Question

ted 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
d the coefficient of variation for each of the two samples, then compare the variation.
coefficient of variation for the amount collected by the security service company is 10.2 %
ound to one decimal place as needed.)
coefficient of variation for the amount collected by the other companies is \\( \square \\% \\)
ound to one decimal place as needed.)

Explanation:

Step1: Calculate the mean for the other companies' data

Let \(x_1 = 1.8,x_2 = 1.9,x_3 = 1.6,x_4 = 1.7,x_5 = 1.8,x_6 = 1.9,x_7 = 1.7,x_8 = 1.5,x_9 = 1.9,x_{10}=1.7\)
The mean formula is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\)
\(\sum_{i=1}^{10}x_i=1.8 + 1.9+1.6+1.7+1.8+1.9+1.7+1.5+1.9+1.7=17.5\)
\(n = 10\), so \(\bar{x}=\frac{17.5}{10}=1.75\)

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

The standard deviation formula is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
\((x_1-\bar{x})^2=(1.8 - 1.75)^2=0.0025\)
\((x_2-\bar{x})^2=(1.9 - 1.75)^2=0.0225\)
\((x_3-\bar{x})^2=(1.6 - 1.75)^2=0.0225\)
\((x_4-\bar{x})^2=(1.7 - 1.75)^2=0.0025\)
\((x_5-\bar{x})^2=(1.8 - 1.75)^2=0.0025\)
\((x_6-\bar{x})^2=(1.9 - 1.75)^2=0.0225\)
\((x_7-\bar{x})^2=(1.7 - 1.75)^2=0.0025\)
\((x_8-\bar{x})^2=(1.5 - 1.75)^2=0.0625\)
\((x_9-\bar{x})^2=(1.9 - 1.75)^2=0.0225\)
\((x_{10}-\bar{x})^2=(1.7 - 1.75)^2=0.0025\)
\(\sum_{i = 1}^{10}(x_i-\bar{x})^2=0.0025+0.0225+0.0225+0.0025+0.0025+0.0225+0.0025+0.0625+0.0225+0.0025 = 0.165\)
\(s=\sqrt{\frac{0.165}{9}}\approx\sqrt{0.0183}\approx0.135\)

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

The coefficient of variation formula is \(CV=\frac{s}{\bar{x}}\times100\%\)
\(CV=\frac{0.135}{1.75}\times100\%\approx7.7\%\)

Answer:

\(7.7\)