Statistics

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$         

(i) & (ii)  ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

=> (α - β) (α - β + 4) = 0

Since   $\alpha \ne \beta so\left|\alpha -\beta \right|=4$

$ForMedian=L+\frac{\frac{N}{2}-c,}{f}xh$

$14=12+\frac{\frac{a+b}{2}+13-\left(a+b\right)}{12}*6$              

->a + b = 18                                   -(ii)

Solving (i) & (ii) a = 8 & b = 10

->(a – b)² = 4

For every a, there  must be a² – 2. So, there will be infinitely many pairs (a, b)

Variance = $\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n}=\frac{\sum \left({x_i}^2+{\overline{x}}^2-2\overline{x}xi\right)}{n}$  

$=\frac{\sum {x_i}^2+{\overline{x}}^2\sum 1-2\overline{x}\sum x_i}{n}$

$=\frac{\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(\frac{n\left(n+1\right)}{2n}\right)}^2.n-\frac{2\left(n+1\right)}{2n}.\frac{n\left(n+1\right)}{2}}{n}$ $=\frac{n^2-1}{12}$

Now, $\frac{n^2-1}{12}=14son=13$

$\sum X=200$

$\sum x^2=2125$

For new data

$\sum x=200-25+35=210$

$\overline{x}=10.5$

$\sum x^2=2725$

${\sigma }^2=\frac{2725}{20}-{\left(10.5\right)}^2=26$

$\frac{3+7+x+y}{4}=5$              

$\Rightarrow x+y=10.........\left(i\right)$

$\frac{1}{4}\left(9+49+x^2+y^2\right)-25=10$              

x² + y² = 82 .(ii)

$\therefore \overline{x}=\frac{\left(3+2x\right)+\left(7+2y\right)+\left(x+y\right)+\left(x-y\right)}{4}$              

$=\frac{10+4x+2y}{4}$              

Solving (i) & (ii), x = 9, y = 1

$\therefore \overline{x}=\frac{48}{4}=12$          

$mean=\frac{n\left(n+1\right)/2}{n}$           

  $=\frac{n+1}{2}$            For n = 2k – 1, k   $\in N$

mean = k

mean deviation = avg. of deviations

$=\frac{2\left(\stackrel{0+}{1}+2+...\left(k-1\right)\right)}{2k-1}=\frac{\left(k-1\right)k}{2k-1}$

$=\frac{5\left(2k\right)}{k-1}\Rightarrow k=11\Rightarrow n=2k-1=21$            

               

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15;\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150;\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance =  $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$

Z distribution and Chi-Squared are some of the most popular distribution patterns of probability, and it is vital to recognise the variations between them and when to use the distribution pattern. A Z table is of no use when the operation revolves around a smaller sample size. On the other hand, the distribution of a sum of independent regular k squares in standard normal variables is the chi-square distribution of k degrees of freedom. The tests are used for the independence of two variables in an incident table and to assess the observable data for