A local electronics firm wants to determine their average daily sales (in dollars.) A sample of the sales for 36 days revealed an average sales of $139,000. Assume that the standard deviation of the population is known to be $12,000.

a) provide a 95% confidence interval estimate for average daily sales.
b) provide a 97% confidence interval estimate for average daily sales.

a) The 95% confidence interval estimate for average daily sales is between $135,080 and $142,920.

b) The 97% confidence interval estimate for average daily sales is between $134,660 and $143,340.

Step-by-step explanation:

a) provide a 95% confidence interval estimate for average daily sales.

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.

The lower end of the interval is the sample mean subtracted by M. So it is 139000 - 3920 = $135,080

The upper end of the interval is the sample mean added to M. So it is 139000 + 3920 = $142,920

The 95% confidence interval estimate for average daily sales is between $135,080 and $142,920.

b) provide a 97% confidence interval estimate for average daily sales.

By the same logic as above, now Z = 2.17.

The lower end of the interval is the sample mean subtracted by M. So it is 139000 - 4340 = $134,660

The upper end of the interval is the sample mean added to M. So it is 139000 + 4340 = $143,340

The 97% confidence interval estimate for average daily sales is between $134,660 and $143,340.

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.