Question

waiting times (in minutes) of customers at a bank where all customers enter a single waiting line and a bank where customers wait in individual lines at three different teller windows are listed below. find the coefficient of variation for each of the two sets of data, then compare the variation.
bank a (single line): 6.5 6.6 6.6 6.8 6.9 7.1 7.4 7.5 7.6 7.7 7.8
bank b (individual lines): 4.1 5.4 5.8 6.3 6.7 7.6 7.7 8.6 9.3 9.8
the coefficient of variation for the waiting times at bank a is 6.5 %
(round to one decimal place as needed.)
the coefficient of variation for the waiting times at the bank b is □%
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean of Bank B's data

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
For Bank B: $n = 10$, $\sum_{i=1}^{10}x_{i}=4.1 + 5.4+5.8 + 6.3+6.7+7.6+7.7+8.6+9.3+9.8=71.3$
$\bar{x}=\frac{71.3}{10}=7.13$

Step2: Calculate the standard deviation of Bank B's data

The formula for the standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$
$(x_{1}-\bar{x})^{2}=(4.1 - 7.13)^{2}=(- 3.03)^{2}=9.1809$
$(x_{2}-\bar{x})^{2}=(5.4 - 7.13)^{2}=(-1.73)^{2}=2.9929$
$(x_{3}-\bar{x})^{2}=(5.8 - 7.13)^{2}=(-1.33)^{2}=1.7689$
$(x_{4}-\bar{x})^{2}=(6.3 - 7.13)^{2}=(-0.83)^{2}=0.6889$
$(x_{5}-\bar{x})^{2}=(6.7 - 7.13)^{2}=(-0.43)^{2}=0.1849$
$(x_{6}-\bar{x})^{2}=(7.6 - 7.13)^{2}=(0.47)^{2}=0.2209$
$(x_{7}-\bar{x})^{2}=(7.7 - 7.13)^{2}=(0.57)^{2}=0.3249$
$(x_{8}-\bar{x})^{2}=(8.6 - 7.13)^{2}=(1.47)^{2}=2.1609$
$(x_{9}-\bar{x})^{2}=(9.3 - 7.13)^{2}=(2.17)^{2}=4.7089$
$(x_{10}-\bar{x})^{2}=(9.8 - 7.13)^{2}=(2.67)^{2}=7.1289$
$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=9.1809+2.9929+1.7689+0.6889+0.1849+0.2209+0.3249+2.1609+4.7089+7.1289 = 29.361$
$s=\sqrt{\frac{29.361}{9}}\approx1.81$

Step3: Calculate the coefficient of variation for Bank B

The formula for the coefficient of variation $CV=\frac{s}{\bar{x}}\times100\%$
$CV=\frac{1.81}{7.13}\times100\%\approx25.4\%$

Answer:

$25.4$