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
other companies: 1.9 1.8 1.6 1.7 1.6 1.6 1.6 1.5 1.5 1.8 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 7.5%.
(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 other companies' data

First, sum up all the values in the other companies' data: \(1.9 + 1.8+1.6 + 1.7+1.6 + 1.6+1.6 + 1.5+1.5 + 1.8+1.7 + 1.9+1.7\)

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There are \(n = 13\) data points. The mean \(\bar{x}=\frac{21.9}{13}\approx1.685\)

Step2: Calculate the standard deviation for the other companies' data

Use the formula \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)

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\(s=\sqrt{\frac{0.197}{13 - 1}}\approx\sqrt{\frac{0.197}{12}}\approx0.129\)

Step3: Calculate the coefficient of variation for the other companies' data

The formula for the coefficient of variation \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{0.129}{1.685}\times100\%\approx7.7\%\)

Answer:

The coefficient of variation for the amount collected by the other companies is \(7.7\%\)