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(C) Given sum of frequencies $N = 5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \implies f_1 + f_2 = 7$. Since median is $6$,the cumulative frequency at $x=6$ mu

Vedclass · Accepted Answer

$(P) \rightarrow (5), (Q) \rightarrow (3), (R) \rightarrow (2), (S) \rightarrow (1)$

Vedclass · Answer

(C) Given sum of frequencies $N = 5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \implies f_1 + f_2 = 7$. Since median is $6$,the cumulative frequency at $x=6$ must be at least $N/2 = 9.5$. Sorted values: $4(5), 5(f_1), 6(1), 8(f_2), 9(2), 11(3), 12(1)$. Cumulative frequencies: $5, 5+f_1, 6+f_1, 6+f_1+f_2, 8+f_1+f_2, 11+f_1+f_2, 12+f_1+f_2$. For median $6$,$5+f_1 Also,$f_1+f_2=7$. Testing values: if $f_1=4, f_2=3$,then $f_1+f_2=7$. Mean $\bar{x} = \frac{\sum x_i f_i}{19} = \frac{4(5) + 5(4) + 6(1) + 8(3) + 9(2) + 11(3) + 12(1)}{19} = \frac{20+20+6+24+18+33+12}{19} = \frac{133}{19} = 7$. $(P) \ 7f_1 + 9f_2 = 7(4) + 9(3) = 28 + 27 = 55$. Mean deviation about mean $\alpha = \frac{\sum f_i |x_i - 7|}{19} = \frac{5|4-7| + 4|5-7| + 1|6-7| + 3|8-7| + 2|9-7| + 3|11-7| + 1|12-7|}{19} = \frac{15+8+1+3+4+12+5}{19} = \frac{48}{19} \implies 19\alpha = 48$. Mean deviation about median $\beta = \frac{\sum f_i |x_i - 6|}{19} = \frac{5|4-6| + 4|5-6| + 1|6-6| + 3|8-6| + 2|9-6| + 3|11-6| + 1|12-6|}{19} = \frac{10+4+0+6+6+15+6}{19} = \frac{47}{19} \implies 19\beta = 47$. Variance $\sigma^2 = \frac{\sum f_i x_i^2}{19} - (\bar{x})^2 = \frac{5(16) + 4(25) + 1(36) + 3(64) + 2(81) + 3(121) + 1(144)}{19} - 49 = \frac{80+100+36+192+162+363+144}{19} - 49 = \frac{1077}{19} - 49 = \frac{1077-931}{19} = \frac{146}{19} \implies 19\sigma^2 = 146$. Thus,$(P)\rightarrow(5), (Q)\rightarrow(3), (R)\rightarrow(2), (S)\rightarrow(1)$.

Value	$4$	$5$	$8$	$9$	$6$	$12$	$11$
Frequency	$5$	$f_1$	$f_2$	$2$	$1$	$1$	$3$

List-$I$	List-$II$
$(P) \ 7f_1+9f_2$ is equal to	$(1) \ 146$
$(Q) \ 19\alpha$ is equal to	$(2) \ 47$
$(R) \ 19\beta$ is equal to	$(3) \ 48$
$(S) \ 19\sigma^2$ is equal to	$(4) \ 145$
	$(5) \ 55$

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