Answer: a) 0.6141
b) 0.9772
Step-by-step explanation:
Given : Mean : [tex]\mu= 62.3\text{ in}[/tex]
Standard deviation : [tex]\sigma = \text{2.4 in}[/tex]
The formula for z -score :
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
a) Sample size = 1
For x= 63 in. ,
[tex]z=\dfrac{63-62.3}{\dfrac{2.4}{\sqrt{1}}}=0.29[/tex]
The p-value = [tex]P(z<0.29)=[/tex]
[tex]0.6140918\approx0.6141[/tex]
Thus, the probability is approximately = 0.6141
b) Sample size = 47
For x= 63 ,
[tex]z=\dfrac{63-62.3}{\dfrac{2.4}{\sqrt{47}}}\approx2.0[/tex]
The p-value = [tex]P(z<2.0)[/tex]
[tex]=0.9772498\approx0.9772[/tex]
Thus , the probability that they have a mean height less than 63 in =0.9772.
