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 [tex]\bar X[/tex] = sample average mileage of the fleet
The z-score probability distribution of sample average is given by;
Z = [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean gas mileage = 31 miles per gallon (mpg)
[tex]\sigma[/tex] = 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([tex]\bar X[/tex] > 30.7 mpg)
P([tex]\bar X[/tex] > 30.7 mpg) = P( [tex]\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{30.7 - 31}{\frac{3}{\sqrt{43} } }[/tex] ) = 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.
