Question

The mean per capita income is 15,451 dollars per annum with a variance of 298,116. What is the probability that the sample mean would differ from the true mean by less than 22 dollars if a sample of 350 persons is randomly selected? Round your answer to four decimal places.

Answer 1

Answer: 0.5467

Step-by-step explanation:

Let X be the random variable that represents the income (in dollars) of a randomly selected person.

Given : $\mu=15451$

$\sigma^2=298116\\\\\Rightarrow\ \sigma=\sqrt{298116}=546$

Sample size : n=350

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

To find the probability that the sample mean would differ from the true mean by less than 22 dollars, the interval will be

$\mu-22,\ \mu+22\\\\=15,451 -22,\ 15,451 +22\\\\=15429,\ 15473$

For x=15429

$z=\dfrac{15429-15451}{\dfrac{546}{\sqrt{350}}}\approx-0.75$

For x=15473

$z=\dfrac{15473-15451}{\dfrac{546}{\sqrt{350}}}\approx0.75$

The required probability :-

$P(15429$

Hence, the required probability is 0.5467.