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

Explanation:

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

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

The formula for z -score :

`z=(x-\mu)/((\sigma)/(√(n)))`

a) Sample size = 1

For x= 63 in. ,

`z=(63-62.3)/((2.4)/(√(1)))=0.29`

The p-value =
`P(z<0.29)=`

`0.6140918\approx0.6141`

Thus, the probability is approximately = 0.6141

b) Sample size = 47

For x= 63 ,

`z=(63-62.3)/((2.4)/(√(47)))\approx2.0`

The p-value =
`P(z<2.0)`

`=0.9772498\approx0.9772`

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