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 <em>normal distribution</em> with mean and standard deviation
, the z-score of a measure X is given by:
- 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 .
The proportion of US women that have a height greater than 69.5 inches is <u>1 subtracted by the p-value of Z when X = 69.5</u>, hence:
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 brainly.com/question/24663213