The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,119. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.

The smallest sample size needed is 49.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income.

This is n for which

So

Rounding up

The smallest sample size needed is 49.

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.