Consider the following frequency distribution:
Value: 4 5 8 9 6 12 11
Frequency: 5 f₁ f₂ 2 1 1 3
Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.
For the given frequency distribution, let $\alpha$ denote the mean deviation about the mean, $\beta$ denote the mean deviation about the median, and $\sigma^2$ denote the variance.
Match each entry in List-I to the correct entry in List-II and choose the correct option.
**List-I**
(P) $7f_1 + 9f_2$ is equal to
(Q) $19\alpha$ is equal to
(R) $19\beta$ is equal to
(S) $19\sigma^2$ is equal to
**List-II**
(1) 146
(2) 47
(3) 48
(4) 145
(5) 55
(P)→(5) (Q)→(3) (R)→(2) (S)→(4)
(P)→(5) (Q)→(2) (R)→(3) (S)→(1)
(P)→(5) (Q)→(3) (R)→(2) (S)→(1)
(P)→(3) (Q)→(2) (R)→(5) (S)→(4)
Step-by-Step Solution
Key Concept: Finding frequencies from median condition; then computing mean, mean deviation, variance
Sum of frequencies: $5+f_1+f_2+2+1+1+3=19 \Rightarrow f_1+f_2=7$.
Median = 6. Arrange values in order: 4(×5), 5(×f₁), 6(×1), 8(×f₂), 9(×2), 11(×3), 12(×1). The 10th value is the median. Cumulative: up to 4: 5; up to 5: 5+f₁; up to 6: 6+f₁. For median=6, the 10th value must fall in the '6' group: $5+f_1 < 10 \leq 6+f_1$, giving $f_1 \geq 4$ and $f_1 < 5$, so $f_1=4$, $f_2=3$.
(P) $7f_1+9f_2=28+27=55$ → (5). ✓
Data: 4(×5), 5(×4), 6(×1), 8(×3), 9(×2), 11(×3), 12(×1). $n=19$.
Mean $\bar{x} = \dfrac{5(4)+4(5)+1(6)+3(8)+2(9)+3(11)+1(12)}{19} = \dfrac{20+20+6+24+18+33+12}{19} = \dfrac{133}{19} = 7$.
$19\alpha = \sum f_i|x_i - 7| = 5(3)+4(2)+1(1)+3(1)+2(2)+3(4)+1(5) = 15+8+1+3+4+12+5=48$ → (3). ✓
Median = 6. $19\beta = \sum f_i|x_i-6| = 5(2)+4(1)+1(0)+3(2)+2(3)+3(5)+1(6)=10+4+0+6+6+15+6=47$ → (2). ✓
$19\sigma^2 = \sum f_i x_i^2 - 19\bar{x}^2/1$... Actually $\sigma^2 = \dfrac{\sum f_i x_i^2}{19} - \bar{x}^2$.
$\sum f_i x_i^2 = 5(16)+4(25)+1(36)+3(64)+2(81)+3(121)+1(144)=80+100+36+192+162+363+144=1077$.
$\sigma^2 = \dfrac{1077}{19}-49 = \dfrac{1077-931}{19}=\dfrac{146}{19}$.
$19\sigma^2=146$ → (1). ✓
Answer: C
Correct Answer: C