JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q8

MathematicsSingle correct+4 / -1Question 8 of 75

The mean deviation about the mean for the data

$x_i$	5	7	9	10	12	15
$f_i$	8	6	2	2	2	6

is equal to:

\dfrac{40}{13}

\dfrac{42}{13}

\dfrac{44}{13}

\dfrac{46}{13}

Answer and worked solution

Official answer: C

Mean. $N=26$ , $\sum f_ix_i=40+42+18+20+24+90=234$ , so $\bar x=9$ .

Deviations $|x_i-9|$ : 4, 2, 0, 1, 3, 6.
$\sum f_i|x_i-9|=32+12+0+2+6+36=88.$

Mean deviation: $\dfrac{88}{26}=\dfrac{44}{13}$ .

(C)

JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 8