Answer: 0.0062
Step-by-step explanation:
Given : Mean : [tex]\mu=\ 31[/tex]
Standard deviation :[tex]\sigma= 4[/tex]
Sample size : [tex]n=100[/tex]
Assume that age of people in the country is normally distributed.
The formula to calculate the z-score :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x = 32
[tex]z=\dfrac{32-31}{\dfrac{4}{\sqrt{100}}}=5[/tex]
The p-value = [tex]P(x\geq32)=P(z\geq5)[/tex]
[tex]=1-P(z<5)=1- 0.9937903\approx0.0062[/tex]
Hence, the the probability that the average age of our sample is at least =0.0062
