Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 35.9 and 2.5 mpg, respectively. [You may find it useful to reference the z table.]

a. What is the probability that a randomly selected passenger car gets more than 37 mpg?

b. What is the probability that the average mpg of three randomly selected passenger cars is more than 37 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

c. If three passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 37 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

µ = 35.9

sd = 2.5

a)

                             

                              = P(Z > 0.44)

                              = 1 - P(Z < 0.44)

                              = 1 - 0.67

                              = 0.33

b) n = 3

                         

                          = P(Z > 0.76)

                          = 1 - P(Z < 0.76)

                          = 1 - 0.7764

                          = 0.2236

c) P(all passenger cars get more than 37 mpg) = 0.33³ = 0.0359