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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
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