Question

The average waist size for teenage males is 29 inches with a standard deviation of 1. 4 inch. If waist sizes are normally distributed, estimate the proportion of teenagers who will have waist sizes greater than 31 inches?.

Answer 1

Using the normal distribution, it is found that 0.0764 = 7.64% of teenagers who will have waist sizes greater than 31 inches.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• 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 $\mu = 29$.

• The standard deviation is of $\sigma = 1.4$.

The proportion of teenagers who will have waist sizes greater than 31 inches is 1 subtracted by the p-value of Z when X = 31, hence:

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

$Z = \frac{31 - 29}{1.4}$

$Z = 1.43$

$Z = 1.43$ 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