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: 1.3 1.7 1.4 1.5 1.6 1.5 1.8 1.7 1.4 1.6
other companies: 1.8 1.9 1.6 1.7 1.8 1.9 1.7 1.5 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 $\bar{x}$ is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
For the security service company's data \(x=\{1.3,1.7,1.4,1.5,1.6,1.5,1.8,1.7,1.4,1.6\}\), \(n = 10\).
\(\sum_{i=1}^{10}x_{i}=1.3 + 1.7+1.4+1.5+1.6+1.5+1.8+1.7+1.4+1.6=15.5\)
\(\bar{x}=\frac{15.5}{10}=1.55\)

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

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(1.3 - 1.55)^{2}=(- 0.25)^{2}=0.0625\)
\((x_{2}-\bar{x})^{2}=(1.7 - 1.55)^{2}=(0.15)^{2}=0.0225\)
\((x_{3}-\bar{x})^{2}=(1.4 - 1.55)^{2}=(-0.15)^{2}=0.0225\)
\((x_{4}-\bar{x})^{2}=(1.5 - 1.55)^{2}=(-0.05)^{2}=0.0025\)
\((x_{5}-\bar{x})^{2}=(1.6 - 1.55)^{2}=(0.05)^{2}=0.0025\)
\((x_{6}-\bar{x})^{2}=(1.5 - 1.55)^{2}=(-0.05)^{2}=0.0025\)
\((x_{7}-\bar{x})^{2}=(1.8 - 1.55)^{2}=(0.25)^{2}=0.0625\)
\((x_{8}-\bar{x})^{2}=(1.7 - 1.55)^{2}=(0.15)^{2}=0.0225\)
\((x_{9}-\bar{x})^{2}=(1.4 - 1.55)^{2}=(-0.15)^{2}=0.0225\)
\((x_{10}-\bar{x})^{2}=(1.6 - 1.55)^{2}=(0.05)^{2}=0.0025\)
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=0.0625+0.0225 + 0.0225+0.0025+0.0025+0.0025+0.0625+0.0225+0.0225+0.0025=0.225\)
\(s=\sqrt{\frac{0.225}{10 - 1}}=\sqrt{\frac{0.225}{9}}=\sqrt{0.025}\approx0.158\)

Step3: Calculate the coefficient of variation

The formula for the coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{0.158}{1.55}\times100\%\approx10.2\%\)

Answer:

\(10.2\)