Question

The mean per capita income is 20,90820,908 dollars per annum with a standard deviation of 407407 dollars per annum. What is the probability that the sample mean would differ from the true mean by less than 2828 dollars if a sample of 5555 persons is randomly selected? Round your answer to four decimal places.

Answer 1

Answer:

$z=\frac{20880-20908}{\frac{407}{\sqrt{55}}}= -0.51$

$z=\frac{20936-20908}{\frac{407}{\sqrt{55}}}= 0.51$

Then we can find the probability of interest with this difference:

$P(-0.51$

And using the normal standard distribution or excel we got:

$P(-0.51$

So then the probability  that the sample mean would differ from the true mean by less than 28 dollars from the sample of 55 is approximately 0.390

Step-by-step explanation:

We define the variable of interest as the per capita income and we know the following properties for this variable:

$\mu=20908$ and $\sigma=407$

We want to find this probability:

$P(20908-28$

We select a sample size of n=55 and we define the z score formula given by:

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

We can find the z score then for 20880 and 20936 and we got:

$z=\frac{20880-20908}{\frac{407}{\sqrt{55}}}= -0.51$

$z=\frac{20936-20908}{\frac{407}{\sqrt{55}}}= 0.51$

Then we can find the probability of interest with this difference:

$P(-0.51$

And using the normal standard distribution or excel we got:

$P(-0.51$

So then the probability  that the sample mean would differ from the true mean by less than 28 dollars from the sample of 55 is approximately 0.390