Question

Personal best finishing times for a particular race in high school track meets are normally distributed with mean 5.27 minutes and standard deviation .42 minutes. If the qualifying time for this event for the regional championship is to be set so that only four percent of all runners qualify, then that qualifying time in minutes is about:

(a) 4.54
(b) 3.52
(c) 4.69
(d) 6.01
(e) 5.46

Answer 1

Answer:

(a) 4.54

Step-by-step 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:

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

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 5.27, \sigma = 0.42$

Fastest 4% qualify.

The smaller the time, the faster the runner. So the qualifying time is times at the 4th percentile or below. The 4th percentile is the value of X when Z has a pvalue of 0.04. So it is X when Z = -1.75.

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

$-1.75 = \frac{X - 5.27}{0.42}$

$X - 5.27 = -1.75*0.42$

$X = 4.54$

The qulifying time is 4.54 minutes, so the correct answer is:

(a) 4.54