The customer service department of a phone company is experimenting with two different systems. On Monday they try the first system which is based on an automated menu system. On Tuesday they try the second system in which each caller is immediately connected with a live agent. A quality control manager selects a sample of seven calls each day. He records the time for each customer to have his or her question answered. The times (in minutes) are listed below. Automated Menu: 11.7 7.4 3.9 2.9 9.2 6.3 5.5 Live agent: 6.2 2.9 4.4 4.1 3.4 5.2 3.27
1. Find the sample mean and sample standard deviation for "Automated Menu"
2. Find the sample mean and sample standard deviation for "Live agent"
3. Find the Coefficient of variation for "Automated Menu" and for "Live agent"
4. Which system has a larger variability between the two.

4) As we can see form the results of part 3 we have more variation for the Automated menu since the variation coefficient for this case of almost two times the live agent Cv.

Step-by-step explanation:

For this case we have the following data:

Automated Menu: 11.7 7.4 3.9 2.9 9.2 6.3 5.5

Live agent: 6.2 2.9 4.4 4.1 3.4 5.2 3.27

In general we can calculate the mean of a sample with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

The standard deviation with this formula:

[tex] s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]

And the coeffcieint of variation for a sample with this formula:

[tex]\hat{Cv}= \frac{s}{\bar X} *100[/tex]

Part 1

Using the formula for Automated menu we have:

[tex] \bar X= 6.7[/tex]

[tex] s= 3.045[/tex]

Part 2

Using the formula for Live Agent we have:

[tex] \bar X= 4.21[/tex]

[tex] s= 1.172[/tex]

Part 3

Automated Menu

[tex] \hat {Cv}= \frac{3.045}{6.7}*100 = 45.44 \%[/tex]

Live Agent

[tex] \hat {Cv}= \frac{1.172}{4.21}*100 = 27.84 \%[/tex]

Part 4

As we can see form the results of part 3 we have more variation for the Automated menu since the variation coefficient for this case of almost two times the live agent Cv.