wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 bull's eye.

Points87654321Arjun(fi)571423111043Karna(fi)58112414853


A

Arjun

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

Karna

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Both of them

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

Insufficient data

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

Arjun


Let xi be score of ith circle strip from center.

Absolute deviation of ith strip= |i-8| = 8-i.

Deviations for Arjun Karna would be

Deviation01234567Arjun(fi)571423111043Karna(fi)58112414853
Sum of Absolute deviation for Karana = 8i=1fi|i8|
=5(0) + 8(1) + 11(2) + 24(3) + 14(4) + 8(5) + 5(6) + 3(7) = 320
Sum of Absolute deviation for Karana = 220
Mean Deviation about score 8 for Arjun = 8i=1fi|i8|8i=1fi=22077=2.85
Mean Deviation about score for Karana = 8i=1fi|i8|8i=1fi=32077=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.


flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mean Deviation about Mean for Discrete Frequency Distributions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon