Let X = {11, 12, 13, ldots, 40, 41} and Y = { | vedclass

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(A) The mean of set $X$ is $\bar{x} = \frac{11+41}{2} = 26$ (number of elements $n_1 = 31$).The mean of set $Y$ is $\bar{y} = \frac{61+91}{2} = 76$ (n

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$603$

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(A) The mean of set $X$ is $\bar{x} = \frac{11+41}{2} = 26$ (number of elements $n_1 = 31$).The mean of set $Y$ is $\bar{y} = \frac{61+91}{2} = 76$ (number of elements $n_2 = 31$).The combined mean $\mu = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2} = \frac{31(26) + 31(76)}{62} = \frac{26+76}{2} = 51$.The variance $\sigma^2$ of the combined set is given by $\sigma^2 = \frac{1}{n_1+n_2} \left( \sum_{i=1}^{31} (x_i - \mu)^2 + \sum_{j=1}^{31} (y_j - \mu)^2 \right)$.For set $X$,$\sum (x_i - \mu)^2 = \sum_{i=11}^{41} (i - 51)^2 = (-40)^2 + (-39)^2 + \ldots + (-10)^2 = \sum_{k=10}^{40} k^2 = \sum_{k=1}^{40} k^2 - \sum_{k=1}^{9} k^2 = \frac{40(41)(81)}{6} - \frac{9(10)(19)}{6} = 22140 - 285 = 21855$.Similarly,for set $Y$,$\sum (y_j - \mu)^2 = \sum_{j=61}^{91} (j - 51)^2 = 10^2 + 11^2 + \ldots + 40^2 = 21855$.Thus,$\sigma^2 = \frac{21855 + 21855}{62} = \frac{43710}{62} = 705$.Finally,$|\bar{x} + \bar{y} - \sigma^2| = |26 + 76 - 705| = |102 - 705| = |-603| = 603$.

List-$I$	List-$II$
$(a) \sum_{i=1}^n(x_i-\bar{x})$	$(i) \text{ Median}$
$(b) \text{ Variance } (\sigma^2)$	$(ii) \text{ Coefficient of variation}$
$(c) \text{ Mean deviation}$	$(iii) \text{ Zero}$
$(d) \text{ Measure used to find the homogeneity of given two series}$	$(iv) \text{ Mean of the absolute deviations from any measure of central tendency}$
	$(v) \text{ Mean of the squares of the deviations from mean}$

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