z < 0.6745) = 0.75$

The third quartile salary:

$= Mean + 0.6745 \times σ \\ = 48600 + 0.6745 \times 14000 \\ = 58043$

Jackie's salary $= 0.83 \times 58043 = 48175.69$