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​ company's employees?
Security Service Company:
1.5
1.8
1.4
1.6
1.6
1.5
1.7
1.6
1.4
1.8
 
Other Companies:
1.7
1.9
1.6
1.8
1.7
1.9
1.7
1.5
1.8
1.6
Find the coefficient of variation for each of the two​ samples, then compare the variation.

To determine if there's evidence of stealing by the security service company's employees, we will compare the variability in the amounts collected by the security service company and other companies. One way to measure this variability is by calculating the coefficient of variation (CV) for each set of data. The CV is a standardized measure of dispersion of a probability distribution or frequency distribution. It is defined as the ratio of the standard deviation to the mean, often expressed as a percentage.

Here are the steps to calculate the CV for both the security service company and the other companies:

1. Calculate the mean (average) of each dataset.
2. Calculate the standard deviation of each dataset.
3. Compute the CV using the formula: \(\text{CV} = \left( \frac{\text{Standard Deviation}}{\text{Mean}} \right) \times 100\%\)

### Security Service Company

**Data:** 1.5, 1.8, 1.4, 1.6, 1.6, 1.5, 1.7, 1.6, 1.4, 1.8

1. **Mean:**

\[
\text{Mean} = \frac{1.5 + 1.8 + 1.4 + 1.6 + 1.6 + 1.5 + 1.7 + 1.6 + 1.4 + 1.8}{10} = \frac{15.9}{10} = 1.59
\]

2. **Standard Deviation:**

\[
\text{Standard Deviation} = \sqrt{\frac{\sum (x_i - \text{Mean})^2}{n-1}}
\]

\[
\begin{aligned}
&= \sqrt{\frac{(1.5 - 1.59)^2 + (1.8 - 1.59)^2 + (1.4 - 1.59)^2 + (1.6 - 1.59)^2 + (1.6 - 1.59)^2 + (1.5 - 1.59)^2 + (1.7 - 1.59)^2 + (1.6 - 1.59)^2 + (1.4 - 1.59)^2 + (1.8 - 1.59)^2}{9}} \\
&= \sqrt{\frac{0.0081 + 0.0441 + 0.0361 + 0.0001 + 0.0001 + 0.0081 + 0.0121 + 0.0001 + 0.0361 + 0.0441}{9}} \\
&= \sqrt{\frac{0.229}{9}} \\
&= \sqrt{0.0254} \\
&\approx 0.159
\end{aligned}
\]

3. **Coefficient of Variation:**

\[
\text{CV} = \left( \frac{0.159}{1.59} \right) \times 100\% \approx 10\%
\]

### Other Companies

**Data:** 1.7, 1.9, 1.6, 1.8, 1.7, 1.9, 1.7, 1.5, 1.8, 1.6

1. **Mean:**

\[
\text{Mean} = \frac{1.7 + 1.9 + 1.6 + 1.8 + 1.7 + 1.9 + 1.7 + 1.5 + 1.8 + 1.6}{10} = \frac{17.2}{10} = 1.72
\]

2. **Standard Deviation:**

\[
\text{Standard Deviation} = \sqrt{\frac{\sum (x_i - \text{Mean})^2}{n-1}}
\]

\[
\begin{aligned}
&= \sqrt{\frac{(1.7 - 1.72)^2 + (1.9 - 1.72)^2 + (1.6 - 1.72)^2 + (1.8 - 1.72)^2 + (1.7 - 1.72)^2 + (1.9 - 1.72)^2 + (1.7 - 1.72)^2 + (1.5 - 1.72)^2 + (1.8 - 1.72)^2 + (1.6 - 1.72)^2}{9}} \\
&= \sqrt{\frac{0.0004 + 0.0324 + 0.0144 + 0.0064 + 0.0004 + 0.0324 + 0.0004 + 0.0484 + 0.0064 + 0.0144}{9}} \\
&= \sqrt{\frac{0.156}{9}} \\
&= \sqrt{0.0173} \\
&\approx 0.1315
\end{aligned}
\]

3. **Coefficient of Variation:**

\[
\text{CV} = \left( \frac{0.1315}{1.72} \right) \times 100\% \approx 7.64\%
\]

### Comparison

- **Security Service Company CV:** \( \approx 10\% \)
- **Other Companies CV:** \( \approx 7.64\% \)

The coefficient of variation for the security service company is higher than that of the other companies. This suggests that there is greater variability in the amounts collected by the security service company. Higher variability might indicate irregularities, which could be consistent with the possibility of theft, but this alone is not definitive proof. Additional analysis and evidence would be necessary to draw a firm conclusion.

Do you have any questions or would you like further details?

Here are five questions that could expand on this information:

1. What other statistical tests can be used to compare the datasets?
2. How would the results differ if we had more data points?
3. What external factors could contribute to the variability in the amounts collected?
4. How can we further investigate the possibility of theft based on this data?
5. What are the limitations of using the coefficient of variation in this context?

**Tip:** Always consider multiple statistical measures and additional evidence when investigating potential irregularities in data.

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​ company's employees? Security Service Company: 1.5, 1.8, 1.4, 1.6, 1.6, 1.5, 1.7, 1.6, 1.4, 1.8. Other Companies: 1.7, 1.9, 1.6, 1.8, 1.7, 1.9, 1.7, 1.5, 1.8, 1.6. Find the coefficient of variation for each of the two​ samples, then compare the variation.

This page explains how to find and compare the coefficient of variation for amounts collected from parking meters by a security service company and other companies. It involves calculating the mean, standard deviation, and CV to investigate possible irregularities. Suitable for students with knowledge of college-level statistics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation