Maths M.C.Q (1 Marks)

$\Rightarrow a+b=17$

$\frac{20}{3}=\frac{7^{2}+10^{2}+11^{2}+15^{2}+a^{2}+b^{2}}{6}-10^{2}$

$a^{2}+b^{2}=145$

Solve $(i)$ and $(ii)$ $\mathrm{a}=9, \mathrm{~b}=8$ or $\mathrm{a}=8, \mathrm{~b}=9$

$|a-b|=1$

mean $=\frac{a+b+c}{3}(=\bar{x})$

$b = a + c$

$\Rightarrow \quad \bar{x}=\frac{2 b}{3}$  $.....(1)$

S.D. $(a+2, b+2, c+2)=$ S.D. $(a, b, c)=d$

$\Rightarrow \quad d ^{2}=\frac{ a ^{2}+ b ^{2}+ c ^{2}}{3}-(\overline{ x })^{2}$

$\Rightarrow \quad d^{2}=\frac{a^{2}+b^{2}+c^{2}}{3}-\frac{4 b^{2}}{9}$

$\Rightarrow 9 d^{2}=3\left(a^{2}+b^{2}+c^{2}\right)-4 b^{2}$

$\Rightarrow \quad b^{2}=3\left(a^{2}+c^{2}\right)-9 d^{2}$

Now $\frac{x_{1}+x_{2}+\ldots+x_{5}+14}{7}=8....(i)$

$\Rightarrow \sum_{i=1}^{5} x_{i}=42$

Also $\frac{x_{1}^{2}+x_{2}^{2}+\ldots x_{5}^{2}+8^{2}+6^{2}}{7}-64=16$

$\Rightarrow \sum_{i=1}^{5} x_{i}^{2}=560-100=460....(ii)$

So variance of $x_{1}, x_{2}, \ldots, x_{5}$

$=\frac{460}{5}-\left(\frac{42}{5}\right)^{2}=\frac{2300-1764}{25}=\frac{536}{25}$

$60+a+b=72$

$a+b=12$

$\text { veriance }=\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)=\frac{37}{4}$

$\sum x_{i}^{2}=6^{2}+10^{2}+7^{2}+13^{2}+a^{2}+b^{2}+12^{2}+12^{2}$

$=a^{2}+b^{2}+642$

$\frac{a^{2}+b^{2}+642}{8}-(9)^{2}=\frac{37}{4}$

$\frac{a^{2}+b^{2}}{8}+\frac{321}{4}-81=\frac{37}{4}$

$\frac{a^{2}+b^{2}}{8}=81+\frac{37}{4}-\frac{321}{4}$

$\frac{a^{2}+b^{2}}{8}=81-71$

$\therefore a^{2}+b^{2}=80$

From $(1)$ $a^{2}+b^{2}+2 a b=144$

$80+2 a b=144 \therefore 2 a b=64$

$(a-b)^{2}=a^{2}+b^{2}-2 a b=80-64=16$

Let age of newly appointed teacher is $N$

$\Rightarrow 1000-60+ N =975$

$\Rightarrow N =35$ years

$\Rightarrow 2 \alpha+3 \beta=16 \quad \ldots \text { (i) }$

$\text { Also, } 4 \times 16+4 \times \alpha+9 \beta=(8+\alpha+\beta) \times 6.8$

$\Rightarrow 640+40 \alpha+90 \beta=544+68 \alpha+68 \beta$

$\Rightarrow 28 \alpha-22 \beta=96$

$\Rightarrow 14 \alpha-11 \beta=48 \quad \ldots (ii)$

from $(i)\, \, (ii)$

$\alpha=5 \, \,\beta=2$

so, new mean $=\frac{32+35+18}{15}=\frac{85}{15}=\frac{17}{3}$

Mean $=\frac{3 a+9 b+180+189+135}{a+b+26}=\frac{309}{22}$

$\Rightarrow 66 a+198 b+11088=309 a+309 b+8034$

$\Rightarrow 243 a+111 b=3054$

$\Rightarrow 81 a+37 b=1018 ....(1)$

Now, Median $=12+\frac{\frac{a+b+26}{2}-(a+b)}{2} \times 6=14$

$\Rightarrow \frac{13}{2}-\left(\frac{a+b}{4}\right)=2$

$\Rightarrow \frac{a+b}{4}=\frac{9}{2}$

$\Rightarrow a+b=18 \rightarrow(2)$

From equation $(1)\, and\,(2)$

$a=8, b=10$

$\therefore(a-b)^{2}=(8-10)^{2}$

Median $(\mathrm{m})=\ell+\left[\frac{\left(\frac{\mathrm{N}}{2}\right)-\mathrm{c}}{\mathrm{f}}\right] \times \mathrm{h}$

$\mathrm{N}=\frac{584}{2}=292$

$\mathrm{~m}=45=40+\left[\frac{292-(\alpha+164)}{30}\right] \times 10$

$45=40+\left(\frac{128-\alpha}{3}\right)$

$15=128-\alpha$

$\alpha=113$

$\beta=277$

$|\alpha-\beta|=|113-277|=164$

$n _{1}=10, n _{2}= n , \sigma_{1}^{2}=2, \sigma_{2}^{2}=1$

$\overline{ x }_{1}=2, \overline{ x }_{2}=3, \sigma^{2}=\frac{17}{9}$

$\frac{17}{9}=\frac{10 \times 2+ n }{ n +10}+\frac{10 n }{( n +10)^{2}}(3-2)^{2}$

$\frac{17}{9}=\frac{(n+20)(n+10)+10 n}{(n+10)^{2}}$

$17 n^{2}+1700+340 n=90 n+9\left(n^{2}+30 n+200\right)$

$8 n^{2}-20 n-100=0$

$2 n^{2}-5 n-25=0$

$(2 n+5)(n-5)=0 \Rightarrow n=\frac{-5}{2} \,(Rejected) , 5$

Hence $n =5$

$\sigma^{2}=\frac{\sum a^{2}+\sum b^{2}}{3 n}-(5)^{2}$

$\Rightarrow \sum a^{2}+\sum b^{2}=87 n$

Now, distribution becomes

$a _{1}+1, a _{2}+1, a _{3}+1, \ldots \ldots a _{2 n }+1, b _{1}-1,b_{2}-1 \ldots \ldots b_{n}-1$

Variance

$=\frac{\sum(a+1)^{2}+\sum(b-1)^{2}}{3 n}-\left(\frac{12 n+2 n+3 n-n}{3 n}\right)^{2}$

$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}$

$=\frac{\left(\sum a^{2}+2 n+2 \sum a\right)+\left(\sum b^{2}+n-2 \sum b\right)}{3 n}-\left(\frac{16}{3}\right)^{2}$

$=\frac{87 n+3 n+2(12 n)-2(3 n)}{3 n}-\left(\frac{16}{3}\right)^{2}$

$\Rightarrow k=\frac{108}{3}-\left(\frac{16}{5}\right)^{2}$

$\Rightarrow 9 k=3(108)-(16)^{2}=324-256=68$

$\bar{x}=\frac{\sum x_{i}}{2 n}=\frac{(a+a+\ldots+a)-(a+a+\ldots+a)}{2 n}$

$\Rightarrow \overline{ x }=0$

and $\sigma_{ x }^{2}=\frac{\sum x _{i}^{2}}{2 n }-(\overline{ x })^{2}=\frac{ a ^{2}+ a ^{2}+\ldots+ a ^{2}}{2 n }-0= a ^{2}$

$\Rightarrow \sigma_{x}=a$

Now, adding a constant $b$ then $\overline{ y }=\overline{ x }+ b =5$

$\Rightarrow b =5$

and $\sigma_{y}=\sigma_{x}$ (No change in S.D.) $\Rightarrow a=20$

$\Rightarrow a^{2}+b^{2}=425$

$=\frac{9+ k ^{2}}{10}-\left(\frac{9+ k }{10}\right)^{2}<10$

$90+10 k^{2}-81-k^{2}-18 k < 1000$

$9 k ^{2}-18 k -991 < 0$

$k^{2}-2 k < \frac{991}{9}$

$( k -1)^{2} < \frac{1000}{9}$

$\frac{-10 \sqrt{10}}{3} <  k -1 < \frac{10 \sqrt{10}}{3}$

$k < \frac{10 \sqrt{10}}{3}+1$

$k \leq 11$

Maximum value of $k$ is $11 .$

$\Rightarrow \sum_{i=1}^{18} x_{i}=18(\alpha+2), \sum_{i=1}^{18} x_{i}^{2}-2 \beta \sum_{i=1}^{18} x_{i}+18 \beta^{2}=90$

Hence $\sum x _{ i }^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$

Given $\frac{\sum x _{ i }^{2}}{18}-\left(\frac{\sum x _{ i }}{18}\right)^{2}=1$

$\Rightarrow 90-18 \beta^{2}+36 \beta(\alpha+2)-18(\alpha+2)^{2}=18$

$\Rightarrow 5-\beta^{2}+2 \alpha \beta+4 \beta-\alpha^{2}-4 \alpha-4=1$

$\Rightarrow(\alpha-\beta)^{2}+4(\alpha-\beta)=0 \Rightarrow|\alpha-\beta|=0$ or $4$

As $\alpha$ and $\beta$ are distinct $|\alpha-\beta|=4$

Mean $(\bar{x})=\frac{\Sigma x_{i}}{20}=10$

or $\Sigma \mathrm{x}_{\mathrm{i}}=200$ (incorrect)

or $200-25+35=210=\Sigma \mathrm{x}_{\mathrm{i}}$ (Correct)

Now correct $\bar{x}=\frac{210}{20}=10.5$

again given $S . D=2.5(\sigma)$

$\sigma^{2}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-(10)^{2}=(2.5)^{2}$

or $\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2125$ (incorrect)

or $\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2125-25^{2}+35^{2}$ $=2725$ (Correct)

$\therefore$ correct $\sigma^{2}=\frac{2725}{20}-(10.5)^{2}$

$\underline{\underline{\sigma}}^{2}=26$

or $\sigma=26$

$\therefore \underline{\alpha}=10.5, \beta=26$

$\operatorname{Var}(x)=10=\frac{3^{2}+7^{2}+x^{2}+y^{2}}{4}-25$

$140=49+9+x^{2}+y^{2}$

$x^{2}+y^{2}=82$

$x+y=10$

$\Rightarrow(x, y)=(9,1)$

Four numbers are $21,9,10,8$

$\text { Mean }=\frac{48}{4}=12$

$\bar{X}=2.3$

$-a+6 b=\frac{9}{10} \ldots (1)$

$\sum P_{i}=\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$a+b=\frac{4}{15} \ldots (2)$

From equation $(1)$ and $(2)$

$a=\frac{1}{10}, b=\frac{1}{6}$

$\sigma^{2}=\Sigma p_{i} x_{i}^{2}-(\bar{X})^{2}$

$\frac{1}{5}(4)+a(1)+\frac{1}{3}(9)+\frac{1}{5}(16)+b(36)-(2.3)^{2}$

$=\frac{4}{5}+a+3+\frac{16}{5}+36 b-(2.3)^{2}$

$=4+a+3+36 b-(2.3)^{2}$

$=7+a+36 b-(2.3)^{2}$

$=7+\frac{1}{10}+6-(2.3)^{2}$

$=13+\frac{1}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{131}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{1310-(23)^{2}}{100}$

$=\frac{1310-529}{100}$

$=\frac{781}{100}$

$100 \sigma^{2}=781$

$\therefore n_{2}=250-100 \Rightarrow n_{2}=150$

$\bar{x}=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$

$15.6=\frac{100(15)+(150)\left(\bar{x}_{2}\right)}{250}$

$\Rightarrow \overline{\mathrm{x}}_{2}=16$

$\overline{\mathrm{X}}_{1}=15 \quad\quad\quad\quad \Rightarrow$

$\sigma_{1}^{2}=V_{1}(x)=9 \quad \sigma^{2}=\operatorname{Var}(x)=13.44$

$\sigma^{2}=\frac{\mathrm{n}_{1} \sigma_{1}^{2}+\mathrm{n}_{2} \sigma_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}$

$\mathrm{n}_{2}=150, \overline{\mathrm{x}}_{2}=16, \mathrm{~V}_{2}(\mathrm{x})=\sigma_{2}$

$13.44=\frac{100 \times 9+150 \times \sigma_{2}^{2}}{250}+\frac{100 \times 150}{(250)^{2}} \times 1$

$\Rightarrow \sigma_{2}^{2}=16 \Rightarrow \sigma_{2}=4$

$\bar{x}_{b}=12 \quad\quad\quad\quad\quad\quad\quad\quad n_{2}=$ no. of girls

$\sigma_{\mathrm{g}}^{2}=2$

$\bar{x}_{g}=\frac{50 \times 15-12 \times \sigma_{b}}{30}=\frac{750-12 \times 20}{30}=17=\mu$

variance of combined series

$\sigma^{2}=\frac{n_{1} \sigma_{b}^{2}+n_{2} \sigma_{g}^{2}}{n_{1}+n_{2}}+\frac{n_{1} \cdot n_{2}}{\left(n_{1}+n_{2}\right)^{2}}\left(\bar{x}_{b}-\bar{x}_{g}\right)^{2}$

$\sigma^{2}=\frac{20 \times 2+30 \times 2}{20+30}+\frac{20 \times 30}{(20+30)^{2}}(12-17)^{2}$

$\sigma^{2}=8$

$\Rightarrow \mu+\sigma^{2}=17+8=25$

$=\frac{9}{10}-\left(\frac{3}{10}\right)^{2}=\frac{81}{100}$

$S.D. =\frac{9}{10}$

Where $M$ and $m$ are upper and lower bounds

of values of any random variable.

$\therefore \quad \sigma^{2}<\frac{1}{4}(10-0)^{2}$

$\Rightarrow 0<\sigma<5$

$\therefore \sigma \neq 6$

$\Rightarrow y_{i}: 15 \quad 25 \quad 35 \;\; or\;\;-10 \quad 0 \quad 10$

$\Rightarrow f_{i}: 2 \quad \;\;\;x \quad \;2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2 \quad \;\;x \quad \;2$

$\Rightarrow \quad$ Variance $\left(\sigma^{2}\right)=\frac{\Sigma x _{ i }^{2} f _{ i }}{\Sigma f _{ i }}-(\overrightarrow{ x })^{2}$

$\Rightarrow \quad 50=\frac{200+0+200}{x+4}-0 \quad\{\bar{x}=0\}$

$\Rightarrow \quad 200+50 x=200+200$

$\Rightarrow \quad x=4$

$=\sqrt{\frac{ na }{ n }-\left(\frac{ n }{ n }\right)^{2}}$

$\left\{\right.$ Given $\left.\sum_{i=1}^{n}\left(x_{i}-a\right)=n \sum_{i=1}^{n}\left(x_{i}-a\right)^{2}=n a\right\}$

$=\sqrt{a-1}$

Mean $=\frac{\sum x_{i} f_{i}}{\sum f_{i}}=\frac{\sum_{x=1}^{n} 2^{x}{ }^{n} C_{x}}{\sum_{x=0}^{n}{ }^{n} C_{r}}$

Mean $=\frac{(1+2)^{n}-{ }^{n} C_{0}}{2^{n}}=\frac{728}{2^{n}}$

$\Rightarrow \frac{3^{n}-1}{2^{n}}=\frac{728}{2^{n}}$

$\Rightarrow 3^{n}=729 \Rightarrow n=6$

$\frac{3+5+7+a+b}{5}=5$

$a+b=10$

S.d. $=2 \Rightarrow \sqrt{\frac{\sum_{i=1}^{5}\left(x_{i}-\bar{x}\right)^{2}}{5}}=2$

$(3-5)^{2}+(5-5)^{2}+(7-5)^{2}+(a-5)^{2}+(b-5)^{2}=20$

$\Rightarrow 4+0+4+(a-5)^{2}+(b-5)^{2}=20$

$a^{2}+b^{2}-10(a+b)+50=12$

$(a+b)^{2}-2 a b-100+50=12$

$a b=19$

Equation is $x^{2}-10 x+19=0$

$x+y=16$

$\frac{\Sigma x^{2}}{n}-\left(\frac{\Sigma x}{n}\right)^{2}=25$

$3^{2}+7^{2}+9^{2}+12^{2}+13^{2}+20^{2}+\mathrm{x}^{2}+\mathrm{y}^{2}=1000$

$x^{2}+y^{2}=148$

$x y=54$

$\Rightarrow n=11$

and variance of first 'm' even natural numbers

$=4\left(\frac{\mathrm{m}^{2}-1}{12}\right) \Rightarrow \frac{\mathrm{m}^{2}-1}{3}=16 \Rightarrow \mathrm{m}=7$

$m+n=18$

$\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{20}-100=4 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^{2}=2080$

Actual mean $=\frac{200-9+11}{20}=\frac{202}{20}$

Variance $=\frac{2080-81+121}{20}-\left(\frac{202}{20}\right)^{2}=3.99$

and $2|p|=1 \Rightarrow p=\pm \frac{1}{2}\ldots(ii)$

so, $\mathrm{p}=-\frac{1}{2}$ and $\mathrm{q}=-20$

$\Rightarrow$ Mean of observation $\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{3}\left(\mathrm{x}_{\mathrm{i}}-5\right)=1$

$\Rightarrow \mu=$ mean of observation $\left(\mathrm{x}_{\mathrm{i}}-3\right)$

$=\left(\text { mean of observation }\left(\mathrm{x}_{\mathrm{i}}-5\right)\right)+2$

$=1+2=3$

Variance of observation

$\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-5\right)^{2}-\left(\mathrm{Mean} \text { of }\left(\mathrm{x}_{\mathrm{i}}-5\right)\right)^{2}=3$

$\Rightarrow \quad \lambda=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-3\right)$

$=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-5\right)=3$ $\therefore \quad(\mu, \lambda)=(3,3)$

$\mu=$ mean

$\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$

$\mu=17$

$\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17$

$\Rightarrow \quad 9 a+b=17$

$\sigma^{2}=216$

$\Rightarrow \quad \frac{\sum_{x=1}^{17}(a x+b-17)^{2}}{17}=216$

$\Rightarrow \frac{\sum_{x=1}^{17} a^{2}(x-9)^{2}}{17}=216$

$\Rightarrow \quad a^{2} 81-18 \times 9 a^{2}+a^{2} 3 \times(35)=216$

$\Rightarrow \quad$ From $(1), b=-10$

So, $a+b=-7$

$\Rightarrow \bar{x}=\frac{63+a+b}{8}=10 \Rightarrow a+b=17$

since, variance is independent of origin. So, we subtract 10 from each observation.

$So , \sigma^{2}=13.5=\frac{79+( a -10)^{2}+( b -10)^{2}}{8}-(10-10)^{2}$

$\Rightarrow a ^{2}+ b ^{2}-20( a + b )=-171$

$\Rightarrow a ^{2}+ b ^{2}=169 \quad \ldots(2)$

From

$(i) and (ii)$ $; a=12 \& b=5$

$x+y=14$

$(\sigma)^{2}=\frac{\sum\left( x _{ i }\right)^{2}}{ n }-\left(\frac{\sum x _{ i }}{ n }\right)^{2}$

$16=\frac{4+16+100+144+196+x^{2}+y^{2}}{7}-8^{2}$

$16+64=\frac{460+x^{2}+y^{2}}{7}$

$560=460+x^{2}+y^{2}$

$x^{2}+y^{2}=100$      (ii)

Clearly by (i) and (ii), $|x-y|=2$

$\because$ $\sum {{f_i}}  = {\left( {x + 1} \right)^2} + \left( {2x - 5} \right) + \left( {{x^2} - 3x} \right) + x = 20$

$ \Rightarrow x = 3, - 4$ (rejected)

$\therefore \bar x = \frac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }} = 2.8$

$ \Rightarrow \frac{{10 + 22 + 26 + 29 + 34 + x + 42 + 67 + 70 + y}}{{10}} = 45$

$ \Rightarrow x + y = 120\,\,\,\,......\left( i \right)$

and median $=35$

$ \Rightarrow \frac{{34 + x}}{2} = 35 \Rightarrow x = 36$

from $(i)$ $y = 84$

$\frac{y}{x} = \frac{{84}}{{36}} = \frac{7}{3}$

$ \Rightarrow \sum {{x_i} = 30\left( {50} \right) + 50 \Rightarrow \frac{{\sum {{x_i}} }}{{50}}}  = 31$

${\sum {\left( {{x_i} - 1} \right)} ^2} = 5n\,\,\,\,\,\,....\left( 2 \right)$

$\left( 1 \right) + \left( 2 \right) \Rightarrow \sum {\left( {x_i^2 + 1} \right)}  = 7n$

$ \Rightarrow \frac{{\sum {x_i^2} }}{n} = 6$

$\left( 1 \right).\left( 2 \right) \Rightarrow 4\sum {{x_i}}  = 4n$

$ \Rightarrow \sum {{x_i} = n} $

$ \Rightarrow \frac{{\sum {{x_i}} }}{n} = 1$

$ \Rightarrow $ variance $=6-1=5$

$ \Rightarrow $standard diviation $ = \sqrt 5 $

$25 = 12 + x + y \Rightarrow x + y = 13\,\,\,\,\,\,\,\,........\left( 1 \right)$

${\sigma ^2} = \frac{{\sum {{{\left( {{x_i} - \mu } \right)}^2}} }}{N}$

$9.2 = \frac{{1 + 9 + 64 + {x^2} + {y^2}}}{5} - 25$

$34.2 \times 5 = 74 + {x^2} + {y^2}$

$171 = 74 + {x^2} + {y^2}$

$97 = {x^2} + {y^2}..........\left( 2 \right)$

${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$

$169 - 97 = 2xy \Rightarrow xy = 36$

$T = 4,9$

So, ratio is $\frac{4}{9}$ or $\frac{9}{4}$

${\left( 3 \right)^2} = \frac{1}{5}\left( {e{x^2} - \frac{{{{\left( {ex} \right)}^2}}}{5}} \right)$

$9 = \frac{1}{5}\left( {\sum {{x^2} - \frac{{2500}}{5}} } \right)$

$\therefore \sum {{x^2} = 545} $

New variable $ = \frac{1}{6}\left( {3045 - \frac{0}{6}} \right) = 507.5$

$\therefore \frac{{1{{\left( { - d} \right)}^2} + 10{{\left( 0 \right)}^2} + 10{{\left( d \right)}^2}}}{{30}} - {\left( {\frac{{10\left( { - d} \right) + 10\left( 0 \right) + 10\left( d \right)}}{{30}}} \right)^2} = \frac{4}{3}$

$\frac{{20{d^2}}}{{30}} = \frac{4}{3}$

$ \Rightarrow {d^2} = 2$

$\left( d \right) = \sqrt 2 $

$\sum {{x_i} = 20} $

${x_4} + {x_5} = 9\,\,\,\,\,\,\,........\left( i \right)$

$\frac{{\sum {x_i^2} }}{x} - {\left( {\bar x} \right)^2} = {\sigma ^2} \Rightarrow \sum {x_i^2}  = 106$

$x_4^2 + x_5^2 = 65\,\,\,\,\,\,\,\,........\left( {ii} \right)$

Using $(i)$ and $(ii)$ ${\left( {{x_4} - {x_5}} \right)^2} = 49$

$\left| {{x_4} - {x_5}} \right| = 7$

$\bar x = 8 \Rightarrow \sum\limits_{i = 1}^7 {{x_i}}  = 56\,\,\,\,\,\,.......\left( 1 \right)$

Also ${\sigma ^2} = 16$

$ \Rightarrow 16 = \frac{1}{7}\left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) - {\left( {\bar x} \right)^2}$

$ \Rightarrow 16 = \frac{1}{7}\left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) - 64$

$ \Rightarrow \left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) = 560\,\,\,\,\,\,\,\,\,.......\left( 2 \right)$

Now, ${x_1} = 2,{x_2} = 4,{x_3} = 10,{x_4} = 12,{x_5} = 14$

$ \Rightarrow {x_6} + {x_7} = 14$ (from $(1)$) and $x_6^2 + x_7^2 = 100$ (from$(2)$)

$\therefore x_6^2 + x_7^2 = {\left( {{x_6} + {x_7}} \right)^2} - 2{x_6}{x_7} \Rightarrow {x_6}{x_7} = 48$

$\bar x = \frac{{\sum x }}{4} = \frac{{ - 1 + 0 + 1 + k}}{4} = \frac{k}{4}$

Now $\sqrt 5 = \sqrt {\frac{{{{\left( { - 1 - \frac{k}{4}} \right)}^2} + {{\left( {0 - \frac{k}{4}} \right)}^2} + {{\left( {1 - \frac{k}{4}} \right)}^2} + {{\left( {k - \frac{k}{4}} \right)}^2}}}{4}} $

$ \Rightarrow 5 \times 4 = 2{\left( {1 + \frac{k}{{16}}} \right)^2} + \frac{{5{k^2}}}{8}$

$ \Rightarrow 18 = \frac{{3{k^2}}}{4}$

$ \Rightarrow {k^2} = 24$

$ \Rightarrow k = 2\sqrt 6 $

Standard deviation $\left( \sigma  \right) = \sqrt {\frac{{\sum {x_i^2} }}{{50}} - {{\left( \mu  \right)}^2}}  = 16$

$ \Rightarrow \left( {256} \right) \times 2 = \frac{{\sum {x_i^2} }}{{50}}$

$\Rightarrow$ New mean

$ = \frac{{\sum {{{\left( {{x_i} - 4} \right)}^2}} }}{{50}} = \frac{{\sum {x_i^2 + 16 \times 50 - 8\sum {{x_i}} } }}{{50}}$

$ = \left( {256} \right) \times 2 + 16 - 8 \times 16 = 400$

${x_5} + ... + {x_{10}} = 96$

$\bar x = 14,\sum {{x_i} = 140} $

Variance $ = \frac{{\sum {x_i^2} }}{n} - {{\bar x}^2} = 4$

Standard deviation $=2$

Given $\bar x = \frac{{\sum {{x_i}} }}{5} = 150\,\,\,\, \Rightarrow \sum\limits_{i = 1}^5 { = 750\,\,\,\,\,\,.....\left( 1 \right)} $

$\frac{{\sum {x_i^2} }}{5} - {\left( {\bar x} \right)^2} = 18 \Rightarrow \frac{{\sum {x_i^2} }}{5} - {\left( {150} \right)^2} = 180$

$ \Rightarrow \sum {x_i^2}  = \left( {22500 + 18} \right) \times 5\,\,\, \Rightarrow \sum\limits_{i = 1}^5 {\,x_i^2 = 112590\,\,\,\,\,.....\left( 2 \right)} $

Heght of new standent $ = 156\left( {Let\,{x_6}} \right)$

Then ${x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = 750 + 156$

${{\bar x}_{new}} = \frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = \frac{{906}}{6} = 151\,\,\,\,\,\,....\left( 3 \right)$

Variance (new) $ = \frac{{\sum {x_i^2} }}{6} - {\left( {{{\bar x}_{new}}} \right)^2}$

from equation $(2)$ and $(3)$

variance (new) $ = \frac{{112590 + {{\left( {156} \right)}^2}}}{6} - {\left( {151} \right)^2} = 2281 - 22801 = 20$

$ \Rightarrow x = 48$

${\sigma ^2} + {48^2} = \frac{1}{6}\left( {{{41}^2} + {{45}^2} + {{54}^2} + {{57}^2} + {{43}^2} + {{48}^2}} \right)$

${\sigma ^2} = \frac{{14024}}{6} - 2304$

$ = \frac{{100}}{3}$

According to the question,

$75\lambda  - 25 = 75 \Rightarrow \lambda  = \frac{4}{3}$.

Also, $\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}}  = 45$

$\sum\limits_{i = 1}^9 {x_i^2}  - 10\sum\limits_{i = 1}^9 {{x_i} + 9\left( {25} \right)}  = 45\,\,\,\,\,...\left( {ii} \right)$

From $(i)$ and $(ii)$ we get,

$\sum\limits_{i = 1}^9 {x_i^2}  = 360$

Since, variance $ = \frac{{\sum {x_i^2} }}{9} - {\left( {\frac{{\sum {{x_i}} }}{9}} \right)^2}$

$ = \frac{{360}}{9} - {\left( {\frac{{54}}{9}} \right)^2} = 40 - 36 = 4$

Standared deviation $ = \sqrt {Variance}  = 2$

           $ \Rightarrow \frac{{54 + \lambda }}{8} = 8 \Rightarrow \lambda  = 10$

Now variance $ = {\sigma ^2}$

$ = \frac{\begin{array}{l}
{\left( {7 - 8} \right)^2} + {\left( {8 - 8} \right)^2} + {\left( {9 - 8} \right)^2} + {\left( {7 - 8} \right)^2} + {\left( {8 - 8} \right)^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {7 - 8} \right)^2} + {\left( {10 - 8} \right)^2} + {\left( {8 - 8} \right)^2}
\end{array}}{8}$

$ \Rightarrow {\sigma ^2} = \frac{{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0}}{8} = \frac{8}{8} = 1$

Hence, the variance is $1$.