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  • hba

Stats help required

Mathematics
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  • hba
1 Attachment
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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Other answers:

  • hba
Sqrt[(Sum(Xi-(a+b/2)/2)]
\[\frac1{250}\cdot\sum_{i=1}^{250}x_i=15.6\]\[\frac1{250}\cdot\left(100\cdot\frac1{100}\sum_{i=1}^{100}x_i+\sum_{i=101}^{250}x_i\right)=15.6\]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.
Ask a concrete question, please.
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?
To find a deviation you need to have a mean of the second group. The last sum is the mean of the second group.
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?
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.
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.
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?
Sure.
  • hba
Gimme a sec
  • hba
This one.
1 Attachment
Variance and standard deviation are the same?
  • hba
No SD=Sqrt(Variance)
Tell me honestly: do you need just answers?
  • hba
No
  • hba
Now please explain
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?
Yes. That is right.
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?
Yes.
  • hba
For part d its 2-(5n/6)
  • hba
Now for the variance part, \[\ 2=\frac{ \sum_{i=1}^{n}(X_i-X(mean) ) }{ n }\] \[\ 1=\frac{ \sum_{i=1}^{n}(Y_i-Y(mean) ) }{ n }\]
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?
Yes.
  • hba
Now Variance?
  • hba
Did i draw the equations correctly?
Yes.
  • hba
So here we know mean of x and y so should we replace it by 12 and 16 in the equations?
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?
This is the mean of \(2x_i-3y_i+6\).
  • hba
How do we get that?
You count this by yourself. Look higher: -18 right?
  • hba
I am not getting it :/
\[\text{variance}=\frac1n\sum_{i=1}^n(x_i-\text{mean}(x_i))\]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
\(\text{mean}(2x_i-3y_i+6)=-18\). Look higher you did it by yourself. There was a formula how to get it.

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