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

klimenkov
 one year ago
Best 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.

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

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

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2You 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay i got the first two equations but the last one i couldn't get.

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

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

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

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2Now 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.

hba
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.2Ok. 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
 one year ago
Best ResponseYou've already chosen the best response.0I know i am just terrible. :/

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.0@klimenkov I have another question Can i post that now?

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

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

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.0\[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]

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

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

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

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

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

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

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

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

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

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

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

klimenkov
 one year ago
Best 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 \(x_i\).

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

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

klimenkov
 one year ago
Best 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.
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