Question

A population is listed as the following: 10 115 19 11. 12. 13. 14. 15. 16. Find μ =? Find Σ(x-μ)-? Find Σ(x-A)-? Find ΣΧ_? Find Σχ2-? Find σ-?

Answer 1

Solution:

Given data :

Population list = 1,3,6,10,15,19

Now, we have to find out the :

11. Find the $\mu$:

We know the ,

$\mu$= mean

The formula for the Mean = ( sum of the all samples ) / (no.of samples )

= 1+3+6+10+15+19

= 54 / 6

= 9

  $\therefore$ Mean  $\mu$ = 9

12. Find the  $\sum ( X- \mu )$ :-

Now,  $\sum ( X- \mu )$ =  ( 1-9) + (3-9) + ( 6-9) + ( 10-9) + ( 15-9) +( 19-9)

= - 8 - 6 - 3 +1 + 6 + 10

= -17 + 17

   = 0

$\therefore$   $\sum ( X- \mu ) =0$

13. Find the $\sum ( X- \mu )^{2}$ :-

Now, $\sum ( X- \mu )^{2}$ = ( 1-9)^2 + (3-9)^2 + ( 6-9)^2 + ( 10-9)^2 + ( 15-9)^2 +( 19-9)^2

$\sum ( X- \mu )^{2}$ =   (- 8)^2 + (- 6 )^2 + (- 3)^2 +( 1)^2 + ( 6 )^2 + (10 )^2

= 64+ 36+ 9+ 1+36+100

= 246

$\therefore$   $\sum ( X- \mu )^{2} = 246$

14. Find the $\sum x$ :-

Now, $\sum x$ = 1+3+6+10+15+19

= 54

$\therefore$ $\sum x = 54$

15. Find the $\sum x^{2}$:-

Now, $\sum x^{2}$  = 1^2 +3^2 + 6^2 + 10^2 + 15^2 +1 9 ^2

= 1+ 9 + 36 +100+ 225+ 361

= 732

$\therefore$  $\sum x^{2} =732$

16. Find the $\sigma$ :-

We know, the formula for the $\sigma$ is

$\sigma$  =  $\sqrt{\frac{\sum (x-\mu )^{2}}{n-1}}$

= $\sqrt{\frac{( 1-9)^2 + (3-9)^2 + ( 6-9)^2 + ( 10-9)^2 + ( 15-9)^2 +( 19-9)^2}{20-1}}$

= $\sqrt{\frac{ (- 8)^2 + (- 6 )^2 + (- 3)^2 +( 1)^2 + ( 6 )^2 + (10 )^2 }{19}}$

= $\sqrt{\frac{ 64+ 36+ 9+ 1+36+100}{19}}$

=  $\sqrt{\frac{246}{19}}$

  = $\sqrt{12.947}$

= 3.5981

$\therefore$  $\sigma$ = 3.5981