Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is a normally distributed random variable with a mean and standard deviation of 33.8 and 3.5 mpg, respectively.

a. What is the probability that a randomly selected passenger car gets more than 35mpg?

b. a random sample of twenty-five passenger cars is selected. Denote Xbar as the sample average mpg of this twenty-five. What is the mean and standard deviation of Xbar?

c. What is the probability that the average mpg of twenty-five randomly selected passenger cars is more than 35 mpg?

Answer 1

µ = 33.8

sd = 3.5

a)

                             

                              = P(Z > 0.34)

                              = 1 - P(Z < 0.34)

                              = 1 - 0.6331

                              = 0.3669

b) n = 25

Mean of samping distribution = µ = 33.8

Standard deviation of sampling distribution = sd / sqrt(n) = 3.5 / sqrt(25) = 0.7

                         

                          = P(Z > 1.71)

                          = 1 - P(Z < 1.71)

                          = 1 - 0.9564

                          = 0.0436