Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.5 in​, and a standard deviation given by sigma equals 2.5 in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in. ​(b) If 35 women are randomly​ selected, find the probability that they have a mean height less than 63 in. ​(​a) The probability is approximately nothing. ​(Round to four decimal places as​ needed.) ​(b) The probability is approximately nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Answer: a) The probability is approximately = 0.5793

b) The probability is approximately=0.8810

Step-by-step explanation:

Given : Mean : $\mu= 62.5\text{ in}$

Standard deviation : $\sigma = \text{2.5 in}$

a) The formula for z -score :

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Sample size = 1

For x= 63 in. ,

$z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{1}}}=0.2$

The p-value = $P(z$

$0.5792597\approx0.5793$

Thus, the probability is approximately = 0.5793

b)  Sample size = 35

For x= 63 ,

$z=\dfrac{63-62.5}{\dfrac{2.5}{\sqrt{35}}}\approx1.18$

The p-value = $P(z$

$= 0.8809999\approx0.8810$

Thus , the probability is approximately=0.8810.