Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 36.5 and 4.4 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 38 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 three randomly selected passenger cars is more than 38 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 38 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

Here we have

(a)

The z-score for X = 38 is

The probability that a randomly selected passenger car gets more than 38 mpg is

P(X > 38) = P(z > 0.34) = 1- P(z <= 0.34) = 0.3669

Answer: 0.3669

(b)

The z-score for is

The probability that the average mpg of three randomly selected passenger cars is more than 38 mpg is

Answer: 0.2776

(c)

Since all passenger are independent from other so the probability that all of the passenger cars get more than 38 mpg is

0.3669 *0.3669* 0.3669 = 0.049390467309

Answer: 0.0494