Question

Arjun and Karna had an archery competition where they had to shoot a target board with 7 concurrent circles dividing the board into 8 concurrent circular strips. The score for each successive outer strip decreases by unity with the centermost circle having a score of 8. Both of them took 78 shots. Dronacharya had difficulty in deciding who the winner was after the scores were recorded. Calculate the mean deviation about the score of innermost circle so as to decide who the winner is. Assume that the winner is less deviated from the bulls eye.

$$\begin{array}{ccccccccc}Points& 8& 7& 6& 5& 4& 3& 2& 1\\ Arjun\left(f_i\right)& 5& 7& 14& 23& 11& 10& 4& 4\\ Karna\left(f_i\right)& 5& 8& 11& 24& 14& 8& 5& 3\end{array}$$

• A. Arjun
• B. Karna
• C. Both of them
• D. Insufficient data

Answer 1

The correct option is B

Karna

The total number of shots is the same for Arjun and Karna which is equal to 78.
Let x_i be score of $i^{th}$ circle strip from center.

Absolute deviation of $i^{th}$ strip= |i-8| = 8-i.

Deviations for Arjun and Karna would be

$$\begin{array}{ccccccccc}Deviation& 0& 1& 2& 3& 4& 5& 6& 7\\ Arjun\left(f_i\right)& 5& 7& 14& 23& 11& 10& 4& 4\\ Karna\left(f_i\right)& 5& 8& 11& 24& 14& 8& 5& 3\end{array}$$
Sum of Absolute deviations for Karna = ${\sum }_{i=1}^8{f}_i|i-8|$
=5(0) + 8(1) + 11(2) + 24(3) + 14(4) + 8(5) + 5(6) + 3(7) = 249
Mean Deviation about score 8 for Karna = $\frac{{\sum }_{i=1}^8{f}_i|i-8|}{{\sum }_{i=1}^8{f}_i}=\frac{249}{78}=3.192$
Mean Deviation about score 8 for Arjun = $\frac{{\sum }_{i=1}^8{f}_i|i-8|}{{\sum }_{i=1}^8{f}_i}=\frac{250}{78}=3.205$

By comparing the mean deviation, we can see that Karna is marginally better than Arjun.