Question

A certain car model has a mean gas mileage of 31 miles per gallon (mpg) with astandard deviation 3 mpg. A pizza delivery company buys 43 of these cars. What is theprobability that the average mileage of the fleet is greater than 30.7 mpg?

Answer 1

Answer:

Probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.

Step-by-step explanation:

We are given that a certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg.

A pizza delivery company buys 43 of these cars.

Let $\bar X$ = sample average mileage of the fleet

The z-score probability distribution of sample average is given by;

                   Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = mean gas mileage = 31 miles per gallon (mpg)

           $\sigma$ = standard deviation = 3 mpg

           n = sample of cars = 43

So, probability that the average mileage of the fleet is greater than 30.7 mpg is given by = P($\bar X$ > 30.7 mpg)

     P($\bar X$ > 30.7 mpg) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{30.7 - 31}{\frac{3}{\sqrt{43} } }$ ) = P(Z > -0.66) = P(Z < 0.66)

                                                                 = 0.7454

Because in z table area of P(Z > -x) is same as area of P(Z < x). Also, the above probability is calculated using z table by looking at value of x = 0.66 in the z table which have an area of 0.7454.

Therefore, probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.