Question

Fuel economy of compact cars is normally distributed with a mean of 32.2 milesper gallon and a standard deviation of 3.7 miles per gallon. The probability that arandomly selected compact car gets at least 40 miles per gallon is about?

Answer 1

The probability that a randomly selected compact car gets at least 40 miles per gallon is about 1.74%

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 32.2, \sigma = 3.7`

The probability that a randomly selected compact car gets at least 40 miles per gallon is about?

This is 1 subtracted by the pvalue of Z when X = 40. So

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

`Z = (40 - 32.2)/(3.7)`

`Z = 2.11`

`Z = 2.11` has a pvalue of 0.9826

1 - 0.9826 = 0.0174

The probability that a randomly selected compact car gets at least 40 miles per gallon is about 1.74%