Question

listed below are amounts (in millions of dollars) collected from parking meters by a security company in a certain city. a larger data set was used to convict 5 members of the company of grand larceny. find the mean and median for each of the two samples and then compare the two sets of results. do the limited data listed here show evidence of stealing by the security companys employees? security company: 1.0 1.4 1.6 1.4 1.7 1.3 1.8 1.3 1.3 1.5 other companies: 1.6 2.1 1.8 2.2 1.8 1.9 2.1 2.2 2.2 2.3 find the means the mean for the security company is $1.40 million and the mean for the other companies is $2.02 million (type integers or decimals rounded to two decimal places as needed.) find the medians the median for the security company is $ million and the median for the other companies is $ million (type integers or decimals rounded to two decimal places as needed.)

Explanation:

Step1: Arrange security - company data in ascending order

1.3, 1.3, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.8, 1.9
Since \(n = 10\) (even number of data - points), the median is the average of the \(\frac{n}{2}\)th and \((\frac{n}{2}+ 1)\)th ordered values. \(\frac{n}{2}=\frac{10}{2}=5\) and \(\frac{n}{2}+1 = 6\). The 5th value is 1.5 and the 6th value is 1.6.
Median\(_{security}=\frac{1.5 + 1.6}{2}=1.55\)

Step2: Arrange other - companies data in ascending order

1.8, 1.9, 2.1, 2.1, 2.2, 2.2, 2.2, 2.3
Since \(n = 8\) (even number of data - points), \(\frac{n}{2}=\frac{8}{2}=4\) and \(\frac{n}{2}+1 = 5\). The 4th value is 2.1 and the 5th value is 2.2.
Median\(_{other}=\frac{2.1+2.2}{2}=2.15\)

Answer:

The median for the security company is \(1.55\) million and the median for the other companies is \(2.15\) million.