hba
  • hba
Stats help required
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hba
  • hba
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klimenkov
  • klimenkov
If I have a set of two number \(\{a,b\}\), what is the mean and what is the standard deviation of those?
hba
  • hba
a+b/2

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hba
  • hba
Sqrt[(Sum(Xi-(a+b/2)/2)]
klimenkov
  • 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
  • hba
I Couldn't get a word.
hba
  • hba
Step by Step would be easier.
klimenkov
  • klimenkov
Ask a concrete question, please.
klimenkov
  • 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
  • hba
Okay i got the first two equations but the last one i couldn't get.
hba
  • hba
Are we subtracting 100 items out of it?
klimenkov
  • 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
  • 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
  • hba
Got it (y).
hba
  • hba
What next?
klimenkov
  • 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
  • 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
  • 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
  • hba
I know i am just terrible. :/
hba
  • hba
I'll do this one later.
hba
  • hba
Thanks anyways.
klimenkov
  • 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
  • hba
@klimenkov I have another question Can i post that now?
klimenkov
  • klimenkov
Sure.
hba
  • hba
Gimme a sec
hba
  • hba
This one.
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klimenkov
  • klimenkov
Variance and standard deviation are the same?
hba
  • hba
No SD=Sqrt(Variance)
klimenkov
  • klimenkov
Tell me honestly: do you need just answers?
hba
  • hba
No
hba
  • hba
Now please explain
klimenkov
  • 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
  • hba
\[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]
hba
  • hba
\[Variance=\frac{ \sum_{i=1}^{n}(X_i-X(mean) )^2}{n}\]
hba
  • hba
@klimenkov Right?
klimenkov
  • klimenkov
Yes. That is right.
klimenkov
  • 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
  • hba
-18 right?
hba
  • hba
76 for the b part?
hba
  • hba
6 for part c?
klimenkov
  • klimenkov
Yes.
hba
  • hba
For part d its 2-(5n/6)
hba
  • 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
  • klimenkov
Write the answer for d), please.
hba
  • hba
The answer for d is 2-5n/6
hba
  • hba
No No wait.
hba
  • hba
The answer to d is 1.167
hba
  • hba
@klimenkov Right?
klimenkov
  • klimenkov
Yes.
hba
  • hba
Now Variance?
hba
  • hba
Did i draw the equations correctly?
klimenkov
  • klimenkov
Yes.
hba
  • hba
So here we know mean of x and y so should we replace it by 12 and 16 in the equations?
klimenkov
  • klimenkov
You have to find the sum\[\sum_{i=1}^n (2x_i-3y_i+6-(-18))^2\]using variances that you have.
hba
  • hba
Where did 18 coming from now?
klimenkov
  • klimenkov
This is the mean of \(2x_i-3y_i+6\).
hba
  • hba
How do we get that?
klimenkov
  • klimenkov
You count this by yourself. Look higher: -18 right?
hba
  • hba
I am not getting it :/
klimenkov
  • 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
  • hba
How are you calculating the mean here was my question
hba
  • hba
Because we have mean x as 12 and mean y as 16
klimenkov
  • 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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