If M and sigma^2 represent respectively the m | vedclass

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(D) The given data is $1, 3, 5, 7, 11, 13, 17, 19, 23$. The number of observations $n = 9$. The mean $\bar{x} = \frac{1+3+5+7+11+13+17+19+23}{9} = \fr

Vedclass · Accepted Answer

$136$

Vedclass · Answer

(D) The given data is $1, 3, 5, 7, 11, 13, 17, 19, 23$. The number of observations $n = 9$. The mean $\bar{x} = \frac{1+3+5+7+11+13+17+19+23}{9} = \frac{99}{9} = 11$. The mean deviation from the mean $M = \frac{\sum |x_i - \bar{x}|}{n} = \frac{|1-11| + |3-11| + |5-11| + |7-11| + |11-11| + |13-11| + |17-11| + |19-11| + |23-11|}{9} = \frac{10+8+6+4+0+2+6+8+12}{9} = \frac{56}{9}$. The variance $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{(-10)^2 + (-8)^2 + (-6)^2 + (-4)^2 + 0^2 + 2^2 + 6^2 + 8^2 + 12^2}{9} = \frac{100 + 64 + 36 + 16 + 0 + 4 + 36 + 64 + 144}{9} = \frac{464}{9}$. Now,$3(\sigma^2 - M) = 3 \left( \frac{464}{9} - \frac{56}{9} \right) = 3 \left( \frac{408}{9} \right) = \frac{408}{3} = 136$.

Class	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Frequency	$2$	$k$	$28$	$54$	$k + 1$	$5$

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

If $M$ and $\sigma^2$ represent respectively the mean deviation from the mean and the variance for the data $1, 3, 5, 7, 11, 13, 17, 19, 23$,then $3(\sigma^2 - M) = $

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