Respuesta :
Using the Normal distribution, it is found that 0.0359 = 3.59% of US women have a height greater than 69.5 inches.
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.
US women’s heights are normally distributed with mean 65 inches and standard deviation 2.5 inches, hence [tex]\mu = 65, \sigma = 2.5[/tex].
The proportion of US women that have a height greater than 69.5 inches is 1 subtracted by the p-value of Z when X = 69.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{69.5 - 65}{2.5}[/tex]
[tex]Z = 1.8[/tex]
[tex]Z = 1.8[/tex] has a p-value of 0.9641.
1 - 0.9641 = 0.0359
0.0359 = 3.59% of US women have a height greater than 69.5 inches.
You can learn more about the Normal distribution at https://brainly.com/question/24663213
