Question

5. In one year, there were 125 major new automobile types available. If you look at the miles per gallon (MPG) for these vehicles, the distributions are roughly normal. The means and standard deviations are given by:

Mean Standard Deviation

City MPG 22.37 4.77

Highway MPG 29.09 5.46


a. The Honda Civic had a rating of 45 highway MPG. What is the probability that your car has a worse highway gas mileage than the Civic?



b. The Lincoln Continental had a rating of 17 city MPG. What percent of cars are likely to have a better city gas mileage than the Continental?

Answer 1

a. 0.0018 = 0.18% probability that your car has a worse highway gas mileage than the Civic

b. 89.06% of cars are likely to have a better city gas mileage than the Continental

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. The Honda Civic had a rating of 45 highway MPG. What is the probability that your car has a worse highway gas mileage than the Civic?

For Highway, we have that
`\mu = 29.09, \sigma = 5.46`

This probability is the pvalue of Z when X = 45. So

`Z = (X - \mu)/(\sigma)`

`Z = (45 - 29.09)/(5.46)`

`Z = 2.91`

`Z = 2.91` has a pvalue of 0.9982

1 - 0.9982 = 0.0018

0.0018 = 0.18% probability that your car has a worse highway gas mileage than the Civic.

b. The Lincoln Continental had a rating of 17 city MPG. What percent of cars are likely to have a better city gas mileage than the Continental?

City means that
`\mu = 22.37, \sigma = 4.37`

This probability is 1 subtracted by the pvalue of Z when X = 17. So

`Z = (X - \mu)/(\sigma)`

`Z = (17 - 22.37)/(4.37)`

`Z = -1.23`

`Z = -1.23` has a pvalue of 0.1094

1 - 0.1094 = 0.8906

0.8906 = 89.06% of cars are likely to have a better city gas mileage than the Continental