Question

The lives of a premium sports car's brakes are normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. What is the probability that such brakes last between 54,000 and 66,000 miles? Group of answer choices

Answer 1

86.64% probability that such brakes last between 54,000 and 66,000 miles

Explanation:

When the distribution is normal, we use 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 question, we have that:

`\mu = 60000, \sigma = 4000`

What is the probability that such brakes last between 54,000 and 66,000 miles?

This is the pvalue of Z when X = 66000 subtracted by the pvalue of Z when X = 54000.

X = 66000

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

`Z = (66000 - 60000)/(4000)`

`Z = 1.5`

`Z = 1.5` has a pvalue of 0.9332

X = 54000

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

`Z = (54000 - 60000)/(4000)`

`Z = -1.5`

`Z = -1.5` has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664

86.64% probability that such brakes last between 54,000 and 66,000 miles