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(C) The sum of frequencies is given by: $\sum f_i = (\alpha+2) + (\alpha-1)^2 + 4 + (\alpha-1) + 2 + \alpha = 20$. Let $y = \alpha-1$. Then $\alpha =

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$2.7$

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(C) The sum of frequencies is given by: $\sum f_i = (\alpha+2) + (\alpha-1)^2 + 4 + (\alpha-1) + 2 + \alpha = 20$. Let $y = \alpha-1$. Then $\alpha = y+1$. The equation becomes: $(y+3) + y^2 + 4 + y + 2 + (y+1) = 20$. $y^2 + 4y + 10 = 20 \implies y^2 + 4y - 10 = 0$. This does not yield an integer. Re-evaluating the sum: $(\alpha+2) + (\alpha^2-2\alpha+1) + 4 + \alpha - 1 + 2 + \alpha = \alpha^2 + \alpha + 8 = 20 \implies \alpha^2 + \alpha - 12 = 0$. $(\alpha+4)(\alpha-3) = 0$. Since frequency must be positive,$\alpha = 3$. The frequencies are: $f_1=5, f_2=4, f_3=4, f_4=2, f_5=2, f_6=3$. Total $N=20$. Mean $\bar{x} = \frac{\sum f_i x_i}{N} = \frac{3(5) + 4(4) + 5(4) + 8(2) + 10(2) + 11(3)}{20} = \frac{15+16+20+16+20+33}{20} = \frac{120}{20} = 6$. Mean deviation about mean $MD = \frac{\sum f_i |x_i - \bar{x}|}{N} = \frac{5|3-6| + 4|4-6| + 4|5-6| + 2|8-6| + 2|10-6| + 3|11-6|}{20}$. $MD = \frac{5(3) + 4(2) + 4(1) + 2(2) + 2(4) + 3(5)}{20} = \frac{15+8+4+4+8+15}{20} = \frac{54}{20} = 2.7$.

$\text{Size}$	$20$	$21$	$22$	$23$	$24$
$\text{Frequency}$	$6$	$4$	$5$	$1$	$4$

Age	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$5$	$8$	$5$	$12$	$X$	$Y$

The following data represents the frequency distribution of $20$ observations.
$x_i$ $3$ $4$ $5$ $8$ $10$ $11$
$f_i$ $\alpha+2$ $(\alpha-1)^2$ $4$ $\alpha-1$ $2$ $\alpha$
Then its mean deviation about the mean is

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The following data represents the frequency distribution of $20$ observations. $x_i$$3$$4$$5$$8$$10$$11$$f_i$$\alpha+2$$(\alpha-1)^2$$4$$\alpha-1$$2$$\alpha$ Then its mean deviation about the mean is