Question

PLEASE HELP! The AHS football team did a weigh-in at the start of training camp. The weights of the players were distributed normally with a mean of 98kg and a standard deviation of 6kg. What percentage of players are under 86kg?

Answer 1

2.28% of players are under 86kg

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 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.

The weights of the players were distributed normally with a mean of 98kg and a standard deviation of 6kg.

This means that
`\mu = 98, \sigma = 6`

What percentage of players are under 86kg?

The proportion is the pvalue of Z when X = 86. So

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

`Z = (86 - 98)/(6)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

0.0228*100% = 2.28%

2.28% of players are under 86kg