Answer:
A. 0.9918
Step-by-step explanation:
Mean Height = u = 63.6 inches
Standard deviation = [tex]\sigma[/tex] = 2.5 inches
Sample size = 100
We have to find the probability that the mean height of sample would be greater than 63 i.e. P(x > 63)
We can do this by converting this value to z-score.
The formula to find z-score given the mean, standard deviation and sample size is:
[tex]z=\frac{x-u}{\frac{\sigma}{\sqrt{n}}}[/tex]
Using the values, we get:
[tex]z=\frac{63-63.6}{\frac{2.5}{\sqrt{100} } }\\\\ z=-2.4[/tex]
So now we have to find the probability that the z score would be greater than -2.4
i.e. P(z > -2.4)
From the z-table this value comes out to be:
P(z > -2.4) = 0.9918
Since,
P ( x > 63.0) is equivalent to P(z > -2.4), we can say:
P( x > 63.0 ) = 0.9918
Thus, the probability that adult Caucasian women have a mean height greater than 63.0 inches is 0.9918
