Consider the given data with frequency distri | vedclass

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(A) First,arrange the data in ascending order of $x_i$:$x_i: 3, 4, 5, 8, 10, 11$$f_i: 5, 4, 4, 2, 2, 3$Total frequency $N = \Sigma f_i = 5 + 4 + 4 + 2

Vedclass · Accepted Answer

$(P) \rightarrow (3), (Q) \rightarrow (2), (R) \rightarrow (4), (S) \rightarrow (5)$

Vedclass · Answer

(A) First,arrange the data in ascending order of $x_i$:$x_i: 3, 4, 5, 8, 10, 11$$f_i: 5, 4, 4, 2, 2, 3$Total frequency $N = \Sigma f_i = 5 + 4 + 4 + 2 + 2 + 3 = 20$.$(P)$ Mean $(\bar{x}) = \frac{\Sigma f_i x_i}{N} = \frac{(3 	imes 5) + (4 	imes 4) + (5 	imes 4) + (8 	imes 2) + (10 	imes 2) + (11 	imes 3)}{20} = \frac{15 + 16 + 20 + 16 + 20 + 33}{20} = \frac{120}{20} = 6$.$(Q)$ Median: Since $N=20$ (even),the median is the average of the $10^{th}$ and $11^{th}$ observations. Cumulative frequencies are $5, 9, 13, 15, 17, 20$. Both $10^{th}$ and $11^{th}$ observations fall in the value $5$. Thus,Median $= 5$.$(R)$ Mean deviation about mean $= \frac{\Sigma f_i |x_i - 6|}{N} = \frac{5|3-6| + 4|4-6| + 4|5-6| + 2|8-6| + 2|10-6| + 3|11-6|}{20} = \frac{5(3) + 4(2) + 4(1) + 2(2) + 2(4) + 3(5)}{20} = \frac{15 + 8 + 4 + 4 + 8 + 15}{20} = \frac{54}{20} = 2.7$.$(S)$ Mean deviation about median $= \frac{\Sigma f_i |x_i - 5|}{N} = \frac{5|3-5| + 4|4-5| + 4|5-5| + 2|8-5| + 2|10-5| + 3|11-5|}{20} = \frac{5(2) + 4(1) + 4(0) + 2(3) + 2(5) + 3(6)}{20} = \frac{10 + 4 + 0 + 6 + 10 + 18}{20} = \frac{48}{20} = 2.4$.Matching: $(P) ightarrow 3, (Q) ightarrow 2, (R) ightarrow 4, (S) ightarrow 5$. Correct option is $(A)$.

List-$I$	List-$II$
$(P)$ The mean of the above data is	$(1) 2.5$
$(Q)$ The median of the above data is	$(2) 5$
$(R)$ The mean deviation about the mean of the above data is	$(3) 6$
$(S)$ The mean deviation about the median of the above data is	$(4) 2.7$
	$(5) 2.4$

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