Statistics & Linear ProgrammingeasyPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Mean Deviation About the Mean = 44/13 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The mean deviation about the mean for the data xi:5,7,9,10,12,15x_i:5,7,9,10,12,15 with frequencies fi:8,6,2,2,2,6f_i:8,6,2,2,2,6 is
A4013\dfrac{40}{13}
B4213\dfrac{42}{13}
C4413\dfrac{44}{13}correct
D4613\dfrac{46}{13}
Solution
Step 1: ∑fi=8+6+2+2+2+6=26\sum f_i=8+6+2+2+2+6=26. ∑fixi=40+42+18+20+24+90=234\sum f_ix_i=40+42+18+20+24+90=234.
xˉ=23426=9.\bar x=\frac{234}{26}=9.
 Step 2: ∣xi−9∣: 4,2,0,1,3,6|x_i-9|:\ 4,2,0,1,3,6.
∑fi∣xi−9∣=8(4)+6(2)+2(0)+2(1)+2(3)+6(6)=32+12+0+2+6+36=88.\sum f_i|x_i-9|=8(4)+6(2)+2(0)+2(1)+2(3)+6(6)=32+12+0+2+6+36=88.
 Step 3: M.D.=8826=4413\text{M.D.}=\dfrac{88}{26}=\dfrac{44}{13}. Correct answer: (3)
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