Question

A new gas- and electric-powered hybrid car has recently hit the market. The distance traveled on 1 gallon of fuel is normally distributed with a mean of 65 miles and a standard deviation of 4 miles. Find the probability of the following events.

a. The car travels more than 70 miles per gallon.

b. The car travels less than 60 miles per gallon.

c. The car travels between 55 and 70 miles per gallon.

Answer 1

a) P [ Z > 70 ] = 0.1075 or 10.75 %

b) P [ Z < 60 ] = 0.1038 or 10.38 %

c) P [ 55 ≤ Z ≤ 70 ] = 0.8882 or 88.82 %

Explanation:

Normal Distribution μ = 65 and σ = 4

a) The probability of the car travels more than 70 miles per gallon is:

P [ Z > 70 ] = (Z-μ) ÷ σ ⇒ P [ Z > 70 ] = (70-65) ÷4 P [ Z > 70 ] = 1.25

the point 1.25 corresponds at values, from left tail up to 70 so we must go and look for the area for the point 1.24 which is 0.8925. Then we have the area or probability of all cars traveling up to 70 miles therefore we have to subtract 1 -0,8925

P [ Z > 70 ] = 0.1075 or 10.75 %

b)The probability of the car travels less than 60 miles per gallon is:

P [ Z < 60 ] = ( 60 - μ ) ÷ σ ⇒ P [ Z < 60 ] = (60-65)÷ 4 P [ Z < 60 ] = -1.25

Again -1.25 corresponds to 60 miles per gallon threfore we move to the left and find for point -1.26 which area is 0.1038 so

P [ Z < 60 ] = 0.1038 10.38 %

c) P [ 55 ≤ Z ≤ 70]

For point Z = 70 or 1.25 (case a above) P [ Z ≤ 70] = (70-65)÷4

P [ Z ≤ 70] = 1.25

In this case we got the whole area from the left tail up to 1.25

P [ Z ≤ 70] = 0.8944 (includes the area of the point from the left tail up to the point assocciated to 55 miles and for that reason we have to subtract that area)

P [ Z ≥ 55 ] = (55-65) ÷ 4 P [ Z ≥ 55 ] = -2.5 and the area is 0.0062

So P [ 55 ≤ Z ≤ 70 ] = 0.8944 - 0.0062 = 0.8882