Question

Assume that the heights of adult caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. if 100 women are randomly​ selected, find the probability that they have a mean height greater than 63.0 inches. round to four decimal places.

Answer 1

The probability that the mean height is greater than 63 in will be given by:
z-score
z=(x-μ)/σ
z=(63.6-63)/2.5
z=0.24
The probability will be:
P(X≥63)=1-(P=z)
=1-0.6331
=0.3669

Answer 2

The probability that they have a mean height greater than 63.0 inches will be 0.9918

Step-by-step explanation

The heights of adult Caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. That means,
`\mu = 63.6` and
`\sigma = 2.5`

As 100 women are randomly​ selected, so the sample size
`(n)` = 100

The formula for finding the z-score is......

`z= (x-\mu)/(((\sigma)/(√(n))))`

So, the z-score of a mean height for 63.0 inches will be......

`z(x=63.0) = (63.0-63.6)/(((2.5)/(√(100))))=(-0.6)/(0.25)=-2.4`

According to the normal distribution table,
`P(z<-2.4)= 0.0082`

So,
`P(x>63.0)=P(z>-2.4)= 1-P(z<-2.4)= 1-0.0082=0.9918`

Thus, the probability that they have a mean height greater than 63.0 inches will be 0.9918