If M_1 and M_2 are the mean deviations from t | vedclass

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(B) The first $15$ even integers are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30$. Mean $\bar{x} = \frac{2+4+6+\dots+30}{15} = \frac{2(1+2

Vedclass · Accepted Answer

$\frac{224}{15}$

Vedclass · Answer

(B) The first $15$ even integers are $2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30$. Mean $\bar{x} = \frac{2+4+6+\dots+30}{15} = \frac{2(1+2+\dots+15)}{15} = \frac{2 	imes \frac{15 	imes 16}{2}}{15} = 16$. Median $	ilde{x} = \left(\frac{15+1}{2}ight)^{	ext{th}}$ term $= 8^{	ext{th}}$ term $= 16$. $M_1 = 	ext{mean deviation about mean} = \frac{\sum_{i=1}^{15} |x_i - \bar{x}|}{15} = \frac{|2-16| + |4-16| + \dots + |30-16|}{15}$. $M_1 = \frac{14+12+10+8+6+4+2+0+2+4+6+8+10+12+14}{15} = \frac{112}{15}$. Since the mean and median are equal,$M_2 = 	ext{mean deviation about median} = M_1 = \frac{112}{15}$. Therefore,$M_1 + M_2 = \frac{112}{15} + \frac{112}{15} = \frac{224}{15}$.

Class Interval	$0-4$	$4-8$	$8-12$	$12-16$
Frequency	$4$	$3$	$2$	$1$

Marks obtained	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
No of students	$2$	$3$	$8$	$14$	$8$	$3$	$2$

If $M_1$ and $M_2$ are the mean deviations from the mean and median of the first $15$ even integers,then $M_1+M_2=$

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