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. Score for each outcome decreases by unity with centermost circle having score of 8. 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 assuming the winner is less deviated from the bulls eye.

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

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

Answer 1

The correct option is A

Arjun

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 Karna would be

$$\begin{array}{ccccccccc}Deviation& 0& 1& 2& 3& 4& 5& 6& 7\\ Arjun\left(f_i\right)& 5& 7& 14& 23& 11& 10& 4& 3\\ karna\left(f_i\right)& 5& 8& 11& 24& 14& 8& 5& 3\end{array}$$
Sum of Absolute deviation for Karan = ${\sum }_{i=1}^8{f}_i|i-8|$
=5(0) + 8(1) + 11(2) + 24(3) + 14(4) + 8(5) + 5(6) + 3(7) = 320
Mean Deviation about score 8 for Arjun = $\frac{{\sum }_{i=1}^8{f}_i|i-8|}{{\sum }_{i=1}^8{f}_i}=\frac{220}{77}=2.85$
Mean Deviation about score for Karna = $\frac{{\sum }_{i=1}^8{f}_i|i-8|}{{\sum }_{i=1}^8{f}_i}=\frac{320}{77}=3.2$

From the mean deviation, we can see Arjun was less deviated from the central spot of the target compared to Karna. Therefore the winner is Arjun.