Answer to Question #183232 in Statistics and Probability for nur farhanah

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? (2 Marks) b. What is the probability that a randomly selected fast-food restaurant will generate an annual profit of more than $80,000? (2 Marks) c. What minimum income does a fast-fast restaurant need to earn to be in the top 5% of incomes? (2 Marks) 3 d. What maximum income does a fast-food restaurant need to earn to be in the bottom 30% of incomes? (2 Marks)

Answer 1

Let x denote the random variable for annual income.

X~N(72000,9000²)

a)P(65000<X<75000)

$z_1=\frac{65000-72000}{9000}$ =-0.78

$z_2=\frac{75000-72000}{9000}$ =0.33

P(-0.78<z<0.33) from the z tables

=0.4116

b)P(X>80000)

$z=\frac{80000-72000}{9000}$ =0.89

p(z>0.89) from the z table

=0.1867

c) To be in the top 5% the ; we need the value; P(X<x)=0.95

the corresponding z value is; $\phi ^{-1}(0.95)$ =1.6449

$1.6449=\frac{X-72000}{9000}$

X=$86804.1

d) P(X<x)=0.30

$\phi^{-1}(0.3)=-0.5244$

$-0.5244=\frac{X-72000}{9000}$

X=$67280.4