Maths M.C.Q (1 Marks)

We know 

Mean $ = \frac{{4 + 7 + 2 + 8 + 6 + a}}{6}$

$ \Rightarrow 7 = \frac{{4 + 7 + 2 + 8 + 6 + a}}{6} \Rightarrow a = 15$

Now, given observations can be written in ascending order which is $2,4,6,7,8,15$

Since, No. of observation is even 

$\therefore $ Median 

$ = \frac{{{{\left( {\frac{6}{2}} \right)}^{th}}observation + {{\left( {\frac{6}{2} + 1} \right)}^{th}}observation}}{2}$

$ = \frac{{{3^{rd}}observation + {4^{th}}observation}}{2}$

$ = \frac{{6 + 7}}{2} = \frac{{13}}{2}$

Now, Mean deviation $ = \frac{{\sum\limits_{i = 1}^6 { \ge \left| {{x_i} - \frac{{13}}{2}} \right|} }}{6}$

Statement $2 : A.M.$ of $2 x_{1}, 2 x_{2}, \ldots ., 2 x_{n}$

$=\frac{2\left(x_{1}+x_{2}+\ldots . .+x_{n}\right)}{n}=2 \bar{x}$

Given $A . M .=4 \bar{x} $

$ \therefore$ Statement $2$ is false.

$M = l + \frac{{\frac{N}{2} - F}}{f} \times C$ where

$l=$ lower limit of the median -class

$f=$ frequency of the median class

$N=$ total frequency

$F=$ cumulative frequency of the class just before the median class

$C=$ legth of median class

Now, given , $M=25,N=100,F=45,C=20-30=10,l=20$.

$\therefore $ By using formula, we have

$25 = 20 + \frac{{50 - 45}}{f} \times 10$

$25 - 20 = \frac{{50}}{f} \Rightarrow 5 = \frac{{50}}{f} \Rightarrow f = 10$

Mean deviation about median $=50$

$\Rightarrow \frac{\Sigma\left|x_{i}-25.5 a\right|}{50}=50$

$\Rightarrow 24.5 \mathrm{a}+23.5 \mathrm{a}+\ldots . .+0.5 \mathrm{a}+0.5 \mathrm{a}+\ldots .+24.5 \mathrm{a}=2500$

$\Rightarrow a+3 a+5 a+\ldots \ldots+49 a=2500$

$\Rightarrow 25 / 2(50 a)=2500 \Rightarrow a=4$

Median $=25.5 \mathrm{a}$

Mean deviation about median $=50$

$\Rightarrow \frac{\Sigma\left|x_{i}-25.5 a\right|}{50}=50$

$\Rightarrow 24.5 \mathrm{a}+23.5 \mathrm{a}+\ldots . .+0.5 \mathrm{a}+0.5 \mathrm{a}+\ldots .+24.5 \mathrm{a}=2500$

$\Rightarrow a+3 a+5 a+\ldots .+49 a=2500$

$\Rightarrow 25 / 2(50 a)=2500 \Rightarrow a=4$

Also given that $\frac{\Sigma x_{i}}{5}=2$ and $\frac{\Sigma y_{i}}{5}=4$

$\Rightarrow \Sigma x_{i}=\bar{x}=10$ and $\Sigma y_{i}=\bar{y}=20$

$\sigma_{x}^{2}=\left(\frac{1}{5} \Sigma x_{i}^{2}\right)-(\bar{x})^{2}$

$=\left(\frac{1}{5} \Sigma x_{i}^{2}\right)-(2)^{2}$      ......$(i)$

$\sigma_{y}^{2}=\frac{1}{5}\left(\Sigma y_{i}^{2}\right)-(\bar{y})^{2}$

$=\frac{1}{5}\left(\Sigma y_{i}^{2}\right)-16$         ..........$(ii)$

Substituting $\sigma_{x}^{2}=4$ in $(i)$ we get

$4=\left(\frac{1}{5} \Sigma x_{i}^{2}\right)-4$

$\Rightarrow 4+4=\frac{1}{5} \Sigma x_{i}^{2}$

$\Rightarrow \Sigma x_{i}^{2}=40$

Similarly by substituting $\sigma_{y}^{2}=5$ in $(ii)$ we have

$5=\frac{1}{5} \Sigma y_{i}^{2}-16$

$\Rightarrow 5+16=\frac{1}{5} \Sigma y_{i}^{2}$

$\Rightarrow 21=\frac{1}{5} \Sigma y_{i}^{2}$

$\Rightarrow \Sigma y_{i}^{2}=105$

Combined varience $=\sigma_{z}^{2}=\frac{1}{10}\left(\Sigma x_{i}^{2}+\Sigma y_{i}^{2}\right)-\left(\frac{\bar{x}+\bar{y}}{2}\right)^{2}$

$=\frac{1}{10}(40+105)-\left(\frac{2+4}{2}\right)^{2}$

$=\frac{145-90}{10}$

$=\frac{55}{10}=\frac{11}{2}$

No. of terms in this series $=101$

Mean of this series $=\bar{x}=\frac{1+(1+d)+(1+2 d)+\ldots+(1+100 d)}{101}$

$1+(1+d)+(1+2 d)+\ldots \ldots+(1+100 d)=\frac{101}{2}(2+(101-1) d)$

$=101(50 d+1)$

$\therefore \bar{x}=\frac{101(50 d+1)}{101}=1+50 d$

Therefore mean deviation from mean

$ = \frac{1}{{101}}\sum\limits_{r = 0}^{100} {\left[ {\left( {1 + rd} \right) - \left( {1 + 50d} \right)} \right]} $

$=\frac{2 d}{101}\left(\frac{50 \times 51}{2}\right)$

$\Rightarrow 255=\frac{50 \times 51 \times d}{101}$

$d=\frac{255 \times 101}{50 \times 51}$

$=10.1$

$\Rightarrow(6-a)^{2}+(6-b)^{2}+4+1+16=34$

$(6-a)^{2}+(6-b)^{2}=34-21$

$(6-a)^{2}+(6-b)^{2}=13$

$(6-a)^{2}+(6-b)^{2}=9+4$

$(6-a)^{2}+(6-b)^{2}=3^{2}+2^{2}$

$(6-a)^{2}=3^{2}(6-b)^{2}=2^{2}$

$6-a=3 \quad 6-b=2$

$-a=-3 \quad-b=-4$

$a=3$

$b=4$

$\therefore 52 x+42 y=50(x+y)$

$52 x+42 y=50 x+50 y$

$52 x-50 x=50 y-42 y$

$2 x=8 y$

$x=4 y$

Total number of students in the class $=x+y$

$=4 y+y$

$=5 y$

Percentage of boys $=\frac{4 y}{5 y} \times 100^{20}$

$=80$

Series $\mathrm{B}=151,152 \ldots \ldots .250$

Here series $\mathrm{B}$ can be obtained if we change the origin of $A$ by $50$ units.

And we know the variance does not change by changing the origin.

So, $\quad V_{A}=V_{B}$

$\Rightarrow \quad \frac{V_{A}}{V_{B}}=1$

Mode = $3$ Median -$2$ Mean = $3(22) -2(21)$

$= 66 -42 = 24.$
 

 $\sqrt {\frac{{\sum\limits_{i = 1}^n {x_i^2} }}{n}} \ge \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n} = \sqrt {\frac{{400}}{n}} \ge \frac{{80}}{n} \Rightarrow n \ge 16$

Hence, possible value of $n = 18.$

$2 = \sqrt {\frac{{n{a^2} + n{a^2}}}{{2n}}} = \sqrt {{a^2}} = \pm a$.

Hence $|a|\; = 2$.

$ = {5^{th\ }} {\rm{ term}}$.

Now, last four observations are increased by $2$.

 The median is $5^{th}$ observation, which is remaining unchanged.

 There will be no change in median.

Increase in $\sum x = 10$, then $\sum x' = 170 + 10 = 180$

Increase in $\sum {x^2} = 900 - 400 = 500$, then

$\sum {x'^2} = 2830 + 500 = 3330$

Variance $ = \frac{1}{n}\sum {x'^2} - {\left( {\frac{{\sum x'}}{n}} \right)^2}$

$ = \frac{{3330}}{{15}} - {\left( {\frac{{180}}{{15}}} \right)^2} = 222 - 144 = 78$.

i.e., $\frac{{7200 - 5250}}{{30}} = x$; 

$x = 65.$

$=\frac{2 \times 60+1 \times 70+1 \times 70+2 \times 80}{2+1+1+2}=70$

MD from median $=\frac{\sum \mid X-\text { median } \mid}{n}$

MD from mode $=\frac{\sum \mid X-\text { mode } \mid}{n}$

Since median $>$ mean $(\bar{X})$ and median $>$ mode.

So, It is clear that the mean deviation from median has the least value.

$=4\left(\frac{25}{5}-\left(\frac{5}{5}\right)^{2}\right)=4(5-1)=16$

$\therefore \mathrm{S} \mathrm{D}=\sqrt{16}=4$

$=\frac{^{20} \mathrm{C}_{10}}{11}-\left(\frac{2^{10}}{11}\right)^{2}$

$=\frac{11 \cdot^{20} \mathrm{C}_{10}-2^{20}}{121}$

$ \Rightarrow \frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2} \geq 0$

$\Rightarrow \quad \frac{400}{n}-\frac{10000}{n^{2}} \geq 0 $

$\Rightarrow n \geq 25$

$\frac{\sum\left|x_{i}-\bar{x}\right|}{100}=5 \Rightarrow \sum\left|x_{i}\right|=500$

$ \Rightarrow \sum {x_i^2}  + 2\sum\limits_{1 \le i < j \le 100} {\left| {{x_i}{x_j}} \right|}  = {(500)^2}$

$\Rightarrow \frac{\sum x_{i}^{2}}{100}=\frac{(500)^{2}-2 \sum\left|x_{i} x_{j}\right|}{100}=2500-1600$

$S. D.=\sqrt{\frac{\sum\left(x_{i}-\bar{x}\right)^{2}}{100}}=\sqrt{900}=30$

$= \text { variance of }\{3,6,9, \ldots \ldots, 93\} $

$= 9(\text { variance of }\{1,2,3, \ldots .31\}) $

${v_2} = {\rm{variance of }}\{ 20,26,32, \ldots .,200\} $

$ = {\rm{ variance of }}\{ 6,12,18, \ldots .,186\} $

$=36 \text { (variance of }\{1,2,3, \ldots . .31\}) $

$ \therefore  \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{4} $

$55 = l.48 + {\rm{k}}$          .........$(i)$

standard deviation of

${\omega _i} = l$ (standard deviation of ${{{\rm{y}}_i}}$)

$15 = l.12$             ...........$(ii)$

$l = 1.25$ and $\mathrm{k}=-5$

$=\frac{45}{18}-\left(\frac{9}{18}\right)^{2}=\frac{5}{2}-\frac{1}{4}=\frac{9}{4}$

then $S.D.$ of $x_{1} \forall i=1,2,3, \ldots . .18$

$=\sqrt{\frac{9}{4}}=\frac{3}{2}$

$\overline{\mathrm{x}}_{1}=25 \quad \overline{\mathrm{x}}_{2}=10$

$\sigma_{1}=3 \quad \sigma_{2}=4$

combined mean $=\frac{25 \times 200+10 \times 300}{500}=16$

$\sigma_{1}^{2}=9=\frac{1}{200}\left(\sum x_{i}^{2}\right)-625$

$126800=\sum x_{i}^{2}$

$\sigma_{2}^{2}=16=\frac{1}{300} \sum y_{1}^{2}-100$

$34800=\sum y_{1}^{2}$

$\sigma^{2}=\frac{1}{500}(126800+34800)-(16)^{2}$

$=323.2-256=67.2$

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

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

Standard deviation of combined series

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

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

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