Answer to Question #181321 in Statistics and Probability for Thanusha

Question

Suppose that the annual income of a fast-food chain is normally distributed with a mean of $72,000 and a standard deviation of $9,000.

a. What is the probability that a randomly selected fast-food restaurant will generate an annual income between $65,000 and $75,000?

b. What is the probability that a randomly selected fast-food restaurant will generate an annual profit of more than $80,000?

c. What minimum income does a fast-fast restaurant need to earn to be in the top 5% of incomes?

d. What maximum income does a fast-food restaurant need to earn to be in the bottom 30% of incomes?

Answer 1

mean $= \mu = 72000$

standard deviation $= \sigma = 9000$

a.

$P(65000<X<75000) = P(\frac{65000-7200}{9000}<Z< \frac{75000-72000}{9000}) \\ = P(-0.777<Z<0.333) \\ = P(Z<0.333) -P(Z< -0.777) \\ = 0.6304 -0.2185 \\ = 0.4119$

b.

$P(X>80000) = 1 -P(X<80000) \\ = 1 - P(Z< \frac{80000-72000}{9000}) \\ = 1 -P(Z<0.888) \\ = 1 -0.8127 \\ = 0.1873$

c.

$P(Z> \frac{x-72000}{9000}) = 0.05 \\ P(Z>1.645) = 0.05 \\ \frac{x-72000}{9000} = 1.645 \\ x -72000 = 14805 \\ x = 86805$

d.

$P(Z< \frac{x-72000}{9000}) = 0.3 \\ P(Z< -0.524) = 0.3 \\ \frac{x-72000}{9000} = -0.524 \\ x -72000 = -4716 \\ x = 67284$