Question

women's heights are normally distributed with a mean 63.1 in and standard deviation of 2.5. A social organization for tall people has a requirement that women must be at least 69in tall. What percentage of woman meet that requirement?

The percentage of woman that are taller than 69in is __%

Answer 1

z = (69-63.1)/ 2.5
z = 5.9/2.5
z = 2.36

Answer 2

z = (69-63.1)/ 2.5
z = 5.9/2.5
z = 2.36

Once you have z-scores, check a z-table for the probability scores. Convert to a percentage.

Answer 3

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Answer 4

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.3in​, and a standard deviation given by sigma equals 2.7
​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 6363 in.
​(b) If 43 women are randomly​ selected, find the probability that they have a mean height less than 63 in.