JEE Advanced 2025 Paper 1 · Q14 · Variance & SD
Frequency distribution: values $4, 5, 8, 9, 6, 12, 11$ with frequencies $5, f_1, f_2, 2, 1, 1, 3$. Sum of frequencies $= 19$, median $= 6$. Let $\alpha = $ mean deviation about the mean, $\beta = $ mean deviation about the median, $\sigma^2 = $ variance. List-I: (P) $7 f_1 + 9 f_2$; (Q) $19\alpha$; (R) $19\beta$; (S) $19\sigma^2$. List-II: (1) $146$; (2) $47$; (3) $48$; (4) $145$; (5) $55$. Choose: (A) P->5,Q->3,R->2,S->4; (B) P->5,Q->2,R->3,S->1; (C) P->5,Q->3,R->2,S->1; (D) P->3,Q->2,R->5,S->4.
Reveal answer + step-by-step solution
Correct answer:C
Solution
Sum of frequencies: $5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 12 + f_1 + f_2 = 19$ $\Rightarrow$ $f_1 + f_2 = 7$. Sort values: $4(5), 5(f_1), 6(1), 8(f_2), 9(2), 11(3), 12(1)$. Cumulative: at value $4$: $5$; at $5$: $5+f_1$; at $6$: $6+f_1$; at $8$: $6+f_1+f_2$; … Median of $19$ = the $10$th term. For median $= 6$ we need cumulative to first reach $\ge 10$ at value $6$, so $5 + f_1 < 10$ and $6 + f_1 \ge 10$ $\Rightarrow$ $f_1 = 4$ and $f_2 = 3$. Then $7 f_1 + 9 f_2 = 28 + 27 = 55$. (P) $\to$ $5$. Mean $= (4 \cdot 5 + 5 \cdot 4 + 6 \cdot 1 + 8 \cdot 3 + 9 \cdot 2 + 11 \cdot 3 + 12 \cdot 1)/19 = (20+20+6+24+18+33+12)/19 = 133/19 = 7$. $19\alpha = \sum f_i |x_i - 7| = 5 \cdot 3 + 4 \cdot 2 + 1 \cdot 1 + 3 \cdot 1 + 2 \cdot 2 + 3 \cdot 4 + 1 \cdot 5 = 15+8+1+3+4+12+5 = 48$. (Q) $\to$ $3$. Median $= 6$. $19\beta = \sum f_i |x_i - 6| = 5 \cdot 2 + 4 \cdot 1 + 1 \cdot 0 + 3 \cdot 2 + 2 \cdot 3 + 3 \cdot 5 + 1 \cdot 6 = 10+4+0+6+6+15+6 = 47$. (R) $\to$ $2$. $19\sigma^2 = \sum f_i (x_i - 7)^2 = 5 \cdot 9 + 4 \cdot 4 + 1 \cdot 1 + 3 \cdot 1 + 2 \cdot 4 + 3 \cdot 16 + 1 \cdot 25 = 45+16+1+3+8+48+25 = 146$. (S) $\to$ $1$. Answer: (C).
Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →