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

Step-by-step 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=(\sigma)/(√(n))=(727)/(√(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<(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