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 440 seconds and a standard deviation of 40 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.A) 0.0107 B) 0.9893 C) 0.5107 D) 0.4893

Answer 1

A) 0.0107

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.

Mean of 440 seconds and a standard deviation of 40 seconds.

This means that
`\mu = 440, \sigma = 40`

Find the probability that a randomly selected boy in secondary school can run the mile in less than 348 seconds.

This is the p-value of Z when X = 348. So

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

`Z = (348 - 440)/(40)`

`Z = -2.3`

`Z = -2.3` has a p-value of 0.0107, and thus, the correct answer is given by option A.