Answer to Question #147976 in Economics for Peeriya

Question

Suppose that the annual household income in a small Midwestern community is normally distributed with a mean of $55,000 and a standard deviation of $4,500.
a. What is the probability that a randomly selected household will have an income between $50,000 and $65,000?
b. What is the probability that a randomly selected household will have an income of more than $70,000?
c. What minimum income does a household need to earn to be in the top 5% of incomes?
d. What maximum income does a household need to earn to be in the bottom 40% of incomes?

Answer 1

a) $\frac{\Phi{\frac{65,000-55,000}{4,500}}-\Phi{\frac{50,000-55,000}{4,500}}}{2}=\frac{0.9736+0.733}{2}=0.8533$

the probability is 85.33%

b) $1-\Phi{\frac{70,000-55,000}{4,500}}=0.0009=0.09\%$

c) $1-\Phi{\frac{x-55,000}{4,500}}=0.05$

$\frac{x-55,000}{4,500}=1.96$

$x=63,820$ - minimum income does a household need to earn to be in the top 5% of incomes

d) $\Phi{\frac{0-55,000}{4,500}}-\Phi{\frac{x-55,000}{4,500}}=0.3969$

$\Phi{\frac{x-55,000}{4,500}}=0.3969+0.5=0.8969$

$\frac{x-55,000}{4,500}=1.63$

$x=47,665$ maximum income does a household need to earn to be in the bottom 40% of incomes