Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 33.4 and 3.4 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 34 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 five randomly selected passenger cars is more than 34 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.) c. If five passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 34 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

The distribution given here is:

a) The probability required here is:

P(X > 34)

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.4286 is the required probability here.

b) For 5 passengers the probability for sample mean is computed as:

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.3483 is the required probability here.

c) Probability that all of the 5 passengers selected get more than 34 mpg is computed as:

= 0.4286*0.4286*..... 5 times

= 0.4286⁵

= 0.0145

Therefore 0.0145 is the required probability here.