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hba
 3 years ago
Stats help required
hba
 3 years ago
Stats help required

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klimenkov
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Step by Step would be easier.

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

klimenkov
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0Are we subtracting 100 items out of it?

klimenkov
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0I know i am just terrible. :/

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

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

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

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

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

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

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

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

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

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

klimenkov
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0How are you calculating the mean here was my question

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

klimenkov
 3 years 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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