Q255 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
The sum and the sum of squares of the observations corresponding to length $X$ (in cm) and weight $Y$ (in gm) of 50 tropical tubers are given as $\Sigma X = 200$, $\Sigma Y = 250$, $\Sigma X^2 = 900$ and $\Sigma Y^2 = 1400$.
Which one of the following statements is correct?
The sum and the sum of squares of the observations corresponding to length $X$ (in cm) and weight $Y$ (in gm) of 50 tropical tubers are given as $\Sigma X = 200$, $\Sigma Y = 250$, $\Sigma X^2 = 900$ and $\Sigma Y^2 = 1400$.
Which one of the following statements is correct?
- Correct answerACoefficient of variation of $X$ is strictly more than coefficient of variation of $Y$.
- BCoefficient of variation of $X$ is strictly less than coefficient of variation of $Y$.
- CCoefficient of variation of $X$ is same as coefficient of variation of $Y$.
- DCoefficient of variation cannot be determined from the given data.
Solution
Mean of $X$, $\bar{X} = \dfrac{\Sigma X}{n} = \dfrac{200}{50} = 4$
Mean of $Y$, $\bar{Y} = \dfrac{\Sigma Y}{n} = \dfrac{250}{50} = 5$
Variance of $X$, $\sigma_X^2 = \dfrac{\Sigma X^2}{n} - (\bar{X})^2 = \dfrac{900}{50} - 4^2 = 18 - 16 = 2$
Standard deviation of $X$, $\sigma_X = \sqrt{2}$
Variance of $Y$, $\sigma_Y^2 = \dfrac{\Sigma Y^2}{n} - (\bar{Y})^2 = \dfrac{1400}{50} - 5^2 = 28 - 25 = 3$
Standard deviation of $Y$, $\sigma_Y = \sqrt{3}$
Coefficient of variation of $X$, $CV_X = \dfrac{\sigma_X}{\bar{X}} \times 100 = \dfrac{\sqrt{2}}{4} \times 100 = 25\sqrt{2} \approx 35.35\%$
Coefficient of variation of $Y$, $CV_Y = \dfrac{\sigma_Y}{\bar{Y}} \times 100 = \dfrac{\sqrt{3}}{5} \times 100 = 20\sqrt{3} \approx 34.64\%$
Since $35.35\% > 34.64\%$, the coefficient of variation of $X$ is strictly more than the coefficient of variation of $Y$.
Answer: Coefficient of variation of $X$ is strictly more than coefficient of variation of $Y$.
Mean of $Y$, $\bar{Y} = \dfrac{\Sigma Y}{n} = \dfrac{250}{50} = 5$
Variance of $X$, $\sigma_X^2 = \dfrac{\Sigma X^2}{n} - (\bar{X})^2 = \dfrac{900}{50} - 4^2 = 18 - 16 = 2$
Standard deviation of $X$, $\sigma_X = \sqrt{2}$
Variance of $Y$, $\sigma_Y^2 = \dfrac{\Sigma Y^2}{n} - (\bar{Y})^2 = \dfrac{1400}{50} - 5^2 = 28 - 25 = 3$
Standard deviation of $Y$, $\sigma_Y = \sqrt{3}$
Coefficient of variation of $X$, $CV_X = \dfrac{\sigma_X}{\bar{X}} \times 100 = \dfrac{\sqrt{2}}{4} \times 100 = 25\sqrt{2} \approx 35.35\%$
Coefficient of variation of $Y$, $CV_Y = \dfrac{\sigma_Y}{\bar{Y}} \times 100 = \dfrac{\sqrt{3}}{5} \times 100 = 20\sqrt{3} \approx 34.64\%$
Since $35.35\% > 34.64\%$, the coefficient of variation of $X$ is strictly more than the coefficient of variation of $Y$.
Answer: Coefficient of variation of $X$ is strictly more than coefficient of variation of $Y$.
Hence, the correct answer is option: 'A'.