Question

is this asking what the variance is????

Answer 1

what?

Answer 2

0

Answer 3

not without the square
if i am not mistaken it must add up to zero

Answer 4

yes. that's correct.

Answer 5

:)
THIS is variance:
\[\Large \sum_{i=1}^6(X_i-\bar X)^2\]

Answer 6

you are subtracting off the mean from each entry, and adding them up
got to get zero when you do that
of course you can always check it

Answer 7

By definition, it is always 0 isnt it?

Answer 8

im soo confused.. can someone work me through it please?

Answer 9

Definition, maybe :)
LOL
If you have 6 elements (this can always be generalised to n elements) and their mean is \(\large \bar X\) Then
\[\Large {\sum_{i=1}^6X_i }=6\bar X\]
right? :D

Answer 10

Right? So that if we divide both sides by 6, on the left side, you get the mean (the average of the 6 elements)

Answer 11

This is not variance. Variance is similar to this; however, to be a variance, you need to divide by the number of terms, 6.

Answer 12

it is possible that there is a typo in the question
and maybe there was supposed to be a square outside the parentheses
but if there is no typo then the answer has to be zero
it it clear what the \(\sum\) means?

Answer 13

Variance is what is called a second moment. E[X^2], where E is a function indicating the expected value of X^2. The mean is a 1st moment, E[X]. All this says is that you sum the values and weight them by the probability of it's occurrence. Since the observations are equally likely in your case you simply weight each term by 1/6. That's why the dividing by 6 makes this a variance.

Answer 14

the answer was 0

Answer 15

how bout this problem?

Answer 16

\[\large \sum_{i=1}^6x_i-\overline {x}\] means
\[x_1-\overline {x}+x_2-\overline {x}+x_3-\overline {x}+x_4-\overline {x}+x_5-\overline {x}+x_2-\overline {x}\]

Answer 17

it has to be zero, because you are subtracting the mean from each entry and then adding

Answer 18

i am soo lost with that.. i moved on

Answer 19

lets look at a simple example
the mean of 10 and 12 is 11
if you compute \(10-11+12-11\) you get \(-1+1=0\)

Answer 20

That it equals zero is not surprising, you are summing the distances from the mean, positive and negative terms will cancel. If you were to square each difference and divide by 6, you would have the variance.

Answer 21

always works that way

Answer 22

@4sodapop
But is the meaning of this
\[\Large \sum_{i=1}^6X_i\] clear to you?

Answer 23

ya

Answer 24

That's what makes the average and average (i.e. \( \bar{x} \) ). The terms above the mean will equal the terms below the mean (in terms of distance). They will cancel.

Answer 25

Well then, that's basically the sum of all the terms of Xi, so if we divide it by 6, we get the mean, right?
\[\Large \frac{\sum X_i}{6}=\bar X\]

Answer 26

is it true or false?

Answer 27

What is true? This?
Remember that the sigma just means you add them up so if you divide it by 6 (the number of terms) then you should get the mean or average...

Answer 28

ohh i thought i posted something else.
i thought i posted this

Answer 29

That's like the generalisation for your previous question D:

Answer 30

huh?
i need to finish thisss, but ive been doing math non stop for 9 hours today.. my brain is fried!!

Answer 31

Take 1 value, x. It's average is x. Subtracting x from x give you zero. Take two values a and b. The average will be (a + b)/2, then a - (a+b)/2 + b - (a+b)/2 = a + b - (a + b) = 0, and so on...

Answer 32

WHAT...?

Answer 33

Let's have a real good think.
\[\Large \sum_{i=1}^n X_i\]
You know what this means, right?

Answer 34

so its false?

Answer 35

We'll find out @4sodapop

Answer 36

no..

Answer 37

You don't know what it means? D:
\[\Large \sum_{i=1}^nX_i\]

It just means the sum of all the terms from \(\large X_1\) to \(\large X_n\) aye? :3

Answer 38

ya

Answer 39

So, if we divide everything by n, we get the mean, or the average \(\Large \bar X\)
\[\Large \frac1n\sum_{i=1}^nX_i = \bar X\]

Answer 40

ok

Answer 41

If we multiply both sides by n, we get...\[\Large \sum_{i=1}^nX_i = n\bar X\]

Answer 42

ok

Answer 43

Now we'll leave that...
Look at this
\[\Large \sum_{i=1}^n(X_i - \bar X)\]
You know that the sigma distributes over addition, and subtraction, so
\[\Large = \sum_{i=1}^nX_i - \sum_{i=1}^n\bar X\]
Now simplify...

Answer 44

okk

Answer 45

how...?

Answer 46

how? We just arrived at a value for this...
\[\LARGE \color{red}{\sum_{i=1}^nX_i}-\sum_{i=1}^n\bar X\]
if you'd just scroll up at what we derived...

Answer 47

but we dont have numbers

Answer 48

don't need them. they're asking for the sum of the deviations which is 0

Answer 49

so its true?

Answer 50

You kept saying 'ok' I thought you understood :(
Or you weren't paying attention? :3

Answer 51

im soo lost..
:(((

Answer 52

variance is this
\[\frac{ \sum_{1}^{n} \left( x-\mu \right)^{2}}{ n-1 }\]

Answer 53

u?

Answer 54

or xbar... xbar for a sample, u for population

Answer 55

ok

Answer 56

there's no xbar symbol so i used u

Answer 57

ok

Answer 58

never mind...

Answer 59

divide by n-1 when your observations come from sampling a population. divide by n when your finding the variance of your population. in this problem, your observations are your population, it doesn't come from a larger population, because this was not mentioned.

Answer 60

they're not looking for the variance, but you're right... n-1 for sampling (it's unbiased) and N for population