Maths Statistics

If each observation is multiplied with p and then q is subtracted
New mean x? = px? - q ⇒ 10 = p(20)-q
and new standard deviations σ? = |p|σ? ⇒ 1 = |p|(2) ⇒ |p|=1/2 ⇒ p=±1/2
If p=1/2, then q=0. If p=-1/2, q=-20.

5+3+7+a+b = 25 ⇒ a+b=10
S.D. = √(((5²+3²+7²+a²+b²)/5) - 5²) = 2
√((a²+b²+83)/5) - 25 = 4 ⇒ a²+b² = 62
⇒ (a+b)² - 2ab = 62 ⇒ ab = 19
So equation whose roots are a and b is x² - 10x + 19 = 0

Mean = (6+10+7+13+a+12+b+12)/8 = (60+a+b)/8 = 9 ⇒ a+b=12.
Variance = (Σx? ²/n) - (mean)² = 37/4.
Σx? ²/8 - 81 = 37/4.
Σx? ² = 8 (81+9.25) = 8 (90.25) = 722.
Σx? ² = 36+100+49+169+a²+144+b²+144 = 642+a²+b².
642+a²+b²=722 ⇒ a²+b²=80.
(a+b)²=144 ⇒ a²+b²+2ab=144 ⇒ 80+2ab=144 ⇒ 2ab=64.
(a-b)² = a²+b²-2ab = 80-64 = 16.

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

${\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2=\frac{9+k^2}{10}-{\left(\frac{9+k}{10}\right)}^2<10$

$\Rightarrow k<\frac{10\sqrt[]{10}}{3}+1\Rightarrow k\le 11$        

$\therefore$ maximum value of k is 11

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

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

->a + b = 18                                   -(ii)

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

->(a – b)² = 4

10 =   $\frac{7+10+11+15+a+b}{6}\Rightarrow a+b=17$  . (i)

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

a² + b² = 145       . (ii)

solving (i) and (ii) we get a = 8, b = 9 or a = 9, b = 8

|a – b|= 1

a, b, c, d, e be 5 unknown

n = 7, mean = 8, variance = 16

sum of observations = 7 * 8 = 56

mean of 5 remaining observation = $\frac{56-8-6}{25}=\frac{42}{5}$

$16=\frac{\sum x_i^2}{7}-64$

$\sum x_i^2=560$

$\begin{array}{l}{a}^2+b^2+c^2+d^2+e^2=460\\ =460-{\left(\frac{42}{5}\right)}^2\end{array}$

$=\frac{536}{25}$                                                

$t_r=\left(3r+4\right)\left(2r+6\right)=6r^2+26r+24$

$\therefore {\displaystyle \sum _{r=1}^{10}{t}_r}=\frac{6*10*11*21}{6}+\frac{26*10*11}{2}+24*10=10*398$

$Mean=\frac{10*398}{10}=398$

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