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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:
- 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:
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