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
find the means.
the mean for the security company is $\square$ million and the mean for the other companies is $\square$ million. (type integers or decimals rounded to two decimal places as needed.)
find the medians.
the median for the security company is $\square$ million and the median for the other companies is $\square$ million. (type integers or decimals rounded to two decimal places as needed.)
compare the results. choose the correct answer below.
a. the mean and median appear to be roughly the same for all collections.
b. the median is lower for the collections performed by other companies, but the mean is lower for the security company.
c. the mean and the median for the collections performed by other companies are both lower than the mean and the median for the security company.

Explanation:

Step1: Calculate mean for security company

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

Step2: Calculate mean for other companies

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

Step3: Calculate median for security company

First, arrange the security - company data in ascending order: $\{1.1,1.2,1.4,1.4,1.4,1.5,1.5,1.7,1.8,1.8\}$. Since $n = 10$ (an even number), the median $M=\frac{x_{\frac{n}{2}}+x_{\frac{n}{2}+1}}{2}=\frac{1.4 + 1.5}{2}=1.45$.

Step4: Calculate median for other companies

Arrange the other - companies data in ascending order: $\{1.5,1.6,1.6,1.7,1.7,1.9,1.9,1.9,2.2,2.3\}$. Since $n = 10$ (an even number), the median $M=\frac{1.7+1.9}{2}=1.80$.

Answer:

The mean for the security company is $1.48$ million and the mean for the other companies is $1.83$ million.
The median for the security company is $1.45$ million and the median for the other companies is $1.80$ million.
The correct answer is: The mean and the median for the collections performed by other companies are both higher than the mean and the median for the security company.