Question

A physical fitness association is including the mile run in its secondary-school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 450 seconds and a standard deviation of 60 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 312 seconds. Group of answer choices

Answer 1

Answer:

0.0107 = 1.07% the probability that a randomly selected boy in secondary school can run the mile in less than 312 seconds.

Step-by-step 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:

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

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 question, we have that:

$\mu = 450, \sigma = 60$

Find the probability that a randomly selected boy in secondary school can run the mile in less than 312 seconds.

This is the pvalue of Z when X = 312. So

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

$Z = \frac{312 - 450}{60}$

$Z = -2.3$

$Z = -2.3$ has a pvalue of 0.0107

0.0107 = 1.07% the probability that a randomly selected boy in secondary school can run the mile in less than 312 seconds.