Question

Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of her race times is normally distributed. We are interested in one of her randomly selected laps.

Find the percent of her laps that are completed in less than 134 seconds. (Round your answer to two decimal places.)



The fastest 5% of her laps are under how many seconds? (Round your answer to two decimal places.)

sec

Answer 1

97% of her laps are completed in less than 134 seconds.

The fastest 5% of her laps are under 125.96 seconds.

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 = 129.71, \sigma = 2.28`

Find the percent of her laps that are completed in less than 134 seconds.

We have to find the pvalue of Z when X = 134. So

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

`Z = (134 - 129.71)/(2.28)`

`Z = 1.88`

`Z = 1.88` has a pvalue of 0.9699, so 97% of her laps are completed in less than 134 seconds.

The fastest 5% of her laps are under how many seconds?

This is the 5th percentile of times, which is X when Z has a pvalue of 0.05, that is, X when Z = -1.645. So

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

`-1.645 = (X - 129.71)/(2.28)`

`X - 129.71 = -1.645*2.28`

`X = 125.96`

The fastest 5% of her laps are under 125.96 seconds.