Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 4.2. Required:
Provide 90%90% and 95%95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Since the Confidence is 0.90 or 90%, the value of and , and the critical value would be

And replacing we got

For the 95% confidence the critical value is

Step-by-step explanation:

Information given

represent the sample mean

population mean (variable of interest)

s¿4.2 represent the sample standard deviation

n¿65 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

  (1)

The degrees of freedom are given by:

Since the Confidence is 0.90 or 90%, the value of and , and the critical value would be

And replacing we got

For the 95% confidence the critical value is

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)