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: 1.8 1.5 1.8 1.5 1.7 1.4 1.2 1.4 1.4 1.1
other companies: 2.3 1.7 1.6 1.9 1.6 2.2 1.7 1.9 1.9 1.5
do the limit data listed here show evidence of stealing by the security companys employees?
a. since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security companys employees.
b. the sample size is not large enough to show any meaningful results.
c. since the security company does not appear to have collected lower revenue than the other companies, there is no evidence of stealing by the security companys employees.
d. since the data is not matched, there is no evidence of stealing by the security companys employees.

Explanation:

Step1: Calculate the mean for the security company

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For the security - company data: $x_1 = 1.8,x_2 = 1.5,\cdots,x_{10}=1.1$. $n = 10$. $\sum_{i=1}^{10}x_{i}=1.8 + 1.5+1.8+1.5+1.7+1.4+1.2+1.4+1.4+1.1=14.8$. So, $\bar{x}_{security}=\frac{14.8}{10}=1.48$.

Step2: Calculate the median for the security company

First, order the data: $1.1,1.2,1.4,1.4,1.4,1.5,1.5,1.7,1.8,1.8$. Since $n = 10$ (an even - numbered data set), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values. The 5th value is $1.4$ and the 6th value is $1.5$. So, $Median_{security}=\frac{1.4 + 1.5}{2}=1.45$.

Step3: Calculate the mean for other companies

For the other - companies data: $x_1 = 2.3,x_2 = 1.7,\cdots,x_{10}=1.5$. $n = 10$. $\sum_{i = 1}^{10}x_{i}=2.3+1.7+1.6+1.9+1.6+2.2+1.7+1.9+1.9+1.5=18.3$. So, $\bar{x}_{other}=\frac{18.3}{10}=1.83$.

Step4: Calculate the median for other companies

Order the data: $1.5,1.6,1.6,1.7,1.7,1.9,1.9,1.9,2.2,2.3$. Since $n = 10$ (an even - numbered data set), the median is the average of the 5th and 6th ordered values. The 5th value is $1.7$ and the 6th value is $1.9$. So, $Median_{other}=\frac{1.7 + 1.9}{2}=1.8$.

Step5: Compare the results

We have $\bar{x}_{security}=1.48<\bar{x}_{other}=1.83$ and $Median_{security}=1.45

Step6: Analyze the evidence of stealing

Since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security company's employees.

Answer:

The mean for the security company is $1.48$ and the median is $1.45$. The mean for other companies is $1.83$ and the median is $1.8$.
E. The mean and the median for the security company are both lower than the mean and the median for the collections performed by other companies.
A. Since the security company appears to have collected lower revenue than the other companies, there is some evidence of stealing by the security company's employees.