Statistics

Kindly go throuigh the solution

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

->(a - b) (a - b + 4) = 0

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

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$  

Variance = $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$  

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$      

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4\lambda or4\lambda +1from.$  

As each square form is $4\lambda or4\lambda +1$  

Variance of 2001, 2003, 2006, 2007, 2009, 2010 is same as variance of 1, 3, 6, 7, 9, 10

mean =  $\frac{1+3+6+7+9+10}{6}=6$

var (x) =  $\frac{1^2+3^2+6^2+7^2+9^2+10^2}{6}?36$

var (x) = 10

  $56=\frac{50n+60m}{m+n}$

56 n + 56 m = 50 n + 60 m

6n = 4m

$\frac{n}{m}=2/3$            

$\therefore 9.\frac{n}{m}=9*\frac{2}{3}=6$            

C.V. = (σ/x? ) * 100 ⇒ σ = (C.V. * x? ) / 100
∴ σ? = (50*30)/100 = 15 and σ? = (60*25)/100 = 15 ⇒ σ? - σ? = 0

C.O.V = σ/x? * 100

mean = Σx? f? /Σf? = (32+8α+9β)/ (8+α+β)=6
⇒2α+3β=16 . (i)
d? =x? −x? =−4,0,2,3
f? d? ²=64,0,4α,9β
Variance σ²=Σf? d? ²/Σf? =6.8
⇒ (64+4α+9β)/ (8+α+β)=6.8
⇒2.8α+ (−2.2β)=9.6
⇒28α−22β=96
14α−11β=48 . (ii)
Solving (i) and (ii),
⇒β=2, α=5
New mean=Σx? f? /Σf? =85/15=17/3

Var (1, 2, ., n) = (Σn²/n) - (Σn/n)² = 10.
(n (n+1) (2n+1)/6n) - (n (n+1)/2n)² = 10.
(n+1) (2n+1)/6 - (n+1)/2)² = 10.
(n+1)/12 * [2 (2n+1) - 3 (n+1)] = 10.
(n+1)/12 * (4n+2 - 3n-3) = 10.
(n+1) (n-1)/12 = 10.
n² - 1 = 120 ⇒ n² = 121 ⇒ n = 11.

Var (2, 4, ., 2m) = Var (2* (1, 2, ., m) = 2² * Var (1, 2, ., m) = 16.
4 * Var (1, 2, ., m) = 16.
Var (1, 2, ., m) = 4.
Using the formula from above: (m²-1)/12 = 4.
m² - 1 = 48 ⇒ m² = 49 ⇒ m = 7.
m + n = 7 + 11 = 18.

The data consists of n values of a and n values of -a.
Mean x? = (n*a + n* (-a) / 2n = 0 / 2n = 0.
Variance σ² = (Σx? ²)/N - x? ² = (n*a² + n* (-a)²) / 2n - 0² = 2na² / 2n = a².
If a value b is added to all observations, the new mean is x? ' = x? + b = 0 + b = b.
We are given the new mean is 5, so b=5.
Adding a constant does not change the variance. The new variance is still a².
We are given the new standard deviation is 20, so the new variance is 20² = 400.
Thus, a² = 400.
The required value is a² + b² = 400 + 5² = 400 + 25 = 425.