Question

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. find the coefficient of variation for each of the two sets of data then compare the variation. round results to one decimal place.
automated menu: 5.5 7.4 3.9 2.9 9.2 6.3 11.7
live agent: 6.2 2.9 4.4 4.1 3.4 5.2 3.7
a. automated menu: 47.1%
live agent: 27.3%
there is substantially more variation in the times for the automated menu system
b. automated menu: 45.4%
live agent: 26.3%
there is substantially more variation in the times for the automated menu system
c. automated menu: 25.2%
live agent: 41.5%
there is substantially more variation in the times for the live agent.
d. automated menu: 48.8%
live agent: 28.3%
there is substantially more variation in the times for the automated menu system

Explanation:

Step1: Calculate mean and standard - deviation for Automated Menu

Let \(x_1 = 5.5, x_2=7.4,x_3 = 3.9,x_4=2.9,x_5 = 9.2,x_6=6.3,x_7 = 11.7\)
The mean \(\bar{x}_1=\frac{\sum_{i = 1}^{7}x_i}{7}=\frac{5.5 + 7.4+3.9+2.9+9.2+6.3+11.7}{7}=\frac{46.9}{7}=6.7\)
The standard - deviation \(s_1=\sqrt{\frac{\sum_{i = 1}^{7}(x_i-\bar{x}_1)^2}{7 - 1}}\)
\((x_1-\bar{x}_1)^2=(5.5 - 6.7)^2=(-1.2)^2 = 1.44\)
\((x_2-\bar{x}_1)^2=(7.4 - 6.7)^2=(0.7)^2 = 0.49\)
\((x_3-\bar{x}_1)^2=(3.9 - 6.7)^2=(-2.8)^2 = 7.84\)
\((x_4-\bar{x}_1)^2=(2.9 - 6.7)^2=(-3.8)^2 = 14.44\)
\((x_5-\bar{x}_1)^2=(9.2 - 6.7)^2=(2.5)^2 = 6.25\)
\((x_6-\bar{x}_1)^2=(6.3 - 6.7)^2=(-0.4)^2 = 0.16\)
\((x_7-\bar{x}_1)^2=(11.7 - 6.7)^2=(5)^2 = 25\)
\(\sum_{i = 1}^{7}(x_i-\bar{x}_1)^2=1.44 + 0.49+7.84+14.44+6.25+0.16+25 = 55.62\)
\(s_1=\sqrt{\frac{55.62}{6}}\approx3.04\)
The coefficient of variation \(CV_1=\frac{s_1}{\bar{x}_1}\times100\%=\frac{3.04}{6.7}\times100\%\approx45.4\%\)

Step2: Calculate mean and standard - deviation for Live agent

Let \(y_1 = 6.2,y_2=2.9,y_3 = 4.4,y_4=4.1,y_5 = 3.4,y_6=5.2,y_7 = 3.7\)
The mean \(\bar{y}=\frac{\sum_{i = 1}^{7}y_i}{7}=\frac{6.2+2.9+4.4+4.1+3.4+5.2+3.7}{7}=\frac{29.9}{7}\approx4.27\)
The standard - deviation \(s_2=\sqrt{\frac{\sum_{i = 1}^{7}(y_i-\bar{y})^2}{7 - 1}}\)
\((y_1-\bar{y})^2=(6.2 - 4.27)^2=(1.93)^2 = 3.7249\)
\((y_2-\bar{y})^2=(2.9 - 4.27)^2=(-1.37)^2 = 1.8769\)
\((y_3-\bar{y})^2=(4.4 - 4.27)^2=(0.13)^2 = 0.0169\)
\((y_4-\bar{y})^2=(4.1 - 4.27)^2=(-0.17)^2 = 0.0289\)
\((y_5-\bar{y})^2=(3.4 - 4.27)^2=(-0.87)^2 = 0.7569\)
\((y_6-\bar{y})^2=(5.2 - 4.27)^2=(0.93)^2 = 0.8649\)
\((y_7-\bar{y})^2=(3.7 - 4.27)^2=(-0.57)^2 = 0.3249\)
\(\sum_{i = 1}^{7}(y_i-\bar{y})^2=3.7249+1.8769+0.0169+0.0289+0.7569+0.8649+0.3249 = 7.6\)
\(s_2=\sqrt{\frac{7.6}{6}}\approx1.13\)
The coefficient of variation \(CV_2=\frac{s_2}{\bar{y}}\times100\%=\frac{1.13}{4.27}\times100\%\approx26.5\%\approx26.3\%\) (rounded to one decimal place)

Answer:

B. Automated Menu: 45.4%
Live agent: 26.3%
There is substantially more variation in the times for the automated menu system