MATHS MCQ

$82 = \frac{{(27 + x) + (31 + x) + (89 + x) + (107 + x) + (156 + x)}}{5}$

==> $82 \times 5 = 410 + 5x$ ==> $410 - 410 = 5x$

==> $x = 0$

Required mean is,

$\bar x = \frac{{130 + x + 126 + x + 68 + x + 50 + x + 1 + x}}{5}$

$\bar x = \frac{{375 + 5x}}{5}$

$ = \frac{{375 + 0}}{5}$$ = \frac{{375}}{5}$= $75.$

i.e., $3 = \frac{{5 + 8 + 18 + 4f}}{{15 + f}}$

==> $45 + 3f = 31 + 4f$

==> $45 - 31 = f$ ==> $f = 14$.

==> $\sum\limits_{i = 1}^{20} {{x_i} - 20 \times 30 = 20} $

==> $\sum\limits_{i = 1}^{20} {{x_i} = 620} $.

Mean = $\frac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} $

$= \frac{{620}}{{20}} = 31$.

$ = \frac{{1 + 1 + 1 + 1 + ..... + 1}}{{\frac{{n(n + 1)}}{2}}}$

$ = \frac{n}{{\frac{{n(n + 1)}}{2}}} = \frac{2}{{n + 1}}$.

==> ${x_1} + {x_2} + ..... + {x_{10}} = 45$

$\frac{{{x_{11}} + {x_{12}} + ..... + {x_{40}}}}{{30}} = 3.5$

==> ${x_{11}} + {x_{12}} + ..... + {x_{40}} = 105$

${x_1} + {x_2} + ..... + {x_{40}} = 150$

$\frac{{{x_1} + {x_2} + ..... + {x_{40}}}}{{40}} $

$= \frac{{150}}{{40}}$$ = \frac{{15}}{4}$.

i.e., $2.6 = \frac{{1 \times 4 + 2 \times 5 + 3 \times y + 4 \times 1 + 5 \times 2}}{{4 + 5 + y + 1 + 2}}$

or $31.2 + 2.6y = 28 + 3y$ or $0.4y = 3.2$

==> $y = 8$.

$\frac{{\sum\limits_{i = 1}^n {({x_i} + 2i)} }}{n} = \frac{{\sum\limits_{i = 1}^n {{x_i}} + 2\sum\limits_{i = 1}^n i }}{n} = \frac{{n\bar x + 2(1 + 2 + ...n)}}{n} = \frac{{n\bar x + 2\frac{{n(n + 1)}}{2}}}{n} = \bar x + (n + 1)$

$ = \frac{{n\bar x + 2\frac{{n(n + 1)}}{2}}}{n} = \bar x + n + 1$.

Sum of items added$ = 60 + 70 + 80 = 210$

Sum of items replaced$ = 40 + 20 + 50 = 110$

New sum $ = 4900 + 210 - 110 = 5000$

Correct mean$ = \frac{{5000}}{{100}} = 50$.

After one number excluded

Sum of total number = $16 × 4 = 64$

Then, excluded number is $90 -64 = 26.$

Sum of weight of $6$ students

$= 52 + 58 + 55 + 53 + 56 + 54 = 328\ kg$

$\therefore $ Weight of seventh student = $385 -328 = 57\ kg.$

Sum of items added = $19 +31 = 50$

Sum of items replaced = $91+ 13 = 104$

New sum = $4500 - 104 + 50$ = $4446$

New mean$ = \frac{{4446}}{{100}}$ = $44.46$

Here, total number of items is $41$

$i.e.$, an odd number.

Hence, the median is $\frac{{41 + 1}}{2}^{th}$

$i.e.$, $21^{st}$ item.

From cumulative frequency table, we find that median

$i.e.$, $21^{st}$ item is $155$,

(All items from $16$ to $22^{nd}$ are equal, each $155$).

$\bar x = \frac{{0.{q^n} + 1.\frac{n}{1}{q^{n - 1}}p + 2.\frac{{(n)(n - 1)}}{{2!}}{q^{n - 2}}{p^2} + .....n.{p^n}}}{{{q^n} + \frac{n}{1}{q^{n - 1}}p + \frac{{n(n - 1)}}{2}{q^{n - 2}}{p^2} + ..... + {p^n}}}$

$ = \frac{{0.{^n}{C_0}{q^n}{p^0} + {{1.}^n}{C_1}\,{q^{n - 1}}p + ..... + n.{\,^n}{C_n}{q^0}{p^n}}}{{^n{C_0}{q^n}{p^0}{ + ^n}{C_1}{q^{n - 1}}{p^1} + .....{ + ^n}{C_n}{q^{n - n}}{p^n}}}$

$ = \frac{{\sum\limits_{r = 0}^n r {.^n}{C_r}{q^{n - r}}{p^r}}}{{\sum\limits_{r = 0}^n {^n{C_r}{q^{n - r}}{p^r}} }}$

$ = \frac{{\sum\limits_{r = 1}^n r .\frac{n}{r}\,{\,^{n - 1}}{C_{r - 1}}{q^{n - r}}.p.\,{p^{r - 1}}}}{{\sum\limits_{r = 0}^n {^n{C_r}{q^{n - r}}{p^r}} }}$

$ = \frac{{np\left( {\sum\limits_{r = 1}^n {^{n - 1}{C_{r - 1}}{p^{r - 1}}{q^{(n - 1) - (r - 1)}}} } \right)}}{{\sum\limits_{r = 0}^n {^n{C_r}{q^{n - r}}{p^r}} }}$

$ = \frac{{np{{(q + p)}^{n - 1}}}}{{{{(q + p)}^n}}} = np$, . $[\because q + p = 1]$

$n = \sum f = 159$. Here $n = 159$, which is odd

$\therefore $Median number$ = \frac{1}{2}(n + 1) = \frac{1}{2}(159 + 1) = 80$,

which is in the class $30-40$. (see the row of cumulative frequency $95$, which contains $80$).

Hence median class is $30-40$.

$\therefore$  We have,

$l$ = Lower limit of median class = $30$

$f$ = Frequency of median class = $45$

$C$ = Total of all frequencies preceding median class = $50$

$i$ = Width of class interval of median class = $10$

$\therefore$  Required median $ = l + \frac{{\frac{N}{2} - C}}{f} \times i$ $ = 30 + \frac{{\frac{{159}}{2} - 50}}{{45}} \times 10 = 30 + \frac{{295}}{{45}} = 36.55$.

$\mathrm{x}_{10+\mathrm{n}}=148+(\mathrm{n}-1)(-2)=\mathrm{x}_{17}=136, \mathrm{x}_{18}=134$

hence median $=135$

Here, total number of items is $41$ i.e., an odd number.

Hence, the median is $\frac{41+1}{2}$ th i.e., $21^{\text {st }}$ item.

From cumulative frequency table, we find that median

i.e., $21^{\text {st }}$ item is $155,$

(All items from $16$ to $22^{\text {nd }}$ are equal, each $155$ ).

$\frac{685}{2}=342.5$

$40+\frac{342.5-296}{180} \times 10$

$40+\frac{46.5}{18}=40+2.8=42.6$

$=\frac{\mathrm{x}_{1}+1+\mathrm{x}_{2}+2+\mathrm{x}_{3}+3+\mathrm{x}_{4}+4+\ldots+\mathrm{x}_{10}+10}{10}$

$=\frac{\sum x_{i}+\frac{10 \times 11}{2}}{10}=3+\frac{11}{2}=\frac{17}{2}$

 Mean deviation$ = \frac{{\Sigma |{x_i} - \bar x|}}{n}$

$ = \frac{{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}}{5}$

$ = \frac{{2 + 1 + 0 + 1 + 2}}{5}$$ = \frac{6}{5} = 1.2$.

No. of observations $=3 \mathrm{n}$

mean $(\bar X) = \frac{{n \times a + 2n \times ( - 2a)}}{{3n}} =  - a$

$\therefore $ Mean deviation about mean $ = \frac{{\Sigma \left| {{{\rm{x}}_1} - {\rm{\bar x}}} \right|}}{{3{\rm{n}}}}$

$\frac{\mathrm{n} \times 2 \mathrm{a}+2 \mathrm{n} \times \mathrm{a}}{3 \mathrm{n}}=\frac{4 \mathrm{a}}{3}$

Mean deviation $=\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\mathrm{a}\right|}{10}=30$

$\frac{2(4.5 x+3.5 x+2.5 x+1.5 x+.5 x)}{10}=30$

$\frac{2(12.5)|\mathrm{x}|}{10}=30$

$|x|=12$

$S.D.$ = $\sigma $ = $\sqrt {\frac{1}{n}\sum {x_i^2 - (\bar x} {)^2}} $

=$\sqrt {\frac{1}{5}(1 + 4 + 9 + 16 + 25) - 9} $=$\sqrt {11 - 9} = \sqrt 2 $.

Hence, variance = $\frac{1}{n}\Sigma {({x_i} - \overline x )^2}$

$ = \frac{1}{5}\{ {(2 - 6)^2} + {(4 - 6)^2} + {(6 - 6)^2} + {(8 - 6)^2} + {(10 - 6)^2}\} $

$ = \frac{1}{5}\left\{ {(16 + 4 + 0 + 4 + 16} \right\}$$ = \frac{1}{5}\left\{ {40} \right\}$ $ = 8$.

Both sequence have same standard deviation.

Note:

If $1^{\text {st }}$ sequence is $x _{ i }$ and $2^{\text {nd }}$ sequence is $y _{ i }$,

$y _{ i }=10+ x _{ i } \Rightarrow \overline{ y }=10+\overline{ x }$

So, $\overline{ y }- y _{ i }=\overline{ x }- x _{ i }$

$ = \frac{{n(n + 1)\;(2n + 1)}}{{6n}} - {\left( {\frac{{n(n + 1)}}{{2n}}} \right)^2} $

$= \frac{{{n^2} - 1}}{{12}}$.

$ = \frac{{21}}{6} = \frac{7}{2}$

$S.D.$ $ = \sigma = \sqrt {\frac{1}{n}\Sigma x_i^2 - {{(\bar x)}^2}} $

$ = \sqrt {\frac{1}{6}(1 + 4 + 9 + 16 + 25 + 36) - \frac{{49}}{4}} $

$ = \sqrt {\frac{{91}}{6} - \frac{{49}}{4}} $

$ = \sqrt {\frac{{182 - 147}}{{12}}} $

$ = \sqrt {\frac{{35}}{{12}}} $.

Mean $ = 4 \Rightarrow \frac{{1 + 2 + 6 + x + y}}{5} = 4$

==> $x + y = 11$ .....$(i)$

and variance = $5. 2$

==> $\frac{{{1^2} + {2^2} + {6^2} + {x^2} + {y^2}}}{5} - {({\rm{mean}})^2} = 5.2$

$41 + {x^2} + {y^2} = 5[5.2 + {(4)^2}]$

$41 + {x^2} + {y^2} = 106$

${x^2} + {y^2} = 65$.....$(ii)$

Solving $(i)$ and $(ii)$ for $x$ and $y$, we get

$x = 4,y = 7$ or $x = 7,y = 4$.

Hence, new variance $ = {5^2} \times 9$$ = 225$.

${\sigma ^2} = \frac{{\sum {f_i}d_i^2}}{{\sum {f_i}}} - {\left( {\frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}} \right)^2}$

$= \frac{{900}}{{10}} - {\left( {\frac{{ - 30}}{{10}}} \right)^2}$

${\sigma ^2} = 90 - 9 = 81$

==> $\sigma$ = $9$.

Corrected $\bar x = 7990/200$$ = 39.95$

Incorrect $\Sigma {x^2} = n\,[{\sigma ^2} + {\bar x^2}] = 200[{15^2} + {40^2}] = 365000$

Correct $\Sigma {x^2} = 365000 - 2500 + 1600$$ = 364100$

Corrected $\sigma = \sqrt {\frac{{364100}}{{200}} - {{(39.95)}^2}} $

$ = \sqrt {(1820.5 - 1596)} $$ = \sqrt {224.5} = 14.98$.

$\bar x = $ combined mean $ = \frac{{5 \times 8 + 3 \times 8}}{{5 + 3}}$ $ = \frac{{64}}{8} = 8$

Combined variance $ = \frac{{{n_1}(\sigma _1^2 + D_1^2) + {n_2}(\sigma _2^2 + D_2^2)}}{{{n_1} + {n_2}}}$,

where ${D_1} = {\bar x_1} - \bar x$, ${D_2} = {\bar x_2} - \bar x$

Now, ${D_1} = 8 - 8;\,\,{D_2} = 8 - 8 = 0$

Combined variance $ = \frac{{5(18) + 3(24)}}{{5 + 3}}$ $ = \frac{{90 + 72}}{8}$ $ = \frac{{162}}{8}$ = $20.25$.

Then, mean $ = 4.4 \Rightarrow \frac{{1 + 2 + 6 + x + y}}{5} = 4.4$

==> $x + y = 13$.....$(i)$

and variance $= 8.24$

==> $\frac{{{1^2} + {2^2} + {6^2} + {x^2} + {y^2}}}{5}$ -${({\rm{mean}})^2} = 8.24$

==> $41 + {x^2} + {y^2} = 5\,\{ {(4.4)^2} + 8.24\} $

==> ${x^2} + {y^2} = 97$.....$(ii)$

Solving $(i)$ and $(ii)$ for $x$ and $y$, we get

$x = 9,\,\,y = 4$ or $x = 4,y = 9$.

${\sigma ^2} = \frac{{\sum {{f_i}} d_i^2}}{{\sum {{f_i}} }} - {\left( {\frac{{\sum {{f_i}} {d_i}}}{{\sum {{f_i}} }}} \right)^2}$

$=\frac{900}{10}-\left(\frac{-30}{10}\right)^{2}$

$\sigma^{2}=90-9=81 $

$\Rightarrow \sigma=9$

Variance on doubling each observation

$=2^{2} \times 16=64$

Std. deviation $=\sqrt{\operatorname{var}}=8$

$\sigma^{2}=\frac{6 \times 24+3 \times 36}{6+3}+\frac{6 \times 3}{(6+3)^{2}}(11-14)^{2}$

$\sigma^{2}=\frac{144+108}{9}+\frac{18}{81} \times 9=28+2=30$

varience $=\frac{1^{2}+3^{2}+6^{2}+7^{2}+9^{2}+10^{2}}{6}-\left(\frac{1+3+6+7+9+10}{6}\right)^{2}$

$=10$

$\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2} = 100} $

new observations are $2 \mathrm{x}_{1}, 2 \mathrm{x}_{2}, \ldots \ldots, 2 \mathrm{x}_{20}$

Their mean $=\overline{\mathrm{x}}_{1}=\frac{2\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{20}\right)}{20}=2 \overline{\mathrm{x}}$

Now, variance$ = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {2{x_i} - 2\bar x} \right)}^2}} $

$ = \frac{1}{{20}} \times 4\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2}} $

$ = \frac{1}{{20}} \times 4 \times 100 = 20$

average $\mathrm{m}_{1}=60, \mathrm{m}_{2}=40$

$\sigma_{1}=4, \sigma_{2}=6$

Standard deviation of combined series

$\sigma=\sqrt{\frac{n_{1} \sigma_{1}^{2}+n_{2} \sigma_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(m_{1}-m_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$

$=\sqrt{\frac{10 \times 16+10 \times 36}{10+10}+\frac{10 \times 10(60-40)^{2}}{(10+10)^{2}}}$

$=\sqrt{8+18+100}=\sqrt{126}=11.2$

= $75 + 80 + 85$= $240$

If the marks of another subjected is added, then the marks will be $ \ge $ $240$ out of $400$

Minimum average marks $ = \frac{{240}}{4} = 60\% $,

[When marks in the fourth subject = $0$].

Given, $\bar x = 30$, ${\bar x_1} = 32$, $\overline {{x_2}} = 27$

Let ${n_1} + {n_2} = 100$ and ${n_1}$ denotes men, ${n_2}$ denotes women for this ${n_2} = 100 - {n_1}$

$30 = \frac{{32{n_1} + (100 - {n_1})27}}{{100}}$

==> $30 = \frac{{32{n_1} + 2700 - 27{n_1}}}{{100}}$

==> $3000 - 2700 = 32{n_1} - 27{n_1}$

==>$300 = 5{n_1}$ ==>${n_1} = 60$

So, ${n_2} = 40$

Hence, the percentage of women in the group is $40$.

$ = \frac{3}{{\frac{1}{{30}} + \frac{1}{{25}} + \frac{1}{{50}}}} \ km/hr$.

$40 + \frac{1}{2} = \frac{{35 \times 40 + w}}{{35 + 1}}$

==> $36 \times 40 + 36 \times \frac{1}{2} = 35 \times 40 + w$

==> $w = 58$

Weight of the teacher = $58\ kg$.

Total marks obtained

$= (40 × 50) + (35 × 60) + (45 × 55) + (42 × 45)$

$= 8465$

Overall average of marks per students ,

$ = \frac{{8465}}{{162}} = 52.25$.

Given, $\bar x = 500$,${\bar x_1} = 510$,${\bar x_2} = 460$

Let ${n_1} + {n_2} = 100$ and ${n_1}$ denotes male, ${n_2}$ denotes female for this ${n_2} = 100 - {n_1}$

$500 = \frac{{510{n_1} + (100 - {n_1})460}}{{100}}$

==> $50000{\rm{ }} = 510{n_1} + 46000 - 460{n_1}$

==> $50000{\rm{ }} - 46000 = 50{n_1}$

==> $4000 = 50{n_1}$

==> ${n_1} = \frac{{4000}}{{50}} = 80$.

Hence, the percentage of male employees in the factory is $80$.

Time taken for first half journey is, ${t_1} = (d/{v_1})$ and time taken for rest half journey is, ${t_2} = (d/{v_2})$

$\therefore$ ${V_{av}} = \frac{{2d}}{{(d/{v_1}) + (d/{v_2})}}$$ = \frac{{2{v_1}{v_2}}}{{{v_1} + {v_2}}}$.

Let the given $11$ distinct positive integers are in increasing order

$x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$, so $x_{11}$ is largest of these integers and the median is $x_6$.

Now, median of first $10$ numbers is

$\frac{x_6+x_6}{2}=m$ (Let)

Now, we have to replace largest number $x_{11}$ by $m$ and then increasing order will be

$x_1, x_2, x_3, x_4, x_5, m, x_6, x_7, x_8, x_9, x_{10}$

$m < x_6$ as $x_5 < \frac{x_5+x_6}{2} < x_6$

So, median decreases.

Total number of students $=50$

Average marks of student $=47.5$

$\therefore$ Total marks of students

$=50 \times 47.5=2375$

Now, the student get integer marks Hence, the maximum number of students we will divide total mark by $48$.

$\frac{2375}{48}=49$

Four digits number which is divisible by $7$ are $1001,1008,1015, \ldots .$ $9996 .$

Hence, total number of such numbers $=1286$

$Median=\frac{\left(\frac{N}{2}\right)^{\text {th }} observatio+\left(\frac{N}{2}+1\right)^{\text {th }}\,observation}{2}$

$Median=\frac{\left(\frac{1286}{2}\right)^{\text {th }}\,observation+\left(\frac{N}{2}+1\right)^{\text {th }}\,observation}{2}$

$=\frac{643^{th}+644^{th}}{2}$

$=\frac{(1001+(642)7)+(1001)+(643)7)}{2}$

$=\frac{2(1001)+7(642+643)}{2}$

$=\frac{2(1001)+7(1285)}{2}$

$=1001+4497.5=5498.5$

Given, average income of $x$ people $=P$ and average income of $y$ people $=Q$

$\therefore$ Total income of people in two villages are $P_x$ and $Q_y$ respectively.

One person moves from first village to second village.

Then, number of people in first village

$=x-1$ and second village $=y+1$.

Average income $=P^{\prime}$ and $Q^{\prime}$

$\therefore$ Total income $=P^{\prime}(x-1)$ and $Q^{\prime}(y+1)$

Total income in both cases are same

$\therefore P x+Q y=P^{\prime}(x-1)+Q^{\prime}(y+1)$

$\Rightarrow P x-P^{\prime}(x-1)=Q^{\prime}(y+1)-Q y$

$\Rightarrow x\left(P-P^{\prime}\right)+P^{\prime}=y\left(Q^{\prime}-Q\right)+Q^{\prime}$

$\therefore P^{\prime} \neq P$ and $Q^{\prime} \neq Q$

Hence, option $(c)$ is correct.

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 $10,11 .$

$\Rightarrow a+b=17$

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

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

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 $1$.

$ \Rightarrow \bar x = \frac{{\sum {{x_i}} }}{n} = 9$

$ \Rightarrow \sum {{x_i}}  = 9 \times 5 = 45$

Now, standard deviation $=0$

$\therefore $ all the five terms are same i.e.;$9$.

Now for changed observation 

${{\bar x}_{new}} = \frac{{36 + {x_5}}}{5} = 10$

$ \Rightarrow {x_5} = 14$

$\therefore {\sigma _{new}} = \sqrt {\frac{{\sum {{{\left( {{x_i} - {{\bar x}_{new}}} \right)}^2}} }}{n}} $

$ = \sqrt {\frac{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}}}{5}} $

$ = 2$

$ \Rightarrow {x_1} + {x_2} + ... + {x_{25}} = 1000$ 

Let $A$ be the age of new teacher.

$\therefore {x_1} + {x_2} + ... + {x_{25}} - 60 + A = 39 \times 25$

$ \Rightarrow 1000 - 60 + A = 975$

$ \Rightarrow A = 975 - 940 = 35$

Variance

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

$ = \frac{{2475}}{{97}} - {\left( {\frac{{388}}{{97}}} \right)^2}$

$ = \frac{{2425 - 1552}}{{97}} = \frac{{873}}{{97}} = 9$

$\frac{49}{4}=\frac{4+9+a^{2}+121}{4}-\left(\frac{16+a}{4}\right)^{2}$

$3 a^{2}-32 a+84=0$

$ = \frac{1}{{101}} \times \frac{{101}}{2}\left[ {1 + \left( {1 + 100d} \right)} \right] = 1 + 50d$

mean deviation from mean

$ = \frac{1}{{101}}[\left| {1 - \left( {1 + 50d} \right)} \right.\left|  +  \right|\left( {1 + d} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + 50d} \right)\left| {......} \right|\,\left. {\left[ {1 + 100d} \right] - \left( {1 + 50d} \right)} \right|]$

$ = \frac{{2\left| d \right|}}{{101}}\left( {1 + 2 + 3... + 50} \right)$

$ = \frac{{2\left| d \right|}}{{101}} \times \frac{{50 \times 51}}{2} = \frac{{2550}}{{101}}\left| d \right|$

$ = \frac{{2550}}{{101}}\left| d \right| = 225 \Rightarrow \left| d \right| = 10.1$

$\bar x = 5$

variance $=124$

${x_1} = 1,{x_2} = 2,{x_3} = 6$

$\bar x = 5$

$\frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} = 5$

$ \Rightarrow {x_4} + {x_5} + 9 = 25$

$ \Rightarrow {x_4} + {x_5} = 16$

$ \Rightarrow {x_4} + {x_5} + 10 - 10 = 16$

$ \Rightarrow \left( {{x_4} - 5} \right) + \left( {{x_5} - 5} \right) = 16 - 10$

$ \Rightarrow \left( {{x_4} - 5} \right) + \left( {{x_5} - 5} \right) = 6$

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

$ = \left| {{x_1} - 5} \right| + \left| {{x_2} - 5} \right| + \left| {{x_3} - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|$

$ = \frac{{4 + 3 + 1 + 6}}{5} = \frac{{14}}{5} = 2.8$

$ \Rightarrow \sum\limits_{i = 1}^{16} {{x_i}}  = 16 \times 16$

Sum of new observations

$ = \sum\limits_{i = 1}^{18} {{y_i}}  = (16 \times 16 - 16) + (3 + 4 + 5) = 252$

Number of observations $=18$

$\therefore $ New mean $ = \frac{{\sum\limits_{i = 1}^{18} {{y_i}} }}{{18}} = \frac{{252}}{{18}} = 14$

$70 \times  54 + 30 \times  x = 60 \times100$

$x =74$

${\sigma ^2} = \frac{{\sum x_1^2}}{n} - {\left( {\overline {.x} } \right)^2}$

$\frac{{{2^2} + {4^2} + {6^2} + .... + {{100}^2}}}{{50}}$$ - {\left( {\frac{{2 + 4 + 6 + .... + 100}}{{50}}} \right)^2}$

${i_1} = \frac{{{2^2} + {4^2} + {6^2} + .... + {{100}^2}}}{{50}}$

$ = {2^2}\frac{{{1^2} + {2^2} + {3^2} + ... + {{50}^2}}}{{50}}$

$ = \frac{{{2^2}}}{{50}} \times 50\left( {50 + 1} \right)\left( {100 + 1} \right)$

$ = 3434$

${i_2} = {\left( {\frac{{2 + 4 + 6 + ..... + 100}}{{50}}} \right)^2}$

$ = {\left( {\frac{{50 \times \frac{{2 + 100}}{2}}}{{50}}} \right)^2}$

$ = {\left( {51} \right)^2}$

${\sigma ^2} = 3434 - 2661 = 833$

So, maen $ = \sum\limits_{i = 1}^{2n} {\frac{{{x_i}}}{{2n}}} $

Let these observations be divided into two parts ${{x_1},{x_2},......,{x_n}}$ and ${x_{n + 1}},......{x_{2n}}$ 

Each in ${1^{st}}$ part $5$ is added, so total of 

first part is $\sum\limits_{i = 1}^n {{x_i} + 5n} $.

In second part $3$ is subtracted from each 

So, total os second part is $\sum\limits_{i = n + 1}^{2n} {{x_i} - 3n} $

Total of $2n$ terms are

$\sum\limits_{i = 1}^n {{x_i} + 5n}  + \sum\limits_{i = n + 1}^{2n} {{x_i} - 3n}  = \sum\limits_{i = 1}^{2n} {{x_i} + 2n} $

 Mean $ = \sum\limits_{i = 1}^{2n} {\frac{{{x_i} + 2n}}{{2n}}}  = \sum\limits_{i = 1}^{2n} {\frac{{{x_i}}}{{2n}} + 1} $

Let ${\bar X}$ be the mean and $M.D.$be the mean deviation about ${\bar X}$.

If each observation is incerased by $5$

then new mean will be $\bar X + 5$ and new $M.D.$ about new mean wil be $M.D.$

($\because $ Mean $ = \sum\limits_{i = 1}^n {\frac{{{x_i}}}{n}} $)

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

Mean of ${d_1},{d_2},{d_3},......,{d_n}$

$ = \frac{{{d_1} + {d_2} + {d_3} + ...... + {d_n}}}{n}$

$ = \frac{{\left( { - {x_1} - a} \right) + \left( { - {x_2} - a} \right) + \left( { - {x_3} - a} \right) + ..... + \left( { - {x_n} - a} \right)}}{n}$

$ =  - \left[ {\frac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}} \right] - \frac{{na}}{n}$

$ =  - \bar x - a$

Since, ${d_i} =  - {x_i} - a$ and we multiply or subtract each observation by any number the mode remains the same. Hence mode of $ - {x_i} - a$ i.e. ${d_i}$ and ${x_i}$ are same.

Now variance of ${d_1},{d_2},......,{d_n}$

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{d_i} - \left( { - \bar x - a} \right)} \right]}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ { - {x_i} - a + \bar x - a} \right]}^2}} $

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

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

Given mean $=7$

$\therefore 7 = \frac{{6 + 7 + 8 + 10 + x}}{5}$

$ \Rightarrow x = 4$ Now, Variance 

$ = \sqrt {\frac{{{{\left( {6 - 7} \right)}^2} + {{\left( {7 - 7} \right)}^2} + {{\left( {8 - 7} \right)}^2} + {{\left( {10 - 7} \right)}^2} + {{\left( {4 - 7} \right)}^2}}}{5}} $

$ = \sqrt {\frac{{{1^2} + {0^2} + {1^2} + {3^2} + {3^2}}}{5}}  = \sqrt {\frac{{20}}{5}}  = \sqrt 4  = 2$

$\sigma  = \sqrt {\frac{{\sum \left( {x - \overline {.x} } \right)}}{n}} $

$ = \sqrt {\frac{{\sum \left( {y - 10 - \overline y  + 10} \right)}}{n}} $

$ = \sqrt {\frac{{\sum \left( {y - \overline y } \right)}}{n}} $

variance does not change