Question

A researcher tests the braking distances of several cars. The braking distance from 60 miles per hour to a complete stop on dry pavement is measured in feet. The braking distances of a sample of cars are normally distributed, with a mean of 129 feet and a standard deviation of 5.18 feet. What is the longest braking distance one of these cars could have and still be in the bottom 1%

Answer 1

The longest braking distance one of these cars could have and still be in the bottom 1% is of 116.94 feet.

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.

The braking distances of a sample of cars are normally distributed, with a mean of 129 feet and a standard deviation of 5.18 feet.

This means that
`\mu = 129, \sigma = 5.18`

What is the longest braking distance one of these cars could have and still be in the bottom 1%?

This is the 1st percentile, which is X when Z has a pvalue of 0.01, so X when Z = -2.327.

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

`-2.327 = (X - 129)/(5.18)`

`X - 129 = -2.327*5.18`

`X = 116.94`

The longest braking distance one of these cars could have and still be in the bottom 1% is of 116.94 feet.