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Home  >>  CBSE XI  >>  Math  >>  Statistics
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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?

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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.
answered Jul 7, 2014 by meena.p
 

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