NCERT Solutions & Concepts: Analysis of Statistics Exercise 15.3 (Class 11)

Understanding Coefficient of Variation (C.V.)

To compare the dispersion of two frequency distributions, standard deviation alone is insufficient if the datasets have different means or are measured in different units (for instance, comparing weight in kilograms with height in centimeters). To overcome this, Karl Pearson introduced a relative measure of dispersion called the Coefficient of Variation (C.V.). It is defined as the ratio of the standard deviation ($\sigma$) to the mean ($\bar{x}$), expressed as a percentage.

$\text{C.V.} = \frac{\sigma}{\bar{x}} \times 100 \quad (\text{where } \bar{x} \neq 0)$

Because both $\sigma$ and $\bar{x}$ share the same physical unit, their ratio is a pure, unitless number. This mathematical property allows direct, objective comparison between any two distributions:

1. Greater C.V.: The series is more variable, more dispersed, or less consistent.
2. Lesser C.V.: The series is less variable, more stable, or more consistent.

In cases where two frequency distributions have identical means ($\bar{x}_1 = \bar{x}_2$), their relative variability depends solely on their standard deviations. The series with the smaller standard deviation is immediately declared more consistent.

Step-by-Step Guide to Compare Two Distributions

1. Step 1: Compute the Means — Calculate the arithmetic mean ($\bar{x}_1$ and $\bar{x}_2$) for both distributions. For grouped frequency distributions, use the step-deviation method or direct method as convenient.
2. Step 2: Calculate Standard Deviations — Determine the standard deviation ($\sigma_1$ and $\sigma_2$) for both groups using the formula: $\sigma = \sqrt{\frac{\sum f_i(x_i - \bar{x})^2}{N}}$ or the shortcut method.
3. Step 3: Calculate Coefficient of Variation (C.V.) — Apply the relative dispersion formula: $\text{C.V.(1)} = \frac{\sigma_1}{\bar{x}_1} \times 100$ and $\text{C.V.(2)} = \frac{\sigma_2}{\bar{x}_2} \times 100$.
4. Step 4: Analyze and Conclude — Compare the two values. The distribution with the higher C.V. value is more variable (less stable), and the one with the lower C.V. value is more consistent (more stable).

Worked Examples on Exercise 15.3 Concepts

• Example 1 (Different Means): Two factories, A and B, pay average daily wages of ₹350 and ₹420 with standard deviations of ₹28 and ₹35 respectively. Which factory has greater variability in individual wages? Step 1: Write down given values. Factory A: Mean $\bar{x}_A = 350$, Standard Deviation $\sigma_A = 28$ Factory B: Mean $\bar{x}_B = 420$, Standard Deviation $\sigma_B = 35$ Step 2: Calculate C.V. for Factory A. $\text{C.V.(A)} = \frac{28}{350} \times 100 = 0.08 \times 100 = 8\%$ Step 3: Calculate C.V. for Factory B. $\text{C.V.(B)} = \frac{35}{420} \times 100 = 0.0833 \times 100 = 8.33\%$ Step 4: Compare. Since $\text{C.V.(B)} > \text{C.V.(A)}$, Factory B shows greater variability in wages.
• Example 2 (Same Mean, Different Dispersion): Two players, X and Y, score runs in 5 matches. Their average score is identical (50 runs). The standard deviation of Player X's scores is 6.2, while that of Player Y's is 4.8. Who is the more consistent scorer? Step 1: Write down given values. Player X: Mean $\bar{x}_X = 50$, $\sigma_X = 6.2$ Player Y: Mean $\bar{x}_Y = 50$, $\sigma_Y = 4.8$ Step 2: Assess relative consistency. Since the means are equal, we can directly compare their standard deviations instead of full C.V. calculation. However, verifying via formula: $\text{C.V.(X)} = \frac{6.2}{50} \times 100 = 12.4\%$ $\text{C.V.(Y)} = \frac{4.8}{50} \times 100 = 9.6\%$ Step 3: Conclusion. Because $\text{C.V.(Y)} < \text{C.V.(X)}$, Player Y is more consistent in performance.

Board Exam Pitfalls & High-Yield Strategy

1. Interpreting the Wordings: Carefully read what the question asks. If it asks for 'more consistent' or 'more stable', choose the group with the lower C.V. If it asks for 'more variable', 'more dispersed', or 'less stable', choose the group with the higher C.V. Students often swap these definitions in stressful exam environments.
2. The 100 Multiplier: Standard Deviation divided by Mean gives relative dispersion in decimals. Remember to multiply by $100$ to represent the Coefficient of Variation as a percentage. Keep calculations accurate up to two decimal points.
3. Variance vs Standard Deviation: Question statements often give 'Variance' ($\sigma^2$). Be extremely careful to take the square root of the variance to obtain the standard deviation ($\sigma$) before substituting it in the C.V. formula.

Practice Questions with Solutions

• Q: The sum of squares of deviations from mean for two groups of observations of size 10 each are 160 and 250. If their means are 16 and 20 respectively, calculate their Coefficients of Variation and identify which group is more consistent. A: Step 1: Compute standard deviation ($\sigma$) for Group 1. Variance $\sigma_1^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{160}{10} = 16 \implies \sigma_1 = 4$. Mean $\bar{x}_1 = 16$. $\text{C.V.(1)} = \frac{4}{16} \times 100 = 25\%$. Step 2: Compute standard deviation ($\sigma$) for Group 2. Variance $\sigma_2^2 = \frac{250}{10} = 25 \implies \sigma_2 = 5$. Mean $\bar{x}_2 = 20$. $\text{C.V.(2)} = \frac{5}{20} \times 100 = 25\%$. Step 3: Compare results. Since both groups have an identical C.V. of 25%, they are equally consistent. Final answer: C.V.(1) = 25%, C.V.(2) = 25%; both are equally consistent.
• Q: For two data series A and B, the standard deviation of A is 12 and its mean is 80. The variance of B is 225 and its mean is 120. Which series shows greater dispersion? A: Step 1: Extract data for Series A. $\sigma_A = 12$, $\bar{x}_A = 80$. $\text{C.V.(A)} = \frac{12}{80} \times 100 = 15\%$. Step 2: Extract data for Series B. Variance $\sigma_B^2 = 225 \implies \sigma_B = \sqrt{225} = 15$. Mean $\bar{x}_B = 120$. $\text{C.V.(B)} = \frac{15}{120} \times 100 = 12.5\%$. Step 3: Compare C.V. values. $\text{C.V.(A)} = 15\%$ and $\text{C.V.(B)} = 12.5\%$. Since $\text{C.V.(A)} > \text{C.V.(B)}$, Series A has greater dispersion. Final answer: Series A has greater dispersion (C.V. = 15%).
• Q: From the prices of Shares X and Y given below, find out which share is more stable in value. Share X Mean = 50, Variance = 25. Share Y Mean = 100, Variance = 64. A: Step 1: Calculate Standard Deviation for both shares. $\sigma_X = \sqrt{25} = 5$ $\sigma_Y = \sqrt{64} = 8$ Step 2: Compute C.V. for Share X. $\text{C.V.(X)} = \frac{5}{50} \times 100 = 10\%$ Step 3: Compute C.V. for Share Y. $\text{C.V.(Y)} = \frac{8}{100} \times 100 = 8\%$ Step 4: Compare values for stability (lower C.V. = more stable). Since $\text{C.V.(Y)} = 8\% < \text{C.V.(X)} = 10\%$, Share Y is more stable. Final answer: Share Y is more stable in value.
• Q: The mean and standard deviation of 100 observations were found to be 40 and 10 respectively. While calculating, one observation was wrongly taken as 30 instead of 70. Correct the mean and standard deviation, and find the correct C.V. A: Step 1: Correct the Mean. Incorrect Sum $\sum x = 100 \times 40 = 4000$. Correct Sum $= 4000 - 30 + 70 = 4040$. Correct Mean $\bar{x}_{new} = \frac{4040}{100} = 40.4$. Step 2: Correct the Standard Deviation. Incorrect Variance $\sigma^2 = 10^2 = 100$. Incorrect $\sum x^2 = 100 \times (100 + 40^2) = 100 \times (100 + 1600) = 170000$. Correct $\sum x^2 = 170000 - 30^2 + 70^2 = 170000 - 900 + 4900 = 174000$. Correct Standard Deviation $\sigma_{new} = \sqrt{\frac{174000}{100} - (40.4)^2} = \sqrt{1740 - 1632.16} = \sqrt{107.84} \approx 10.38$. Step 3: Calculate correct C.V. $\text{C.V.} = \frac{10.38}{40.4} \times 100 \approx 25.69\%$. Final answer: Correct Mean = 40.4, Correct S.D. = 10.38, Correct C.V. = 25.69%.

Frequently Asked Questions