Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 32.6 and 4.9 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 36 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.) b. What is the probability that the average mpg of two randomly selected passenger cars is more than 36 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.) c. If two passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 36 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

µ = 32.6

sd = 4.9

a)

                             

                              = P(Z > 0.69)

                              = 1 - P(Z < 0.69)

                              = 1 - 0.7549

                              = 0.2451

b) n = 2

                         

                          = P(Z > 0.98)

                          = 1 - P(Z < 0.98)

                          = 1 - 0.8365

                          = 0.1635

c) P(all 2 passenger cars get more than 36 mpg) = 0.2451² = 0.0601