Question

listed below are amounts (in millions of dollars) collected from parking meters by a security service company and other companies during similar time periods. do the limited data listed here show evidence of stealing by the security service companys employees?
security service company: 1.4 1.6 1.7 1.6 1.5 1.5 1.8 1.6 1.3 1.7
other companies: 1.6 1.9 1.7 1.8 1.7 1.9 1.7 1.5 1.8 1.7
find the coefficient of variation for each of the two samples, then compare the variation.
the coefficient of variation for the amount collected by the security service company is
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean for the security service company

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
For the security service company data \(x=\{14,16,17,16,15,15,18,16,13,17\}\), \(n = 10\).
\(\sum_{i=1}^{10}x_{i}=14 + 16+17+16+15+15+18+16+13+17=157\)
\(\bar{x}=\frac{157}{10}=15.7\)

Step2: Calculate the standard deviation for the security service company

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(14 - 15.7)^{2}=(-1.7)^{2}=2.89\)
\((x_{2}-\bar{x})^{2}=(16 - 15.7)^{2}=(0.3)^{2}=0.09\)
\((x_{3}-\bar{x})^{2}=(17 - 15.7)^{2}=(1.3)^{2}=1.69\)
\((x_{4}-\bar{x})^{2}=(16 - 15.7)^{2}=0.09\)
\((x_{5}-\bar{x})^{2}=(15 - 15.7)^{2}=(-0.7)^{2}=0.49\)
\((x_{6}-\bar{x})^{2}=(15 - 15.7)^{2}=0.49\)
\((x_{7}-\bar{x})^{2}=(18 - 15.7)^{2}=(2.3)^{2}=5.29\)
\((x_{8}-\bar{x})^{2}=(16 - 15.7)^{2}=0.09\)
\((x_{9}-\bar{x})^{2}=(13 - 15.7)^{2}=(-2.7)^{2}=7.29\)
\((x_{10}-\bar{x})^{2}=(17 - 15.7)^{2}=1.69\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=2.89+0.09 + 1.69+0.09+0.49+0.49+5.29+0.09+7.29+1.69=20.1\)
\(s=\sqrt{\frac{20.1}{10 - 1}}=\sqrt{\frac{20.1}{9}}\approx1.49\)

Step3: Calculate the coefficient of variation for the security service company

The formula for the coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{1.49}{15.7}\times100\%\approx9.5\%\)

Answer:

\(9.5\)