A machine used to fill beverage cans is supposed to put exactly 12 ounces of beverage in each can, but the actual amount varies. The standard deviation of fill volumes is 0.05 ounce. How large a sample of cans must be selected to estimate the population mean fill volume with a 95% confidence interval that has a margin of error of 0.008 ounce? Round your answer UP to the nearest integer.

A sample of 151 cans must be selected.

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 .

That is z with a pvalue of , so Z = 1.96.

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.

The standard deviation of fill volumes is 0.05 ounce.

This means that

How large a sample of cans must be selected to estimate the population mean fill volume with a 95% confidence interval that has a margin of error of 0.008 ounce?

We need a sample of n.

n is found when M = 0.008. So

Rounding up:

A sample of 151 cans must be selected.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)