Maths M.C.Q (1 Marks)

Given,

$\mu=\frac{\sum x_i}{n}$

$\sigma=\sqrt{\frac{\sum x_1^2}{n}-(\mu)^2}$

$\hat{\mu}=\frac{\Sigma y_i}{n}$

$=\frac{\frac{x_1+x_2}{2}+\frac{x_1+x_2}{2}+x_3+x_4+\ldots+x_n}{n}$

$\hat{\mu}=\frac{x_1+x_2+x_3 \ldots+x_n}{n}=\frac{\Sigma x_i}{n}=\mu$

$\sigma=\sqrt{\frac{\sum y_1^2}{n}-\left(\mu^{\prime}\right)^2}=\sqrt{\frac{\sum y_1^2}{n}-\mu^2}$

$\sum x_1^2=x_1^2+x_2^2+x_3^2+\ldots+x_n^2$

$\sum y_1^2=$

$\frac{\left(x_1+x_2\right)^2}{4}+\frac{\left(x_1+x_2\right)^2}{4}+x_3^2+x_4^2+\ldots+x_n^2$

$\Sigma x_1^y-\Sigma y_1^2=x_1^2+x_2^2-2 x_1 x_2=\left(x_1-x_2\right)^2 \geq 0$

$\sum x_1^2 \geq \Sigma y_1^2$

We have,

$\operatorname{Mean}(\mu)=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$

$\mu=\frac{\Sigma x_i}{n}$

Standard deviation $\sigma=\sqrt{\frac{\Sigma x_i^2}{n}-(\mu)^2}$

Mean of other observations

$\operatorname{Mean}\left(\mu^{\prime}\right)=\frac{y_1+y_2+y_3+\ldots+y_{n-1}+y_n}{n}$

$=\frac{0+x_2+x_3+\ldots+x_{n-1}+x_1+x_n}{n}$

$=\frac{\Sigma x_i}{n}=\mu$

$\mu^{\prime} =\mu$

$\mu^{\prime} =\sqrt{\frac{\sum y_i^2}{n}-\left(\mu^{\prime}\right)^2}$

$\sigma^{\prime}=\sqrt{\frac{0+x_2^2+x_3^2+\ldots+x_{n-1}^2}{+\left(x_1+x_n\right)^2}-\mu}$

$\quad \Sigma x_1^2=x_1^2+x_2^2+\ldots+x_n^2$
$\quad \Sigma y_1^2=0+x_2^2+\ldots+x_{n-1}^2+x_1^2+x_n^2+2 x_1 x_n$
$\text { Clearly, } \Sigma y_1^2 \geq \Sigma x_1^2$
$\therefore \quad \sigma^{\prime} \geq \sigma$
Hence, option (a) is correct.

Let total number of people whose salary less than $10000\,Rupees$ per annum $=x$ and annual salary of each person $=a$

$\therefore$ Total salary $=a x$

and total number of people whose salary more than $10000\,Rupees$ per annum $=y$ and annual salary of each person $=b$

$\therefore$ Total salary $=b x$

When $5 \%$ increase of salary of people $x$ i.e. $\quad x(a+5 \%$ of $a)=\frac{105 a x}{100}$

and $5 \%$ decrease of salary of people $y$ i.e. $y(b-5 \%$ of $b)=\frac{95 b y}{100}$

$\frac{\text { Average salary after }}{\text { Average salary before }} =\frac{105 a x+95 b y}{a x+b y}$

$=1+\frac{5}{100}\left(\frac{a x-b y}{a x+b y}\right)$

$a x-b y < 0$

$\therefore$ Average salary af ter be decreases.

$ \text { Median }=18=\mathrm{M} $

$ \text { M.D. }=\frac{\sum \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\sum \mathrm{f}_{\mathrm{i}}} $

$ 1.25=\frac{36+\mathrm{x}+2 \mathrm{y}}{40} $

$ \mathrm{x}+2 \mathrm{y}=14 \ldots \ldots \ldots .(1)$

$ \text { by (1) \& (2) } $

$ x=6, y=4 $

$ \Rightarrow 4 x+5 y=24+20=44$

$ \alpha+\beta=10 $

$ \alpha^2+\beta^2=52$

solving we get $\alpha=4, \beta=6$

$\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{6}=\frac{5+2+5+8+2+4}{6}=\frac{13}{3}$

$ \sigma^2=66.2 $

$ \Rightarrow \frac{\alpha^2+\beta^2+25678}{10}-(56)^2=66.2 $

$ \therefore \alpha^2+\beta^2=6344$

$ \therefore \quad \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=176$

$ \text { So } \overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{176}{22}=8 $

$ \text { for } \sigma^2=\frac{1}{\mathrm{~N}} \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-(\overline{\mathrm{x}})^2 $

$ \quad=\frac{1}{22} \times 2048-(8)^2$

$ \quad=93.090964 $

$\quad=29.0909$

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10 $

$ \Sigma \mathrm{x}_{\mathrm{i}}=10 \times 20=200$

If $8$ is replaced by $12$ , then $\Sigma x_1=200-8+12=204$

$\therefore$ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}$

$=\frac{204}{20}=10.2$

$ \because$ Standard deviation $=2$

$ \therefore$ Variance $=( S.D.)^2=2^2=4 $

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^2=4 $

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-(10)^2=4 $

$ \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}=104 $

$ \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2080$

Now, replaced $'8'$ observations by $'12'$

$\text { Then, } \Sigma \mathrm{x}_{\mathrm{i}}^2=2080-8^2+12^2=2160$

$\therefore$ Variance of removing observations

$ \Rightarrow \frac{\Sigma x_i^2}{20}-\left(\frac{\Sigma x_i}{20}\right)^2 $

$ \Rightarrow \frac{2160}{20}-(10.2)^2 $

$ \Rightarrow 108-104.04 $

$ \Rightarrow 3.96$

Correct standard deviation

$=\sqrt{3.96}$

$\bar{x}=\frac{(2+2+3+4+5+6) C}{7}=\frac{22 C}{7}$

$ \operatorname{Var}(\mathrm{x})=\frac{\mathrm{c}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7} $

$ -\left(\frac{22 c}{7}\right)^2 $

$ =\frac{92 c^2}{7}-\mathrm{c}^2 \times \frac{484}{49} $

$ =\frac{(644-484) c^2}{49}=\frac{160 c^2}{49} $

$ 160=\frac{160 \times c^2}{49} \Rightarrow c=7$

$ \mathrm{M}=1+\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\right) \mathrm{h} $

$ \mathrm{M}=8+\frac{18-12}{10} \times 4 $

$ \mathrm{M}=10.4 $

$ 20 \mathrm{M}=208$

Mean deviation about

Median $=$ $\frac{0+45+60+20+40+170-a+170-b}{7}=\frac{205}{7}$

$\Rightarrow \mathrm{a}+\mathrm{b}=300$

Mean=$\frac{50+175-a+175-b+5+15+35+55}{7}=30$

Mean deviation

About mean $=$    $\frac{50+175-a+175-b+5+15+35+55}{7}=30$

$ \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50$     $.........(i)$

$ \sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=1100 $         $...........(ii)$

$ \text { If } \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50$ .

$ \left(\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}\right)^2=2500 $

$\Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2+2 \sum_{\mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=2500$

$ \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=2500-2(1100) $

$ \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=300, \text { Standard deviation ' } \sigma \text { ' } $

$ \frac{\sum^{\frac{a_i^2}{2}}}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2}$

$ =\sqrt{30-25}=\sqrt{5}$

We have

$\mu^{\prime}=\frac{\Sigma x_i}{15}=12 \Rightarrow \Sigma x_i=180$

As per given information correct $\Sigma x_i=180-10+12$

$\Rightarrow \mu(\text { correct mean })=\frac{182}{15}$

Also

$ \sigma^{\prime}=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295 $

$\text { Correct } \Sigma \mathrm{x}_{\mathrm{i}}^2=2295-100+144=2339 $

$ \sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}$

Required value

$ =15\left(\mu+\mu^2+\sigma^2\right) $

$ =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) $

$ =15\left(\frac{182}{15}+\frac{2339}{15}\right) $

$ =2521$

Let first four observation be $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$

Here, $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$.

Also, $\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}$

$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14$

Now from eqn $-1$

$\mathrm{x}_5$=$10$

Now, $\sigma^2=\frac{194}{25}$

$ \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} $

$ \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54$

Now, variance of first $4$ observations

Var $=\frac{\sum_{i=1}^4 x_i^2}{4}-\left(\frac{\sum_{i=1}^4 x_i}{4}\right)^2$

$ =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}$

Mean $=55$       $a>b$

Variance $=194$    $a+3 b$

$\frac{a+b+68+44+48+60}{6}=55$

$\Rightarrow 220+a+b=330$

$\therefore a+b=110 \ldots . .(1)$

Also,

$\sum \frac{\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2}{\mathrm{n}}=194 $

$\Rightarrow(\mathrm{a}-55)^2+(\mathrm{b}-55)^2+(68-55)^2+(44-55)^2$

$+(48-55)^2+(60-55)^2=194 \times 6$

$\Rightarrow(\mathrm{a}-55)^2+(\mathrm{b}-55)^2+169+121+49+25=1164$

$\Rightarrow(\mathrm{a}-55)^2+(\mathrm{b}-55)^2=1164-364=800$

$\mathrm{a}^2+3025-110 \mathrm{a}+\mathrm{b}^2+3025-110 \mathrm{~b}=800$

$\Rightarrow \mathrm{a}^2+\mathrm{b}^2=800-6050+12100$

${a}^2+\mathrm{b}^2=6850 \ldots \ldots .(2)$

Solve $(1) \& (2);$

$a=75, b=35$

$\therefore$ $a+3 b=75+3(35)=75+105=180$

$ \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\alpha\right)=2 \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}-10 \alpha=2 $

$ \text { Mean } \mu=\frac{6}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{10} $

$ \therefore \quad \sum_{\mathrm{i}}=12 $

$ \quad 10 \alpha+2=12 \quad \therefore \alpha=1 $

$ \text { Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2=40 \text { Let } \mathrm{y}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\beta $

$ \therefore \sigma_{\mathrm{y}}^2=\frac{1}{10} \sum \mathrm{y}_{\mathrm{i}}^2-(\overline{\mathrm{y}})^2 $

$ \sigma_{\mathrm{x}}^2=\frac{1}{10} \sum\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2-\left(\frac{\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)}{10}\right)^2 $

$ \frac{84}{25}=4-\left(\frac{12-10 \beta}{10}\right)^2 $

$ \therefore\left(\frac{6-5 \beta}{5}\right)^2=4-\frac{84}{25}=\frac{16}{25} $

$ 6-5 \beta= \pm 4 \Rightarrow \beta=\frac{2}{5} \text { (not possible) or } \beta=2$

Hence $\frac{\beta}{\alpha}=2$

$ \mathrm{~b}=3 \cdot \frac{{ }^9 \mathrm{C}_4}{{ }^{12} \mathrm{C}_5} $

$ \mathrm{c}=3 \cdot \frac{{ }^9 \mathrm{C}_3}{{ }^{12} \mathrm{C}_5} $

$ \mathrm{~d}=1 \cdot \frac{{ }^9 \mathrm{C}_2}{{ }^{12} \mathrm{C}_5} $

$ \mathrm{u}=0 \cdot \mathrm{a}+1 \cdot \mathrm{b}+2 \cdot \mathrm{c}+3 \cdot \mathrm{d}=1.25 $

$ \sigma^2=0 \cdot \mathrm{a}+1 \cdot b+4 \cdot c+9 \mathrm{~d}-\mathrm{u}^2 $

$ \sigma^2=\frac{105}{176}$

Ans. $176-105=71$

$ \bar{x}=\text { mean }=\frac{9+25+a+b+c}{5}=18 $

$ a+b+c=56 $

$ \text { Mean deviation }=\frac{\sum\left|x_i-\bar{x}\right|}{n}=4 $

$ =9+7+|18-a|+|18-b|+|18-c|=20 $

$ =|18-a|+|18-b|+|18-c|=4 $

$ \text { Variance }=\frac{\Sigma\left|x_i-\bar{x}\right|^2}{n}=\frac{136}{5} $

$ =81+49+|18-a|^2+|18-b|^2+|18-c|^2=136 $

$ =(18-a)^2+(18-b)^2+(18-c)^2=6 $

$ \text { Possible values }(18-a)^2=1, \quad(18-b)^2=1 \quad(18-c)^2=4 $

$ \mathrm{a}<\mathrm{b}<\mathrm{c} \quad 18-\mathrm{a}=1 \quad 18-b=-1 \quad 18-\mathrm{c}=-2 $

$ \text { so } \quad \quad \quad \mathrm{a}=17 \quad \mathrm{~b}=19 \quad \mathrm{c}=20 $

$ \mathrm{a}+\mathrm{b}+\mathrm{c}=56 \quad 2 \mathrm{a}+\mathrm{b}-\mathrm{c} \quad 34=19-20=33$

$10=\frac{1+4+16+25+ x ^2+ y ^2}{6}-25$

$x ^2+ y ^2=164 \ldots \ldots \text { (ii) }$

By solving $(i)$ and $(ii)$

$x =8, y =10$

$\text { M.D. }(\overline{ x })=\frac{\sum\left| x _{ i }-\overline{ x }\right|}{6}=\frac{8}{3}$

$a_1+a_2+a_3+a_4+a_5+a_6=57$

$\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57$

$\Rightarrow a_1+a_6=19$

$\Rightarrow 2 a_1+5 d=19 \text { and } a_1+d=5$

$\Rightarrow a_1=2, d=3$

$\text { Numbers }: 2,5,8,11,14,17$

$\text { Variance }=\sigma^2=\text { mean of squares }-\text { square of mean }$ $=\frac{2^2+5^2+8^2+(11)^2+(14)^2+(17)^2}{6}-\left(\frac{19}{2}\right)^2 ~\\ =\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$

$8 \sigma^2=210$

$\text { Now } \frac{\sum \limits_{ i =1}^{ n } x _{ i }^2}{20} x _{ i }-8+12=(10.2) n \quad \therefore n =20$

$\frac{\sum \limits_{ i =1}^{20} x _{ i }2-8^2+12^2}{20}-4 \Rightarrow \sum \limits_{ i =1}^{20} x _{ i }^2=2080$

$=108-104.04=3.96$

$\overline{ y }=\frac{\sum \limits_{ j =61}^{91} j }{31}=\frac{61+91}{2}=76 \quad \text { (31 elements) }$

$\text { Combined mean, }$

$\mu =\frac{31 \times 26+31 \times 76}{31+31}$

$=\frac{26+76}{2}=51$

$\sigma^2=\frac{1}{62} \times\left(\sum_{i=1}^{31}\left(x_i-\mu\right)^2+\sum_{i=1}^{31}\left(y_i-\mu\right)^2\right)=705$

Since, $x _{ i } \in X$ are in $A.P.$ with $31$ elements and common difference $1$,same is $y _{ i } \in y$, when written 

in increasing order.

$\therefore \sum \limits_{i=1}^{31}\left(x_i-\mu\right)^2=\sum \limits_{i=1}^{31}\left(y_i-\mu\right)^2$

$=10^2+11^2+\ldots . .+40^2$

$=\frac{40 \times 41 \times 81}{6}-\frac{9 \times 10 \times 19}{6}=21855$

$\therefore \left|\bar{x}+\bar{y}-\sigma^2\right|=|26+76-705|=603$

$\frac{x_1+x_2+x_3 \ldots .+x_6+14}{7}=8$

$\Rightarrow x_1+x_2+\ldots .+x_6=42$

$\therefore \frac{x_1+x_2 \ldots .+x_6}{6}=\frac{42}{6}=7=a$

$\frac{\sum x_i^2}{7}-8^2=16$

$\Rightarrow x^2=560$

$\Rightarrow x_1^2+x_2^2+\ldots+x_6^2=364$

$b=\frac{x_1^2+x_2^2+\ldots . .+x_6^2}{6}-7^2$

$=\frac{364}{6}-49$

$b=\frac{70}{6}$

$a+3 b-5=7+3 \times \frac{70}{6}-5$

$=37$

$a_1, a_1+1, a_1+2, \ldots ., a_1+99 \text { are values of } a_i ' S$

$\bar{x}=\frac{a_1+\left(a_1+1\right)+\left(a_1+2\right)+\ldots . .+a_1+99}{100}$

$=\frac{100 a_1+(1+2+\ldots . .+99)}{100}=a_1+\frac{99 \times 100}{2 \times 100}$

$=a_1+\frac{99}{2}$

$\text { Mean deviation about mean }=\frac{\sum \limits_{i=1}^{100}\left|x_i-\bar{x}\right|}{100}$

$=\frac{2\left(\frac{99}{2}+\frac{97}{2}+\frac{95}{2}+\ldots .+\frac{1}{2}\right)}{100}$

$=\frac{1+3+\ldots .+99}{100}$

$=\frac{\frac{50}{2}[1+99]}{100}$

$=25$

So, it is true for every natural no. ' $a_1{ }^{\prime}$

$\overline{ x }=\frac{100 \times 40+55 n }{100+ n }$

$5000+50 n =4000+55 n$

$1000=5 n$

$n =200$

$\sigma_1{ }^2=\frac{\sum x _{ i }^2}{100}-40^2$

$\sigma_2{ }^2=\frac{\sum x _{ j }^2}{100}-55^2$

$350=\sigma^2=\frac{\sum x _{ i }^2+\sum x _{ j }^2}{300}-(\overline{ x })^2$

$350=\frac{\left(1600+\alpha^2\right) \times 100+\left[(30-\alpha)^2+3025\right] \times 200}{300}-(50)^2$

$2850 \times 3=\alpha^2+2(30-\alpha)^2+1600+6050$

$8550=\alpha^2+2(30-\alpha)^2+7650$

$\alpha^2+2(30-\alpha)^2=900$

$\alpha^2-40 \alpha+300=0$

$\alpha=10,30$

$\sigma_1^2+\sigma_2^2=10^2+20^2=500$

$a+b=16 \ldots \ldots(1)$

$\sigma^2=\frac{\sum x_1^2}{5}-\left(\frac{\sum x}{5}\right)^2$ $8=\frac{1^2+3^2+5^2+a^2+b^2}{5}-25$

$a^2+b^2=130 \ldots \ldots(2)$

$b y(1),(2)$

$a=7, b=9$

$9,9+d, 9+2 d, \ldots \ldots .9+6 d$

$0, d, 2 d, \ldots \ldots \cdot 6$

$\bar{x}_{\text {new }}=\frac{21 d }{7}=3 d$

$16=\frac{1}{7}\left(0^2+1^2+\ldots \ldots+6^2\right) d^2-9 d^2$

$=\frac{1}{7}\left(\frac{6 \times 7 \times 13}{6}\right) d ^2-9 d ^2$

$16=4 d^2$

$d^2=4$

$d=2$

$\bar{x}+x_6=6+9+10+9$

$\sum f_i x_i=360+6 \alpha+12 \beta$

$\sum f _{ i } x _{ i }^2=3904+36 \alpha+144 \beta$

$\operatorname{Mean}(\overline{ x })=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}=9$

$\Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta)$

$3 \alpha=3 \beta \Rightarrow \alpha=\beta$

$\sigma^2=\frac{\sum f _{ i } x _1^2}{\sum f _{ i }}-\left(\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}\right)^2$

$\Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(\overline{ x })^2=15.08$

$\Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}-(9)^2=15.08$

$\Rightarrow \alpha=5$

Now, $\alpha^2+\beta^2-\alpha \beta=\alpha^2=25$

For variance

$x-9, y-9,3,3,1,-5,-1,-3$

$\bar{x}=0$

$\therefore \frac{(x-9)^2+(y-9)^2+54}{8}-0^2=9.25$

$(x-9)^2+(11-x)^2=20$

$x=7 \text { or } 13 \therefore y=13,7$

$3 x-2 y=3 \times 13-2 \times 7=25$

$\sum x =54$

$\frac{\Sigma x ^2}{12}-\left(\frac{9}{2}\right)^2=4$

$\sum x ^2=291$

$\sum x _{\text {new }}=54-(9+10)+7+14=56$

$\sum x _{\text {new }}^2=291-(81+100)+49+196=355$

$\sigma_{\text {new }}^2=\frac{355}{12}-\left(\frac{56}{12}\right)^2$

$\sigma_{\text {new }}^2=\frac{281}{36}=\frac{ m }{ n }$

$m + n =317$

$\frac{2 \times 5+3 \times 15+x \times 25+5 \times 35+4 \times 45}{14+x}=28$

$x =6$

$\text { Variance }=\left(\frac{\sum x_i^2 f_i}{\sum f_i}\right)-(\text { mean })^2$

$\text { Variance }==\frac{2 \times 5^2+3 \times 15^2+6 \times 25^2+5 \times 35^2+4 \times 45^2}{20}-(28)^2$

$=151$

$3 k ^2+16 k -12 k -64=0$

$k =\text { or }-\frac{16}{3}(\text { rejected) }$

$\mu=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}$

$\mu=\frac{8+2(15)+3(15)+4(17)+5}{62}=\frac{156}{62}$

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

$=\frac{8 \times 1^2+15 \times 13+17 \times 16+25}{62}-\left(\frac{156}{62}\right)^2$

$\sigma^2=\frac{500}{62}-\left(\frac{156}{62}\right)^2$

$\sigma^2+\mu^2=\frac{500}{62}$

${\left[\sigma^2+\mu^2\right]=8}$

$B=\left\{b_1, b_2, b_3, b_4, b_5\right\}$

$\text { Given, } \sum_{ i =1}^3 ai =25, \sum_{ i =1}^3 bi =40$

$\frac{\sum_{ i =1}^5 a _{ i }^2}{5}-\left(\frac{\sum_{ i =1}^5 a _{ i }}{5}\right)^2=12, \frac{\sum_{ i =1}^5 b _{ i }^2}{5}-\left(\frac{\sum_{ i =1}^5 b _{ i }}{5}\right)^2=20$

$\sum_{ i =1}^5 a _{ i }^2=185 \quad, \quad \sum_{ i =1}^5 b _{ i }^2=420$

$\text { Now, } C =\left\{ C _1, C _2, \ldots . C _{10}\right\}$

$\text { s.f. } C_i=a_i=3 \text { or } b_i+2$

$\therefore \text { Mean of } C , \overline{ C }=\frac{\left(\sum a _{ i }-15\right)+\left(\sum b _{ i }+10\right)}{10}$

$\overline{ C }=\frac{10+50}{10}=6$

$\therefore \quad \sigma^2=\frac{\sum \limits_{ i =1}^{10} C _{ i }^2}{10}=(\overline{ C })^2$

$=\frac{\sum\left( a _{ i }-3\right)^2+\sum\left( b _{ i }+2\right)^2}{10}-(6)^2$

$=\frac{\sum a _{ i }{ }^2+\sum b _{ i }{ }^2-6 \sum a _{ i }+4 \sum b _{ i }+65}{10}-36$

$=\frac{185+420-150+160+65}{10}-36$

$=32$

$\therefore \quad$ Mean + Variance $=\overline{ C }+\sigma^2=6+32=38$

$\Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16$

$\text { M.. }(\bar{x})=\frac{\sum f _{ i }\left| x _{ i }-\overline{ x }\right|}{\sum f _{ i }} \text { where } \sum f _{ i }=64+16=80$

$\text { M.D. }(\bar{x})=\frac{4 \times 4+24 \times 2+28 \times 0+16 \times 2+8 \times 4}{80}$

$=\frac{8}{5}$

$\text { Variance }=\frac{\sum f _{ i }\left( x _{ i }-\overline{ x }\right)^2}{\sum f _{ i }}$

$=\frac{4 \times 16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{352}{80}$

$\therefore \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{128}{80}+\frac{352}{80}}=8$

$\sum x_i=500$

$\sum x_{i \text { correct }}=500+20+25-45-50=450$

$\sigma^2=144$

$\frac{\sum x_i^2}{10}-(50)^2=144$

$\sum x_{i c o r r e c t}^2=\left(144+(50)^2\right) \times 10-(45)^2-(50)^2+(20)^2+(25)^2$

$22940$

Correct variance $=\frac{\sum\left(x_{\text {icorrect }}\right)^2}{10}-\left(\frac{\sum x_{\text {icorrect }}}{10}\right)^2$

$=2294-(45)^2$

$=2294-2025=269$

$\mu_{\text {Corrected }}=\frac{200-50+40}{10}=19$

$\sigma^2=\frac{1}{10} \sum x_i^2-20^2$

$(64+400) 10=\sum x_i^2$

$\sigma_{\text {Corrected }}^2=\frac{1}{10}[(64+400) 10-2500+1600]-19^2$

$=374-361$

$=13$

$13=\frac{15.14+15 \cdot \sigma^2}{30}+\frac{15.15(12-14)^2}{30 \times 30}$

$13=\frac{14+\sigma^2}{2}+\frac{4}{4}$

$\sigma^2=10$

mean $(\bar{x})=15$

Standard deviation $(\sigma)=2$

Let incorrect observation is $x_{1}$ and correct observation is $\left( x _{1}^{\prime}\right)$

Given $x_{1}+x_{1}^{\prime}=70$

$\bar{x}=\frac{x_{1}+x_{2}+\ldots+x_{\text {s0 }}}{50}=15(\text { given })$

$\Rightarrow x_{1}+x_{2}+\ldots . x_{50}=750$           $\ldots(i)$

Now

Mean of correct observation is $16$

$\frac{x_{1}^{\prime}+x_{2}+\ldots+x_{50}}{50}=16$

$x_{1}^{\prime}+x_{2}+x_{3}+\ldots x_{s 0}=16 \times 50$           $\ldots(ii)$

eq. $(ii)$ - eq. $(i)$

$\Rightarrow x_{1}^{\prime}-x_{1}=16 \times 50-15 \times 50$

$x_{1}^{\prime}-x_{1}=50 and  x_{1}+x_{1}^{\prime}=70$

$x_{1}^{\prime}=60$

$x_{1}=10$

$\Rightarrow 4=\frac{x_{1}^{2}+x_{2}^{2}+\ldots .+x_{50}^{2}}{50}-15^{2}$           $\ldots(iii)$

$\Rightarrow \sigma^{2}=\frac{x_{1}^{\prime 2}+x_{2}^{2}+\ldots . x_{50}^{2}}{50}-16^{2}$           $\ldots(iv)$

from $(iii)$

$\Rightarrow 4=\frac{(10)^{2}}{50}+\frac{x_{2}^{2}+x_{3}^{2}+\ldots .+x_{50}^{2}}{50}-225$

$\Rightarrow 4=2-225+\frac{\left(x_{2}^{2}+x_{3}^{2}+\ldots .+x_{90}^{2}\right)}{50}$

$\Rightarrow 227=\frac{\left(x_{2}^{2}+x_{3}^{2}+\ldots x_{50}^{2}\right)}{50}$

$\text { From }( iv )$

$\sigma^{2}=\frac{(60)^{2}}{50}+\left(\frac{x_{2}^{2}+x_{3}^{2}+\ldots+x_{90}^{2}}{50}\right)-(16)^{2}$

$\sigma^{2}=\frac{60 \times 60}{50}+227-256$

$\sigma^{2}=72+227-256$

$\sigma^{2}=43$

$\therefore n =21$

$\Rightarrow\left| x _{1}-62\right|^{2}+\left| x _{2}-62\right|^{2}+\ldots .+\left| x _{7}-62\right|^{2}=140$

$If$ $x _{1}=49$

$|49-62|^{2}=169$

then, $\left| x _{2}-62\right|^{2}+\ldots .+\left| x _{7}-62\right|^{2}=$ Negative Number which is not possible, therefore, no student can fail.

$\text { Variance }=\frac{\sum\limits_{ r =1}^{15} x _{ r }^{2}}{15}-\left(\frac{\sum\limits_{ r =1}^{15} x _{ r }}{15}\right)^{2}$

Now, as per information given in equation

$\frac{\sum x _{ r }^{2}}{15}-8^{2}=3^{2} \Rightarrow \sum x _{ T }^{2}=\log 5$

Now, the new $\sum x _{ r }^{2}=\log 5-5^{2}+20^{2}=1470$

And, new $\sum x _{ r }=(15 \times 8)-5+(20)=135$

Variance $=\frac{1470}{15}-\left(\frac{135}{15}\right)^{2}=98-81=17$

Variance $=\frac{9+49+144+ a ^{2}+(43- a )^{2}}{5}-13^{2} \in N$

$\Rightarrow \frac{2 a^{2}-a+1}{5} \in N$

$\Rightarrow 2 a^{2}-a+1-5 n=0$ must have solution as natural numbers

its $D=40 n-7$ always has $3$ at unit place

$\Rightarrow D$ can't be perfect square

So, a can't be integer.

Mean deviation $=\sum \frac{\left|x_{i}-M\right|}{n}=6$

$(k+3)+(k+1)+(k-1)+(6-k)+(6-k)$

$\frac{+(10-k)+(15-k)+(18-k)}{8}$

$\therefore \quad \frac{58-2 k}{8}=6$

$k=5$

Median $=\frac{2 \times 5+12}{2}=11$

$6^{2}=\frac{\sum x_{i}^{2}}{10}-(\bar{x})^{2}=15$

$\sum_{i=1}^{10} x_{i}=150$

$\sum_{i=1}^{9} x_{i}+25=150$

$\sum_{i=1}^{9} x_{i}=125$

$\sum_{i=1}^{9} x_{i}+15=140$

Actual mean $=\frac{140}{10}=14=\bar{x}_{\text {nev }}$

$\sum_{i=1}^{9} \frac{x_{i}^{2}+25^{2}-15^{2}}{10}=15$

$\sum_{i=1}^{9} x_{i}^{2}+625=2400$

$\sum_{i=1}^{9} x_{i}^{2}=1775$

$\sum_{i=1}^{9} x_{i}^{2}+15^{2}=2000=\left(\sum x_{i}^{2}\right)_{\text {acnaal }}$

$6_{\text {actual }}^{2}=\frac{\left(\sum x_{i}^{2}\right)_{\text {actual }}-\left(\bar{x}_{\text {new }}\right)^{2}}{10}$

$=\frac{2000}{10}-14^{2}$

$=200-196=4$

$(\text { S.D })_{\text {attul }}=6=2$

$\Rightarrow x+y=48-36=12$

Variance

$=\frac{1}{8}\left(16+25+36+36+49+64+x^{2}+y^{2}\right)-36=\frac{9}{4}$

$\Rightarrow x^{2}+y^{2}=80$

$\therefore x=4 ; y=8$

$x^{4}+y^{2}=256+64=320$

Mean $=6$

$\frac{a+b+8+5+10}{5}=6$

$a+b=7$

$b=7-a$

$6.8=\frac{(a-6)^{2}+(b-6)^{2}+(8-6)^{2}+(5-6)^{2}+(10-6)^{2}}{5}$

$34=(a-6)^{2}+(7-a-6)^{2}+4+1+18$

$a^{2}-7 a+12=0 \Rightarrow a=4$ or $a=3$

$a=4 \quad a=3$

$b=3 \quad b=4$

$M=\frac{\sum\limits_{i=1}^{5}\left|x_{i}-x\right|}{n}$

$M=\frac{|a-6|+|b-6|+|8-6|+|5-6|+|10-6|}{5}$

when $a =3, b =4 \quad$ 

$M =\frac{3+2+2+1+4}{5}$

$M =\frac{12}{5}$

when $a =4, b =3$

$ M =\frac{2+3+2+1+7}{5}$

$M =\frac{12}{5}$

$25\;M =25 \times \frac{12}{5}=60$

$\sigma^{2}=\frac{\sum x_{i}^{2}}{5}-\left(\frac{24}{5}\right)^{2}=\frac{194}{25}$

$\Rightarrow \sum x_{i}^{2}=154$

$x_{1}+x_{2}+x_{3}+x_{4}=14$

$\Rightarrow x_{5}=10$

$\sigma^{2}=\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}{4}-\frac{49}{4}=a$

$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4 a+49$

$x_{5}^{2}=154-4 a-49$

$\Rightarrow 100=105-4 a \Rightarrow 4 a=5$

$4 a+x_{5}=15$

Wrong $S.D$ $=\sigma_{1}=5$

$\frac{\sum x _{ i }}{40}=30$

$\sum x _{ i }=1200$

$\sigma_{1}^{2}=25$

$\frac{\sum x _{ i }^{2}}{40}-30^{2}=25$

$\sum x _{ i }^{2}=925 \times 40=37000$

New sum $=\sum x _{ i }^{\prime}=1200-10-12=1178$

New mean $=\mu_{1}^{\prime}=\frac{1178}{38}=31$

New $\sum x _{ i }^{2}=37000-(10)^{2}-(12)^{2}=36756$

New $S.D$, $\sigma_{1}^{\prime}=\sqrt{\frac{36756}{38}-(31)^{2}}=\sigma$

$36756-(31)^{2} \times 38=38 \sigma^{2}$

$38 \sigma^{2}=238$

$\frac{\sum x_{1}^{2}}{20}-(15)^{2}=9$

$\sum x_{1}^{2}=234 \times 20=4680$

$\frac{\sum\left(x_{1}+\alpha\right)^{2}}{20}=178 \Rightarrow \sum\left(x_{1}+\alpha\right)^{2}=3560$

$\Rightarrow \sum x_{1}^{2}+2 \alpha \sum x_{1}+\sum \alpha^{2}=3560$

$4680+600 \alpha+20 \alpha^{2}=3560$

$\Rightarrow \alpha^{2}+30 \alpha+56=0$

$\Rightarrow(\alpha+28)(\alpha+2)=0$

$\alpha=-2,-28$

Square of maximum value of $\alpha$ is $4$

Now,

$10.25=\frac{\left(4+16+25+49+a^{2}+(21-a)^{2}\right)}{6}$

(Using formula for variance)

$\Rightarrow 6(10.25)+6(6.5)^{2}=94+a^{2}+(21-a)^{2}$

$\Rightarrow a 2+\left(21-a^{2}\right)=221$

$\therefore a=10 \text { and }(21-a)=21-10=11$

so, remaining two observations are $10,11 .$