Answer to Question #148659 in Statistics and Probability for PYIONA

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?

Answer 1

$(a) \; P( 50000 < x < 65000) = P(\frac{50000 – 55000}{4500} < \frac{x – μ}{σ} < \frac{65000 – 55000}{4500}) \\ = P(-1.111 < z < 2.222) \\ = P(z < 2.222) - P(z < -1.111) \\ = 0.9868 - 0.1332 \\ = 0.8533$

$(b) \; P(x > 70000) = P(\frac{x – μ}{σ} > \frac{70000 – 55000}{4500}) \\ = P(z > 3.333) \\ = 1 - P(z ≤ 3.333) \\ = 1 - 0.99957\\ = 0.00043$