Respuesta :
Using the normal distribution, it is found that 0.1587 = 15.87% of men are taller than 6 feet (72 inches).
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Normal Probability Distribution
In a normal distribution with mean and standard deviation , 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:
- Mean of 69.1 inches, thus [tex]\mu = 69.1[/tex].
- Standard deviation of 2.9 inches, thus [tex]\sigma = 2.9[/tex].
- The proportion above 72 inches is 1 subtracted by the p-value of Z when X = 72, then:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{72 - 69.1}{2.9}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.8413.
1 - 0.8413 = 0.1587.
0.1587 = 15.87% of men are taller than 6 feet (72 inches).
A similar problem is given at https://brainly.com/question/22934264
