Question

The mean per capita income is $21,604 per annum with the standard deviation of $727 per annum. What is the probability that the sample mean will be less than $21,635 if a sample of 193 persons is randomly selected? Round your answer to four decimal places

Answer 1

Explanation

In the question, we are given that

$\begin{gathered} \mu=21,604 \\ \sigma= 727 \\ n=193 \end{gathered}$

First, we will get the standard deviation of the sample mean as

$\sigma_x=\frac{\sigma}{\sqrt{n}}=\frac{727}{\sqrt{193}}=52.3306$

Then, we can find the probability that the sample mean will be less than $21,635 for a sample of 193 persons

$\begin{gathered} P(\bar{X}<21635)=P(z<\frac{\bar{X}-\mu}{\sigma_x})=P(z<\frac{21635-21604}{52.3306}) \\ =P(z<0.59238) \end{gathered}$

Therefore, using the z score calculator

$P(z<0.59238)=0.7232$

Answer: 0.7232