The weekly earnings of students in one age group are normally distributed with a standard deviation of 88 dollars. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 98 percent confidence that the sample mean will not differ from the population mean by more than 2 dollars.

The sample size needed is of 10,484.

Step-by-step explanation:

We have 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 Z-table as such z has a p-value of .

That is z with a p-value of , so Z = 2.327.

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 weekly earnings of students in one age group are normally distributed with a standard deviation of 88 dollars.

This means that

Find the sample size needed to assure with 98 percent confidence that the sample mean will not differ from the population mean by more than 2 dollars.

This is n for which M = 2. So

Rounding up:

The sample size needed is of 10,484.

9 x 8 divided by (5-3)=36

the final amount you will pay is $68.04.

the tip is determined by the pre-tax amount, so we calculate the tip based on the original bill, not after we've added sales tax.

to find the tip:

we can divide 54.00 by 100 to get 1% of 54. then we multiply that by 18.

54 / 100 = .54

.54 x 18 = 9.72

tip = 9.72$

to find the tax:

we can divide 54.00 by 100 to get 1% of 54. then we can multiply that by 8 to get 8%.

54 / 100 = .54

.54 x 8 = 4.32

tax= 4.32

now we can add the tip and tax to the bill to get our total.

54 + 9.72 + 4.32 = 68.04

total: $68.04

step-by-step explanation: