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

Answer:

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

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$-2.327 = \frac{X - 129}{5.18}$

$X - 129 = -2.327*5.18$

$X = 116.94$

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