Statistics

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}{\displaystyle \sum _{}^{}{x}_i}=1+2+3+4+\dots +n=\frac{n\left(n+1\right)}{2}\\ {\displaystyle \sum _{}^{}{x}_i^2}=1^2+2^2+3^2+4^2+\dots +n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\ \therefore S.D.\left(\sigma \right)=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2}\\ =\sqrt[]{\frac{n\left(n+1\right)\left(2n+1\right)}{6n}-\frac{n^2{\left(n+1\right)}^2}{4n^2}}\\ =\sqrt[]{\frac{\left(n+1\right)\left(2n+1\right)}{6}-\frac{{\left(n+1\right)}^2}{4}}\\ =\sqrt[]{\frac{2n^2+3n+1}{6}-\frac{n^2+2n+1}{4}}\\ =\sqrt[]{\frac{4n^2+6n+2-3n^2-6n-3}{12}}=\sqrt[]{\frac{n^2-1}{12}}\\ Hence,therequiredSD=\sqrt[]{\frac{n^2-1}{12}}.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Firstnnaturalnumbersare1,2,3,4,5,6,\dots ,n.Here,niseven.\\ \therefore Mean\overline{x}=\frac{1+2+3+4+\dots +n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n+1}{2}\\ \therefore MD=\frac{1}{n}\left[\begin{array}{l}\left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|+\dots +\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|\frac{n}{2}-\frac{n+1}{2}\right|+\\ \left|\frac{n+2}{2}-\frac{n+1}{2}\right|+\dots +\left|n-\frac{n+1}{2}\right|\end{array}\right]\\ =\frac{1}{n}\left[\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+\dots +\left|\frac{-3}{2}\right|+\left|-\frac{1}{2}\right|+\left|\frac{1}{2}\right|+\dots +\left|\frac{n-1}{2}\right|\right]\\ =\frac{1}{n}\left[\frac{1}{2}+\frac{3}{2}+\dots +\frac{n-1}{2}\right]\left(\frac{n}{2}\right)terms\\ =\frac{1}{n}{\left(\frac{n}{2}\right)}^2=\frac{1}{n}.\frac{n^2}{4}=\frac{n}{4}\left[?Sumoffirstoddnnaturalnumbers=n^2\right]\\ Hence,therequiredMD=\frac{n}{4}.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Firstnnaturalnumbersare1,2,3,\dots ,n.Here,nisodd.\\ \therefore Mean\overline{x}=\frac{1+2+3+\dots +n}{n}=\frac{\frac{n\left(n+1\right)}{2}}{n}=\frac{n+1}{2}\\ Thedeviationsofnumbersfrommean\left(\frac{n+1}{2}\right)are\\ 1-\frac{n+1}{2},2-\frac{n+1}{2},3-\frac{n+1}{2},\dots ,n-\frac{n+1}{2}\\ i.e.,-\frac{n-1}{2},-\frac{n-3}{2},\dots ,-2,-1,0,1,2,\dots ,\frac{n-1}{2}.\\ Theabsolutevaluesofdeviationfromthemeani.e.,\left|x_i-\overline{x}\right|are\\ \frac{n-1}{2},\frac{n-3}{2},\dots ,2,1,0,1,2,\dots ,\frac{n-1}{2}.\\ Thesumofabsolutevaluesofdeviationfromthemeani.e.,\left|x_i-\overline{x}\right|are\\ =2\left(1+2+3+\dots to\frac{n-1}{2}terms\right)\\ =2.\frac{\frac{n-1}{2}\left(\frac{n-1}{2}+1\right)}{2}=\frac{n-1}{2}.\frac{n+1}{2}=\frac{n^2-1}{4}.\\ \therefore Meandeviationaboutthemean=\frac{{\displaystyle \sum _{}^{}\left|x_i-\overline{x}\right|}}{n}=\frac{\frac{n^2-1}{4}}{n}=\frac{n^2-1}{4n}.\end{array}$

$\bar{x}=10$

$\Rightarrow \stackrel{?}{\mathrm{x}}=\frac{63+\mathrm{a}+\mathrm{b}}{8}=10$

$\Rightarrow a+b=17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

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

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

$\Rightarrow a^2+b^2=169$

From (1) and (2) ; $a=12$ and $b=5$

 $\overline{x}=15\Rightarrow {\displaystyle \sum _{i=1}^{20}{x}_i=300}$

$\frac{{\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2}}{20}=9\Rightarrow {\displaystyle \sum _{i=1}^{20}{\left(x_i-15\right)}^2=180}$

${\displaystyle \sum _{i=1}^{20}{\left(x_i+\alpha \right)}^2=178*20=3560}$

$4680+2\alpha \left(300\right)+20{\alpha }^2=3560$

${\alpha }^2+30\alpha +234-178=0$

= 2, 28

${\alpha }_{max}2={\left(-2\right)}^2=4$

 Since,  ${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x}$ exist $\Rightarrow f\left(0\right)=0$

Now,  $f^{\mathrm{\text{'}}}\left(x\right)={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)+x{h}^2+x^{2}h}{h}($ take $y=h)$

$={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)}{h}+{\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to .0}?\left(xh\right)+x^2$

$\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1+0+{\mathrm{x}}^2$

$\Rightarrow f^{\mathrm{\text{'}}}\left(3\right)=10$

 $=\frac{4+5+6+6+7+8+x+y}{8}=6$

$\therefore x+y=12$ …. (i)

And variance

$=\frac{2^2+1^2+0^2+0^2+1^2+2^2+{\left(x-6\right)}^2+{\left(y-6\right)}^2}{8}$

$=\frac{9}{4}$

$\therefore {\left(x-6\right)}^2+{\left(y-6\right)}^2=8$ ……. (ii)

From (i) and (ii)

x = 4 and y = 8

$\therefore x^4+y^2=320$

34. Given, n = 100.

incorrect mean ( $\overline{x}$ ) = 20.

incorrect standard deviation (σ) = 3

We know that,

$\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{n}_i}$

$\Rightarrow 20=\frac{1}{100}{\displaystyle \sum _{i=1}^n{n}_i}$

$\Rightarrow {\displaystyle \sum _{i=1}^n{n}_i}=2000.$

So, incorrect sum of observation = 2000

Correct sum of observation $=2000-21-21-18$

= 1940

33. C.V in mathematics = $\frac{12}{42}*100=28.57.$

C.V in Physics = $\frac{15}{32}*100=46.87.$

C.V in chemistry = $\frac{20}{40.9}*100=48.89$

$\therefore$  Chemistry has the highest variability and mathematics has the lowest variability.

32.  (i) Given, n = 20.

Incorrect mean $(\overline{x})=10$

Incorrect standard deviation $(\sigma )=2$

We know that,

$\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$

$\Rightarrow 10=\frac{1}{20}{\displaystyle \sum _{i=1}^{20}{x}_i}$

$\Rightarrow {\displaystyle \sum _{i=1}^{20}{x}_i}=200$

So, incorrect sum of observation = 200.

correct sum of observation =200 – 8 = 192

And correct mean $=\frac{192}{19}=10.1$