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 1.6 1.3 1.7
other companies: 1.6 1.9 1.7 1.8 1.7 1.9 1.8 1.6 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 9.5%
(round to one decimal place as needed.)
the coefficient of variation for the amount collected by the other companies 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 \(x=\{1.4,1.6,1.7,1.6,1.5,1.5,1.6,1.3,1.7\}\), \(n = 9\).
\(\sum_{i=1}^{9}x_{i}=1.4 + 1.6+1.7+1.6+1.5+1.5+1.6+1.3+1.7=13.9\)
\(\bar{x}_{1}=\frac{13.9}{9}\approx1.5\)

Step2: Calculate the standard deviation for the security service company

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(1.4 - 1.5)^{2}=0.01\), \((x_{2}-\bar{x})^{2}=(1.6 - 1.5)^{2}=0.01\), \((x_{3}-\bar{x})^{2}=(1.7 - 1.5)^{2}=0.04\), \((x_{4}-\bar{x})^{2}=(1.6 - 1.5)^{2}=0.01\), \((x_{5}-\bar{x})^{2}=(1.5 - 1.5)^{2}=0\), \((x_{6}-\bar{x})^{2}=(1.5 - 1.5)^{2}=0\), \((x_{7}-\bar{x})^{2}=(1.6 - 1.5)^{2}=0.01\), \((x_{8}-\bar{x})^{2}=(1.3 - 1.5)^{2}=0.04\), \((x_{9}-\bar{x})^{2}=(1.7 - 1.5)^{2}=0.04\)
\(\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=0.01+0.01 + 0.04+0.01+0+0+0.01+0.04+0.04=0.16\)
\(s_{1}=\sqrt{\frac{0.16}{9 - 1}}=\sqrt{\frac{0.16}{8}}=\sqrt{0.02}\approx0.14\)
The coefficient of variation \(CV_{1}=\frac{s_{1}}{\bar{x}_{1}}\times100=\frac{0.14}{1.5}\times100\approx9.3\%\)

Step3: Calculate the mean for the other companies

For the other companies data \(x=\{1.6,1.9,1.7,1.8,1.7,1.9,1.8,1.6,1.8\}\), \(n = 9\)
\(\sum_{i=1}^{9}x_{i}=1.6+1.9+1.7+1.8+1.7+1.9+1.8+1.6+1.8 = 15.8\)
\(\bar{x}_{2}=\frac{15.8}{9}\approx1.8\)

Step4: Calculate the standard deviation for the other companies

\((x_{1}-\bar{x})^{2}=(1.6 - 1.8)^{2}=0.04\), \((x_{2}-\bar{x})^{2}=(1.9 - 1.8)^{2}=0.01\), \((x_{3}-\bar{x})^{2}=(1.7 - 1.8)^{2}=0.01\), \((x_{4}-\bar{x})^{2}=(1.8 - 1.8)^{2}=0\), \((x_{5}-\bar{x})^{2}=(1.7 - 1.8)^{2}=0.01\), \((x_{6}-\bar{x})^{2}=(1.9 - 1.8)^{2}=0.01\), \((x_{7}-\bar{x})^{2}=(1.8 - 1.8)^{2}=0\), \((x_{8}-\bar{x})^{2}=(1.6 - 1.8)^{2}=0.04\), \((x_{9}-\bar{x})^{2}=(1.8 - 1.8)^{2}=0\)
\(\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=0.04+0.01+0.01+0+0.01+0.01+0+0.04+0=0.12\)
\(s_{2}=\sqrt{\frac{0.12}{9 - 1}}=\sqrt{\frac{0.12}{8}}=\sqrt{0.015}\approx0.12\)
The coefficient of variation \(CV_{2}=\frac{s_{2}}{\bar{x}_{2}}\times100=\frac{0.12}{1.8}\times100\approx6.7\%\)

Answer:

The coefficient of variation for the amount collected by the security service company is \(9.3\%\). The coefficient of variation for the amount collected by the other companies is \(6.7\%\)