Question #147976
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?
1
Expert's answer
2020-12-03T10:51:18-0500

a) Φ65,00055,0004,500Φ50,00055,0004,5002=0.9736+0.7332=0.8533\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) 1Φ70,00055,0004,500=0.0009=0.09%1-\Phi{\frac{70,000-55,000}{4,500}}=0.0009=0.09\%

c) 1Φx55,0004,500=0.051-\Phi{\frac{x-55,000}{4,500}}=0.05

x55,0004,500=1.96\frac{x-55,000}{4,500}=1.96

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

d) Φ055,0004,500Φx55,0004,500=0.3969\Phi{\frac{0-55,000}{4,500}}-\Phi{\frac{x-55,000}{4,500}}=0.3969

Φx55,0004,500=0.3969+0.5=0.8969\Phi{\frac{x-55,000}{4,500}}=0.3969+0.5=0.8969

x55,0004,500=1.63\frac{x-55,000}{4,500}=1.63

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


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