Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 30.9 and 2.7 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 32 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 four randomly selected passenger cars is more than 32 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

c. If four passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 32 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

Solution :

Given that ,

a.

P(x > 32) = 1 - P(x < 32)

= 1 - P[(x - ) / < (32 - 30.9) / 2.7]

= 1 - P(z < 0.38)

= 1 - 0.648

= 0.3520

Probability = 0.3520

b.

= / n = 2.7/ 4 = 1.35

P( > 32) = 1 - P( < 32)

= 1 - P[( - ) / < (32 - 30.9) / 1.35]

= 1 - P(z < 0.81)

= 1 - 0.791

= 0.2090

Probability = 0.2090

c.

= / n = 2.7/ 4 = 1.35

P( > 32) = 1 - P( < 32)

= 1 - P[( - ) / < (32 - 30.9) / 1.35]

= 1 - P(z < 0.81)

= 1 - 0.791

= 0.2090

Probability = 0.2090