The probability that they have a mean height greater than 63.0 inches will be 0.9918
Explanation
The heights of adult Caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. That means, [tex]\mu = 63.6[/tex] and [tex]\sigma = 2.5[/tex]
As 100 women are randomly selected, so the sample size[tex](n)[/tex] = 100
The formula for finding the z-score is......
[tex]z= \frac{x-\mu}{(\frac{\sigma}{\sqrt{n}})}[/tex]
So, the z-score of a mean height for 63.0 inches will be......
[tex]z(x=63.0) = \frac{63.0-63.6}{(\frac{2.5}{\sqrt{100}})}=\frac{-0.6}{0.25}=-2.4[/tex]
According to the normal distribution table, [tex]P(z<-2.4)= 0.0082[/tex]
So, [tex]P(x>63.0)=P(z>-2.4)= 1-P(z<-2.4)= 1-0.0082=0.9918[/tex]
Thus, the probability that they have a mean height greater than 63.0 inches will be 0.9918
