Question

In a certain country, the average age is 31 years old and the standard deviation is 4 years. If we select a simple random sample of 100 people from this country, what is the probability that the average age of our sample is at least 32?

Answer 1

Answer: 0.0062

Step-by-step explanation:

Given : Mean : $\mu=\ 31$

Standard deviation :$\sigma= 4$

Sample size : $n=100$

Assume that age of people in the country is normally distributed.

The formula to calculate the z-score :-

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

For x = 32

$z=\dfrac{32-31}{\dfrac{4}{\sqrt{100}}}=5$

The p-value = $P(x\geq32)=P(z\geq5)$

$=1-P(z$

Hence, the the probability that the average age of our sample is at least =0.0062