# Following are the marks obtained , out of 100 by two students Ravi and Hashina in 10 tests: Who is more intelligent and who is more consistent?

Toolbox:
• The formulae used to solve are Mean $\bar{x} =A+ \large\frac{ \sum d_1}{n}$
• Standard deviation $\sigma= \sqrt { \large\frac{\sum d_1^2}{n} - \bigg( \frac{\sum d_1}{n}\bigg)^2}$
• Coefficient of variation $= \large\frac{ \sigma}{\bar{x}} $$\times 100 Step 1: Let the assumed mean (A) =42 For Ravi Step 2: Mean \bar{X} =A+ \large\frac{ \sum d_1}{n} \qquad=42+\large\frac{21}{10} \qquad= 42+2.1 \qquad= 44.1 Step 3: Standard deviation \sigma= \sqrt { \large\frac{\sum d_1^2}{n} - \bigg( \frac{\sum d_1}{n}\bigg)^2} \qquad= \sqrt {\large\frac{1735}{10} - \bigg( \large\frac{21}{10}\bigg)^2} \qquad= \sqrt { \large\frac{ 17350- 441}{100}} \qquad= \sqrt { \large\frac{16909}{100}} \qquad= \sqrt {169.09} \qquad= 13.003 Step 4: CV of Ravi =\large\frac{\sigma}{\bar{X}}$$ \times 100$
$\qquad=\large\frac{ 13.003}{44.1} $$\times 100 \qquad= 29.485 For Hashina Step 1: Let the assumed mean by 55 Step 2: Mean \bar{Y} =A+ \large\frac{ \sum d_2}{n} \qquad= 55- \large\frac{20}{10} \qquad= 55-2 \qquad= 53 Step 3: Standard deviation \sigma= \sqrt { \large\frac{\sum d_1^2}{n} - \bigg( \frac{\sum d_1}{n}\bigg)^2} \qquad= \sqrt {\large\frac{5968}{10} - \bigg( \frac{-20}{10} \bigg)^2 } \qquad= \sqrt{\large\frac{59680-400}{100}} \qquad= \sqrt {592.8} \qquad= 24.347 Step 4: CV of Hashina =\large\frac{\sigma}{\bar{X}}$$ \times 100$
$\qquad= \large\frac{24.347}{53}$$\times 100$
$\qquad= 45.28$
COMPARISON:
As CV of Ravi < CV of Hashina
$\therefore$ Ravi is more intelligent and consistent.