What is the probability that a randomly selected hybrid gets more than 60 miles per gallon? b) What is the probability that a randomly selected hybrid gets 53?

Question

The mean gas mileage for a hybrid car is 56 miles per gallon.  Suppose that the gasoline mileage is approximately normally distributed with a standard deviation of 3.3 miles per gallon.  a) what is the probability that a randomly selected hybrid gets more than 60 miles per gallon?  b) What is the probability that a randomly selected hybrid gets 53 miles per gallon or less?  c) What is the probability that a randomly selected hybrid gets between 58 and 61 miles per gallon?  d) What is the probability that a randomly selected hybrid gets less than 46 miles per gallon? 

Answer 1

Obtain a z-score chart

http://www.utdallas.edu/dept/abp/zscoretable.pdf

 

z = (X - μ) / σ, where X = score, μ = mean, σ = standard deviation.

 

For 60 mpg

z = (60 - 56) / 3.3

z = 4 / 3.3 

z = 1.212

For z = 1.212 the area below is 0.88726 or 88.726%

 

Do the same for 53 mpg and 46 mpg

 

For c) find the area below 61 and subtract the area below 58.