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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?.

Sagot :

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • 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 [tex]\mu = 29[/tex].
  • The standard deviation is of [tex]\sigma = 1.4[/tex].

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:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{31 - 29}{1.4}[/tex]

[tex]Z = 1.43[/tex]

[tex]Z = 1.43[/tex] 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 https://brainly.com/question/24663213