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Following Are the Marks Obtained,Out of 100 by Two Students Ravi and Hashina in 10 Tests: - Mathematics

Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests: 

Ravi: 25 50 45 30 70 42 36 48 35 60
Hashina: 10 70 50 20 95 55 42 60 48 80


Who is more intelligent and who is more consistent? 

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Solution

For Ravi

 

Marks
 

\[\left( x_i \right)\]
\[d_i = x_i - 45\]
 

\[d_i^2\]
25 −20 400
50 5 25
45 0 0
30 −15 225
70 25 625
42 −3 9
36 −9 81
48 3 9
35 −10 100
60 15 225
 
 
\[\sum_{} d_i = - 9\]
 

\[\sum_{} d_i^2 = 1699\]
\[X_R = A + \frac{\sum_{} d_i}{10} = 45 + \frac{\left( - 9 \right)}{10} = 44 . 1\]
Standard deviation,
\[\sigma_R = \sqrt{\frac{\sum_{} d_i^2}{10} - \left( \frac{\sum_{} d_i}{10} \right)^2} = \sqrt{\frac{1699}{10} - \left( \frac{- 9}{10} \right)^2} = \sqrt{169 . 09} = 13 . 003\]
Coefficicent of variation = \[\frac{\sigma_B}{X_B} \times 100 = \frac{110}{770} \times 100 = 14 . 29\]
For Hashina
Marks
\[\left( x_i \right)\]
\[d_i = x_i - 55\]
\[d_i^2\]
10 −45 2025
70 15 625
50 −5 25
20 −35 1225
95 40 1600
55 0 0
42 −13 169
60 5 25
48 −7 49
80 25 625
 
 

\[\sum_{} d_i = - 20\]
 

\[\sum_{} d_i^2 = 6368\]

Mean,

\[X_H = A + \frac{\sum_{} d_i}{10} = 55 + \frac{\left( - 20 \right)}{10} = 53\]
Standard deviation,
\[\sigma_H = \sqrt{\frac{\sum_{} d_i^2}{10} - \left( \frac{\sum_{} d_i}{10} \right)^2} = \sqrt{\frac{6368}{10} - \left( \frac{- 20}{10} \right)^2} = \sqrt{632 . 8} = 25 . 16\]
Coefficicent of variation = \[\frac{\sigma_H}{X_H} \times 100 = \frac{25 . 16}{53} \times 100 = 47 . 47\]
Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.
Concept: Statistics - Statistics Concept
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.7 | Q 12 | Page 49
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