Using the normal distribution, it is found that 0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- It 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, which is the percentile of X.
In this problem:
- The mean is of
.
- The standard deviation is of
.
The proportion of teenagers who will have waist sizes greater than 31 inches is <u>1 subtracted by the p-value of Z when X = 31</u>, hence:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{31 - 29}{1.4}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B31%20-%2029%7D%7B1.4%7D)
![Z = 1.43](https://tex.z-dn.net/?f=Z%20%3D%201.43)
has a p-value of 0.9236.
1 - 0.9236 = 0.0764.
0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.
More can be learned about the normal distribution at brainly.com/question/24663213
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