Question

Assume that​ women's heights are normally distributed with a mean given by mu equals 62.3 in​, and a standard deviation given by sigma equals 2.4 in.​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in.​(b) If 47 women are randomly​ selected, find the probability that they have a mean height less than 63 in.

Answer 1

Answer: a)  0.6141

b) 0.9772

Step-by-step explanation:

Given : Mean : $\mu= 62.3\text{ in}$

Standard deviation : $\sigma = \text{2.4 in}$

The formula for z -score :

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

a) Sample size = 1

For x= 63 in. ,

$z=\dfrac{63-62.3}{\dfrac{2.4}{\sqrt{1}}}=0.29$

The p-value = $P(z$

$0.6140918\approx0.6141$

Thus, the probability is approximately = 0.6141

b)  Sample size = 47

For x= 63 ,

$z=\dfrac{63-62.3}{\dfrac{2.4}{\sqrt{47}}}\approx2.0$

The p-value = $P(z$

$=0.9772498\approx0.9772$

Thus , the probability that they have a mean height less than 63 in =0.9772.