Question

Assume that women's heights are normally distributed with a mean 64.5 in, and a standard deviation of 2.3 in.

a. if 1 woman is randomly selected, find the probability that her height is less than 65 in.

b. if 46 women are randomly selected, find the probability that they have a mean height less than 65 in.

Answer 1

Solution:

Given, Population mean, $\mu=64.5\ in \quad$

And Population standard deviation, $\sigma=2.3\ in \quad$

(a) To find: $P(x<65\ in )=?$

The z -score is :

$\begin{array}{l} z_{0} \quad=\frac{x_{0}-\mu}{\sigma} \\ \ \ \ \ \ \ \ \ \ =\frac{65-64.5}{2.3} =\frac{0.5}{2.3} =0.22 \end{array}$

$\Rightarrow P(x<65\ in )=P(z<0.22)$

Finally, the required probability is

$\begin{array}{l} P(x<65\ { in })=0.5871 =58.71 \% \end{array}$ [Using tables]

(b): Now, we must consider that we have a sample with size greater than one:

Given Population mean, $\mu=64.5\ in \quad$

And Population standard deviation, $\sigma=2.3\ in \quad$

Sample size, n=46

To find$: P(\bar{x}<65\ in )=?$

The z -score is :

$\begin{aligned} z_{0} &=\frac{\bar{x}_{2}-\mu}{\sigma / \sqrt{n}} \\ &=\frac{65-64.5}{2.3 / \sqrt{46}} \\ &=1.47 \end{aligned}$

$\Rightarrow P(\bar{x}<65\ in )=P(z<1.47)$

So, the required probability is 0.9292 [Using tables]