Stats help required

- hba

Stats help required

- chestercat

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- hba

##### 1 Attachment

- klimenkov

If I have a set of two number \(\{a,b\}\), what is the mean and what is the standard deviation of those?

- hba

a+b/2

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## More answers

- hba

Sqrt[(Sum(Xi-(a+b/2)/2)]

- klimenkov

\[\frac1{250}\cdot\sum_{i=1}^{250}x_i=15.6\]\[\frac1{250}\cdot\left(100\cdot\frac1{100}\sum_{i=1}^{100}x_i+\sum_{i=101}^{250}x_i\right)=15.6\]The first addend in parenthesis is \(100\cdot15=1500\). Name the second addend \(M\) and find it. This will be the mean you need to find.

- hba

I Couldn't get a word.

- hba

Step by Step would be easier.

- klimenkov

Ask a concrete question, please.

- klimenkov

You have a set of numbers \(\{x_1,x_2,\ldots,x_{250}\}\). You know that\[\frac1{250}\cdot\sum_{i=1}^{250}x_i=15.6\] and\[\frac1{100}\cdot\sum_{i=1}^{100}x_i=15\]The question is: what equals the sum?\[\frac1{150}\cdot\sum_{i=101}^{250}x_i\]

- hba

Okay i got the first two equations but the last one i couldn't get.

- hba

Are we subtracting 100 items out of it?

- klimenkov

To find a deviation you need to have a mean of the second group. The last sum is the mean of the second group.

- klimenkov

Actually the set \(\{x_1,\ldots,x_{250}\}\) is divided into 2 groups:
\(\{x_1,\ldots,x_{100}\}\) and \(\{x_{101},\ldots,x_{250}\}\).

- hba

Got it (y).

- hba

What next?

- klimenkov

Now try to do the same thing with deviation.
\[\sum_{i=1}^{100}(x_i-15)^2=\sum_{i=1}^{100}x_i^2-30\sum_{i=1}^{100}x_i-1500=9\]\[\sum_{i=1}^{250}(x_i-15,6)^2=\ldots=13.44^2\]Find the value of \(\sum_{i=101}^{250}x_i^2\) and then try to get the value of \(\sum_{i=101}^{250}(x_i-M)^2\), where \(M\) is the mean of the second group. Did you get it?
Actually, this method is a bit complicated. There may be an easier method.

- hba

I guess i would get it if you would explain me the general forms and then let me do some work and help me step by step.

- klimenkov

Ok.
Lets make an example which will look like your problem.
Lets take a concrete set \(\{1,3,5,9\}\). And now divide it into 2 groups:
\(\{1,3\}\) and \(\{5,9\}\).
Find mean and standard deviation of all of these 3 groups.

- hba

I know i am just terrible. :/

- hba

I'll do this one later.

- hba

Thanks anyways.

- klimenkov

If you really want to solve this, and not just to get an answer, think about it by yourself. It is not so difficult task. It is a pure math problem and not stats. Hope you will.
Anyway you can ask any question anytime.
Best wishes.

- hba

@klimenkov I have another question Can i post that now?

- klimenkov

Sure.

- hba

Gimme a sec

- hba

This one.

##### 1 Attachment

- klimenkov

Variance and standard deviation are the same?

- hba

No
SD=Sqrt(Variance)

- klimenkov

Tell me honestly: do you need just answers?

- hba

No

- hba

Now please explain

- klimenkov

If you really want to get this, you have to get what is mean and variance.
What is mean for \(\{x_1,\ldots,x_n\}\) ?

- hba

\[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]

- hba

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

- hba

@klimenkov Right?

- klimenkov

Yes. That is right.

- klimenkov

Now you have\[\frac1n\sum_{i=1}^n x_i=12, \quad\frac1n\sum_{i=1}^n y_i=16\]Find\[\frac1n\sum_{i=1}^n (2x_i-3y_i+6)\]

- hba

-18 right?

- hba

76 for the b part?

- hba

6 for part c?

- klimenkov

Yes.

- hba

For part d its 2-(5n/6)

- hba

Now for the variance part,
\[\ 2=\frac{ \sum_{i=1}^{n}(X_i-X(mean) ) }{ n }\]
\[\ 1=\frac{ \sum_{i=1}^{n}(Y_i-Y(mean) ) }{ n }\]

- klimenkov

Write the answer for d), please.

- hba

The answer for d is 2-5n/6

- hba

No No wait.

- hba

The answer to d is 1.167

- hba

@klimenkov Right?

- klimenkov

Yes.

- hba

Now Variance?

- hba

Did i draw the equations correctly?

- klimenkov

Yes.

- hba

So here we know mean of x and y so should we replace it by 12 and 16 in the equations?

- klimenkov

You have to find the sum\[\sum_{i=1}^n (2x_i-3y_i+6-(-18))^2\]using variances that you have.

- hba

Where did 18 coming from now?

- klimenkov

This is the mean of \(2x_i-3y_i+6\).

- hba

How do we get that?

- klimenkov

You count this by yourself. Look higher:
-18 right?

- hba

I am not getting it :/

- klimenkov

\[\text{variance}=\frac1n\sum_{i=1}^n(x_i-\text{mean}(x_i))\]Now you put \(2x_i-3y_i+6\) instead of \(x_i\).

- hba

How are you calculating the mean here was my question

- hba

Because we have mean x as 12 and mean y as 16

- klimenkov

\(\text{mean}(2x_i-3y_i+6)=-18\). Look higher you did it by yourself. There was a formula how to get it.

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