1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The mean deviation about the median for the following data Marks0âˆ’1010âˆ’2020âˆ’3030âˆ’4040âˆ’5050âˆ’60Number of students68141642 is

A
9.35
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
10.34
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
11.35
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
12.33
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B 10.34ClassMid points(xi)fic.f.|xi−M|fi|xi−M|0−1056622.85137.110−201581412.85102.820−302514282.8539.930−403516447.15114.440−504544817.1568.650−605525027.1554.3 The class interval containing N2th or 25th term is 20−30. Therefore 20−30 is the median class. Median =l+N2−f1f×h =20+25−1414×10 =20+1114×10 =20+7.85=27.85 Mean deviation about median is given by 1N6∑1fi|xi−M| =150×517.1=10.34

Suggest Corrections
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program