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

Solution:

Let X be a random variable which represents the mileage of a certain car model

Given that, X ~ N(31, 3²)

i.e.  μ = 31 mpg and σ = 3 mpg

We have to obtain the probability that average mileage of 43 of these cars is greater than 30.7 mpg.

i.e. We have to obtain P(x̄ > 30.7 mpg).

We know that if X ~ N(μ, σ^{​​​​​​2}) then x̄ ~ N(μ ,σ^{​​​​​​2}/n)

And if x̄ ~ N(μ, σ^{​​​​​​2}/n) then,

(Where, x̄ is sample mean and n is sample size.)

We have, μ = 31 mpg, σ = 3 mpg and n = 43

Using "pnorm" function of R we get, P(Z > -0.6557) = 0.7440

Hence, the probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7440.

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