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klimenkov Group TitleBest ResponseYou've already chosen the best response.2
If I have a set of two number \(\{a,b\}\), what is the mean and what is the standard deviation of those?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Sqrt[(Sum(Xi(a+b/2)/2)]
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
I Couldn't get a word.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Step by Step would be easier.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Ask a concrete question, please.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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?\[\frac1{150}\cdot\sum_{i=101}^{250}x_i\]
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Okay i got the first two equations but the last one i couldn't get.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Are we subtracting 100 items out of it?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
To find a deviation you need to have a mean of the second group. The last sum is the mean of the second group.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Actually the set \(\{x_1,\ldots,x_{250}\}\) is divided into 2 groups: \(\{x_1,\ldots,x_{100}\}\) and \(\{x_{101},\ldots,x_{250}\}\).
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Now try to do the same thing with deviation. \[\sum_{i=1}^{100}(x_i15)^2=\sum_{i=1}^{100}x_i^230\sum_{i=1}^{100}x_i1500=9\]\[\sum_{i=1}^{250}(x_i15,6)^2=\ldots=13.44^2\]Find the value of \(\sum_{i=101}^{250}x_i^2\) and then try to get the value of \(\sum_{i=101}^{250}(x_iM)^2\), 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.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
I know i am just terrible. :/
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
I'll do this one later.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Thanks anyways.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
@klimenkov I have another question Can i post that now?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Variance and standard deviation are the same?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
No SD=Sqrt(Variance)
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Tell me honestly: do you need just answers?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Now please explain
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
If you really want to get this, you have to get what is mean and variance. What is mean for \(\{x_1,\ldots,x_n\}\) ?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
\[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
\[Variance=\frac{ \sum_{i=1}^{n}(X_iX(mean) )^2}{n}\]
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
@klimenkov Right?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Yes. That is right.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Now you have\[\frac1n\sum_{i=1}^n x_i=12, \quad\frac1n\sum_{i=1}^n y_i=16\]Find\[\frac1n\sum_{i=1}^n (2x_i3y_i+6)\]
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
76 for the b part?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
For part d its 2(5n/6)
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Now for the variance part, \[\ 2=\frac{ \sum_{i=1}^{n}(X_iX(mean) ) }{ n }\] \[\ 1=\frac{ \sum_{i=1}^{n}(Y_iY(mean) ) }{ n }\]
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Write the answer for d), please.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
The answer for d is 25n/6
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
The answer to d is 1.167
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
@klimenkov Right?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Did i draw the equations correctly?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
So here we know mean of x and y so should we replace it by 12 and 16 in the equations?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
You have to find the sum\[\sum_{i=1}^n (2x_i3y_i+6(18))^2\]using variances that you have.
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Where did 18 coming from now?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
This is the mean of \(2x_i3y_i+6\).
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
How do we get that?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
You count this by yourself. Look higher: 18 right?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
I am not getting it :/
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
\[\text{variance}=\frac1n\sum_{i=1}^n(x_i\text{mean}(x_i))\]Now you put \(2x_i3y_i+6\) instead of \(x_i\).
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
How are you calculating the mean here was my question
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Because we have mean x as 12 and mean y as 16
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
\(\text{mean}(2x_i3y_i+6)=18\). Look higher you did it by yourself. There was a formula how to get it.
 one year ago
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