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 1.7 1.5 1.6 1.7 1.6 1.6 1.5 1.5 1.8
other companies: 1.9 1.8 1.6 1.7 1.6 1.8 1.5 1.7 1.9 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 of the security service company's data

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
For the security service company's data \(x=\{1.4,1.7,1.5,1.6,1.7,1.6,1.6,1.5,1.5,1.8\}\), \(n = 10\).
\(\sum_{i=1}^{10}x_{i}=1.4 + 1.7+1.5+1.6+1.7+1.6+1.6+1.5+1.5+1.8=15.9\)
\(\bar{x}=\frac{15.9}{10}=1.59\)

Step2: Calculate the standard deviation of the security service company's data

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(1.4 - 1.59)^{2}=(- 0.19)^{2}=0.0361\)
\((x_{2}-\bar{x})^{2}=(1.7 - 1.59)^{2}=(0.11)^{2}=0.0121\)
\((x_{3}-\bar{x})^{2}=(1.5 - 1.59)^{2}=(-0.09)^{2}=0.0081\)
\((x_{4}-\bar{x})^{2}=(1.6 - 1.59)^{2}=(0.01)^{2}=0.0001\)
\((x_{5}-\bar{x})^{2}=(1.7 - 1.59)^{2}=(0.11)^{2}=0.0121\)
\((x_{6}-\bar{x})^{2}=(1.6 - 1.59)^{2}=(0.01)^{2}=0.0001\)
\((x_{7}-\bar{x})^{2}=(1.6 - 1.59)^{2}=(0.01)^{2}=0.0001\)
\((x_{8}-\bar{x})^{2}=(1.5 - 1.59)^{2}=(-0.09)^{2}=0.0081\)
\((x_{9}-\bar{x})^{2}=(1.5 - 1.59)^{2}=(-0.09)^{2}=0.0081\)
\((x_{10}-\bar{x})^{2}=(1.8 - 1.59)^{2}=(0.21)^{2}=0.0441\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=0.0361+0.0121 + 0.0081+0.0001+0.0121+0.0001+0.0001+0.0081+0.0081+0.0441=0.139\)
\(s=\sqrt{\frac{0.139}{10 - 1}}=\sqrt{\frac{0.139}{9}}\approx0.124\)

Step3: Calculate the coefficient of variation

The formula for the coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{0.124}{1.59}\times100\%\approx7.8\%\)

Answer:

\(7.8\%\)