Sales personnel for skillings distributors submit weekly reports listing the customer contacts made during the week. a sample of 68 weekly reports showed a sample mean of 17.5 customer contacts per week. the sample standard deviation was 4.2. provide a 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. (round your answers to two decimal places.)

(16.65, 18.35)

Step-by-step explanation:

By considering the given information , we have

mm

s=  4.2 mm

n= 68

Significance level

Since population standard deviation is not given ,

Formula :

Confidence interval for population mean is given :-

, where n = sample size.

= Two-tailed t-value for significance level of and degree of freedom df= n-1.

= sample standard deviation.

Two-tailed t-value for significance level of and degree of freedom df= 67:

95% Confidence interval for population mean:

Hence, the 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. = (16.65, 18.35)

41*67/2=1373.5

step-by-step explanation:

i just used a calculator

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8