Question

Stats help required

Answer 1

Remember the definition of variance?

Answer 2

\[
\text{Var}[X] = E[X-E[X]]
\]Isn't this it?

Answer 3

Whoops, it's: \[
\text{Var}[X] = E[(X-E[X])^2]
\]

Answer 4

Never really studied stats in my life.

Answer 5

Okay, first of all, what is the mean of the first n natural numbers?

Answer 6

Why are you answering stats questions if you don't know anything about stats?

Answer 7

Well i have studied it but never as a separate subject.

Answer 8

Okay

Answer 9

The mean of first n natural numbers is (n+1)/2

Answer 10

Now we want to find \((n - \mu)^2\) where \(\mu\) is the mean.

Answer 11

What does that get you?

Answer 12

It will get you some other sequence. The variance is the average of that sequence.

Answer 13

Well i know that

\[\sigma^2=(n-\mu)^2\]

Answer 14

No.

Answer 15

\[
\sigma^2 = \text{variance} = \text{average} [ (x -\mu)^2]
\]

Answer 16

Oh yeah. :/

Answer 17

so first let's just find the sequence \[
(n-\mu)^2 = \left(n-\frac{n+1}{2}\right)^2 = ?
\]

Answer 18

Then we can find the average of that sequence.

Answer 19

Okay So Variance is actually the Measure of SD

and

\[Variance=\frac{[ \sum_{}^{}(X_i-X(mean)]}{n}\]

Answer 20

Umm, that's what I've been trying to say dude.

Answer 21

Haha lol.

Answer 22

yes you got it . just put value of \[\Large \mu\] in the expression and simplify .

Answer 23

^ Wut?

Answer 24

Okay so basically, \[
E[X] = \frac{1}{n}\sum_{i=1}^nx_i
\]

Answer 25

\[
E[X] = \mu
\]

Answer 26

\[
\text{Var}[X] = \sigma^2
\]

Answer 27

\[
\text{Var}[X] = E[(X-\mu)^2]
\]

Answer 28

\[
\text{Var}[X] = E[(X-\mu)^2] = \frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2
\]

Answer 29

\[Variance=\frac{\sum_{i=1}^{n} X_i^2}{n}-(Mean X)^2\]

Answer 30

In this case \(x_i=n\) and we know already \(\mu=(n+1)/2\) So: \[
\text{Var}[X] = \frac{1}{n}\sum_{i=1}^n\left(i-\frac{n+1}{2}\right)^2
\]

Answer 31

How do we know That mean=(n+1)/2

Answer 32

I should have said \(x_i=i\)

Answer 33

You already said that yourself!!!!

Answer 34

Yeah that was understood.

Answer 35

What next?

Answer 36

I have no idea how to help you because I have no idea what you don't get.

Answer 37

\[
\text{Var}[X] = \frac{1}{n}\sum_{i=1}^n\left(i-\frac{n+1}{2}\right)^2
\]Simplify this summation!

Answer 38

lol :P

Thanks :D

Answer 39

What will i get after i expand it?

Answer 40

\[
\text{Var}[X] = \frac{1}{n}\sum_{i=1}^n\left(i-\frac{n+1}{2}\right)^2 = \frac{1}{n}\sum_{i=1}^n\left(i^2-i(n+1)+\frac{(n+1)^2}{4} \right)
\]

Answer 41

Thanks a lot :)

Answer 42

What about the sigma rules used its confusing me please help @wio @sami-21

Answer 43

Oh my god really?

Answer 44

No not all of them Only the middle one.

Answer 45

\[

\frac{1}{n}\sum_{i=1}^n\left(i^2-i(n+1)+\frac{(n+1)^2}{4} \right) \\
= \frac{1}{n}\sum_{i=1}^ni^2- \frac{1}{n}\sum_{i=1}^ni(n+1)+ \frac{1}{n}\sum_{i=1}^n\frac{(n+1)^2}{4} \\
= \frac{1}{n}\sum_{i=1}^ni^2- \frac{1}{n}(n+1)\sum_{i=1}^ni+ \frac{1}{n}\frac{(n+1)^2}{4}\sum_{i=1}^n1
\]