Question

The coefficient of variation for the following data is

• A. \( \frac{8\sqrt{22}}{3} \)
• B. \( \frac{8\sqrt{110}}{\sqrt{3}} \)
• C. \( \frac{4\sqrt{110}}{\sqrt{3}} \)
• D. \( \frac{4\sqrt{22}}{3} \)

Hint:

The coefficient of variation (CV) standardizes the standard deviation by expressing it as a percentage of the mean. This allows for comparison of variability between datasets with different means. Remember the formula is \( (\sigma / \mu) \times 100% \).

Answer 1

The Correct Option is B

To calculate the coefficient of variation (CV) for the given data, we follow these steps:

1. Determine the midpoints (x) of the class intervals.
2. Calculate the mean (\(\bar{x}\)) using the formula: \(\bar{x} = \frac{\sum{f_ix_i}}{\sum{f_i}}\).
3. Calculate the variance (\(\sigma^2\)) and standard deviation (\(\sigma\)) using: \(\sigma^2 = \frac{\sum{f_i(x_i-\bar{x})^2}}{\sum{f_i}}\).
4. Calculate the coefficient of variation (CV) using: \(\text{CV} = \frac{\sigma}{\bar{x}} \times 100\).

Let's perform the calculations:

Class Interval	Frequency (\(f_i\))	Midpoint (\(x_i\))	\(f_ix_i\)	\(f_i(x_i-\bar{x})^2\)
0-2	2	1	2	2 \((1-\bar{x})^2\)
2-4	3	3	9	3 \((3-\bar{x})^2\)
4-6	5	5	25	5 \((5-\bar{x})^2\)
6-8	3	7	21	3 \((7-\bar{x})^2\)
8-10	2	9	18	2 \((9-\bar{x})^2\)
Total	15		75

1. Calculating \(\bar{x}\):

\(\bar{x} = \frac{75}{15} = 5\)

2. Calculating variance (\(\sigma^2\)):

• Substitute the midpoint and frequency values into the formula \(\sigma^2 = \frac{\sum{f_i(x_i-\bar{x})^2}}{\sum{f_i}}\) to find variance.

3. Calculating Standard Deviation (\(\sigma\)):

• \(\sigma = \sqrt{\sigma^2}\)

Finally, calculate the CV:

• \(\text{CV} = \frac{\sigma}{\bar{x}} \times 100\)

According to the given options, the correct answer is:

\(\frac{8\sqrt{110}}{\sqrt{3}}\)