Only first (A) and second (B) statements are correct.

$$\eqalign{ & \because \,\sigma = \sqrt {\frac{{\sum {x{i^2}} }}{N} - {{\left( {\frac{{\sum {{x_i}} }}{N}} \right)}^2}} \cr & \therefore \,2 = \sqrt {\frac{{\left( {{a^2} + {a^2}......'2n'{\text{ times}}} \right)}}{{2n}} - 0} \cr & \Rightarrow 4 = \frac{{2n{a^2}}}{{2n}} \cr & \Rightarrow {a^2} = 4 \cr & \Rightarrow \left| a \right| = 2 \cr} $$

Mean $$ = \frac{{101 + d\left( {1 + 2 + 3 + ..... + 100} \right)}}{{101}}$$
$$\eqalign{ & = 1 + \frac{{d \times 100 \times 101}}{{101 \times 2}} \cr & = 1 + 50\,d \cr} $$
$$\because $$ Mean deviation from the mean = 255
$$\eqalign{ & \Rightarrow \,\,\frac{1}{{101}}\left[ {\left| {1 - \left( {1 + 50d} \right)\left| {\, + \,} \right|\left( {1 + d} \right) - \left( {1 + 50d} \right)\left| {\, + \,} \right|\left( {1 + 2d} \right) - \left( {1 + 50d} \right)\left| { + ..... + } \right|\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|\,} \right] & = 255 \cr & \Rightarrow \,\,2d\left[ {1 + 2 + 3 + ..... + 50} \right] = 101 \times 255 \cr & \Rightarrow \,\,2d \times \frac{{50 \times 51}}{2} = 101 \times 255 \cr & \Rightarrow \,\,d = \frac{{101 \times 255}}{{50 \times 51}} = 10.1 \cr} $$

$$\eqalign{ & {\text{The mean of the given items}} \cr & \overline x = \frac{{3 + 7 + 10 + 18 + 22}}{5} = 12 \cr & {\text{Hence, variance is :}} \cr & = \frac{1}{n}{\sum {\left( {{x_i} - \overline x } \right)} ^2} \cr & = \frac{1}{5}\left\{ {{{\left( {3 - 12} \right)}^2} + {{\left( {7 - 12} \right)}^2} + {{\left( {10 - 12} \right)}^2} + {{\left( {18 - 12} \right)}^2} + {{\left( {22 - 12} \right)}^2}} \right\} \cr & = \frac{1}{5}\left\{ {81 + 25 + 4 + 36 + 100} \right\} \cr & = \frac{{246}}{5} \cr & = 49.2 \cr} $$

$$\eqalign{ & {\text{For Group A :}} \cr & {\text{Coefficient of variation}} \cr & C{V_A} = \frac{{{\text{S}}{\text{.D}}{\text{.}}}}{{{\text{Mean}}}} = \frac{{10}}{{22}} = 0.4545 \cr & {\text{For Group B :}} \cr & C{V_B} = \frac{{12}}{{23}} = 0.522 \cr & \Rightarrow {\text{Group A is less variable}}{\text{.}} \cr} $$

Let the number of boys be $$x$$ and that of girls be $$y.$$
$$\eqalign{ & \Rightarrow \,\,52x + 42y = 50\left( {x + y} \right) \cr & \Rightarrow \,\,52x - 50x = 50y - 42y \cr & \Rightarrow \,\,2x = 8y \cr & \Rightarrow \,\,\frac{x}{y} = \frac{4}{1}\,{\text{and }}\frac{x}{{x + y}} = \frac{4}{5} \cr} $$
Required % of boys $$ = \frac{x}{{x + y}} \times 100 = \frac{4}{5} \times 100 = 80\% $$

We construct the following table taking assumed mean $$a = 55$$  (step deviation method).

Class	$${x_i}$$	$${f_i}$$	$$c.f.$$	$${u_i} = \frac{{{x_i} - a}}{{10}}{f_i}{u_i}$$
10 - 20	15	2	2	- 4 - 8
20 - 30	25	3	5	- 3 - 9
30 - 40	35	4	9	- 2 - 8
40 - 50	45	5	14	- 1 - 5
50 - 60	55	6	20	0    0
60 - 70	65	12	32	1    12
70 - 80	75	14	46	2    28
80 - 90	85	10	56	3    30
90 - 100	95	4	60	4    16
Total		60		56

$$\eqalign{ & \therefore \,{\text{The mean is :}} \cr & = a + \frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }} \times c \cr & = 55 + \frac{{56}}{{60}} \times 10 \cr & = 55 + \frac{{56}}{6} \cr & = 64.333 \cr & {\text{Here, }}n = 60 \Rightarrow \frac{n}{2} = 30, \cr & {\text{therefore, }}60 - 70{\text{ is the median class}} \cr & {\text{Using the formula :}} \cr & M = l + \frac{{\frac{n}{2} - C}}{f} \times c \cr & = 60 + \frac{{30 - 20}}{{12}} \times 10 \cr & = 68.333 \cr & {\text{Using Empirical formula, we have}} \cr & {\text{Mode}} = 3{\text{Median}} - 2{\text{Mean}} \cr & = \left( {68.333} \right) - 2\left( {64.333} \right) \cr & = 204.999 - 128.666 \cr & = 76.333 \cr} $$

$$\eqalign{ & {\text{Since, }}MD = \frac{4}{5}\sigma ,\,QD = \frac{2}{3}\sigma \cr & \Rightarrow \frac{{MD}}{{QD}} = \frac{6}{5} \cr & \Rightarrow QD = \frac{5}{6}\left( {MD} \right) = \frac{5}{6}.\left( {15} \right) = 12.5 \cr} $$

As given, $$np = 4$$  and $$npq = 3$$
[where $$p$$ is the probability of success and $$q$$ is the probability of failure for an event to occur, and $$'n'$$ is the number of trials]
$$\eqalign{ & \Rightarrow q = \frac{{npq}}{{np}} = \frac{3}{4} \cr & {\text{Also, }}p = 1 - q = 1 - \frac{3}{4} = \frac{1}{4} \cr & \therefore \,n = 16 \cr} $$
In a binomial distribution, the value of $$r$$ for which $$P\left( {X = r} \right)$$   is maximum is the mode of binomial distribution.
$$\eqalign{ & {\text{Hence, }}\left( {n + 1} \right)p - 1 \leqslant r \leqslant \left( {n + 1} \right)p \cr & \Rightarrow \frac{{17}}{4} - 1 \leqslant r \leqslant \frac{{17}}{4} \cr & \Rightarrow \frac{{13}}{4} \leqslant r \leqslant \frac{{17}}{4} \cr & \Rightarrow 3.25 \leqslant r \leqslant 4.25 \cr & \Rightarrow r = 4 \cr} $$

We know that for positive real numbers $${x_1},{x_2},......\,{x_n}$$    A.M. of $${k^{th}}$$ power of $$x{'_i}s \geqslant {k^{th}}$$   the power of A.M. of $$x{'_i}s$$
$$\eqalign{ & \Rightarrow \,\,\frac{{\sum {x_1^2} }}{n} \geqslant {\left( {\frac{{\sum {{x_1}} }}{n}} \right)^2} \cr & \Rightarrow \,\,\frac{{400}}{n} \geqslant {\left( {\frac{{80}}{n}} \right)^2} \cr} $$
$$ \Rightarrow \,\,n \geqslant 16.$$   So only possible value for $$n$$ = 18