Normal Distribution

Question

An automobile manufacturer introduces a new model that averages 27 miles per gallon in the city. A person who plans to purchase one of these new cars wrote the manufacturer for the details of the tests, and found out that the standard deviation is 3 miles per gallon. Assume that in-city mileage is approximately normally distributed.

 

What is the probability that the person will purchase a car that averages less than 20 miles per gallon for in-city driving?

 

What is the probability that the person will purchase a car that averages between 25 and 29 miles per gallon for in-city driving?

Answer 1

Hi James,

 

Since we have a normal distribution, 

         X - μ

Z = --------

           σ

 

μ is the average (27 mpg) and σ is the standard deviation (3)

 

Prob(mileage < 20)

Z = (20-27)/3 = -2.3333

Look up this value in a z table

value = 0.0098

 

So the probability that the mileage is less than 20 mpg is .98%

 

For the second question we need to find two probabilities

<29 and <25 and take the difference

 

For Prob(mileage<29)

Z = (29-27)/3 = 0.667
Look up this value in a z table
value = 0.71468

 

For Prob(mileage<25)
Z = (25-27)/3 = -0.667
Look up this value in a z table
value = 0.25364

 

.71468 - .25364 = 0.46104 ≈ 46.1%

Probability that actual mileage is between 25 and 27 mpg is 46.1%