Delete
Share
This Question is Closed
hba
Best Response
You've already chosen the best response.
0
klimenkov
Best Response
You'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?
hba
Best Response
You've already chosen the best response.
0
a+b/2
hba
Best Response
You've already chosen the best response.
0
Sqrt[(Sum(Xi-(a+b/2)/2)]
klimenkov
Best Response
You'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
Best Response
You've already chosen the best response.
0
I Couldn't get a word.
hba
Best Response
You've already chosen the best response.
0
Step by Step would be easier.
klimenkov
Best Response
You've already chosen the best response.
2
Ask a concrete question, please.
klimenkov
Best Response
You'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?\[\frac1{150}\cdot\sum_{i=101}^{250}x_i\]
hba
Best Response
You've already chosen the best response.
0
Okay i got the first two equations but the last one i couldn't get.
hba
Best Response
You've already chosen the best response.
0
Are we subtracting 100 items out of it?
klimenkov
Best Response
You'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.
klimenkov
Best Response
You'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 \(\{x_{101},\ldots,x_{250}\}\).
hba
Best Response
You've already chosen the best response.
0
Got it (y).
hba
Best Response
You've already chosen the best response.
0
What next?
klimenkov
Best Response
You've already chosen the best response.
2
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
Best Response
You'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.
klimenkov
Best Response
You'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.
hba
Best Response
You've already chosen the best response.
0
I know i am just terrible. :/
hba
Best Response
You've already chosen the best response.
0
I'll do this one later.
hba
Best Response
You've already chosen the best response.
0
Thanks anyways.
klimenkov
Best Response
You'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.
hba
Best Response
You've already chosen the best response.
0
@klimenkov I have another question Can i post that now?
klimenkov
Best Response
You've already chosen the best response.
2
Sure.
hba
Best Response
You've already chosen the best response.
0
Gimme a sec
hba
Best Response
You've already chosen the best response.
0
This one.
klimenkov
Best Response
You've already chosen the best response.
2
Variance and standard deviation are the same?
hba
Best Response
You've already chosen the best response.
0
No
SD=Sqrt(Variance)
klimenkov
Best Response
You've already chosen the best response.
2
Tell me honestly: do you need just answers?
hba
Best Response
You've already chosen the best response.
0
No
hba
Best Response
You've already chosen the best response.
0
Now please explain
klimenkov
Best Response
You'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 \(\{x_1,\ldots,x_n\}\) ?
hba
Best Response
You've already chosen the best response.
0
\[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]
hba
Best Response
You've already chosen the best response.
0
\[Variance=\frac{ \sum_{i=1}^{n}(X_i-X(mean) )^2}{n}\]
hba
Best Response
You've already chosen the best response.
0
@klimenkov Right?
klimenkov
Best Response
You've already chosen the best response.
2
Yes. That is right.
klimenkov
Best Response
You'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\[\frac1n\sum_{i=1}^n (2x_i-3y_i+6)\]
hba
Best Response
You've already chosen the best response.
0
-18 right?
hba
Best Response
You've already chosen the best response.
0
76 for the b part?
hba
Best Response
You've already chosen the best response.
0
6 for part c?
klimenkov
Best Response
You've already chosen the best response.
2
Yes.
hba
Best Response
You've already chosen the best response.
0
For part d its 2-(5n/6)
hba
Best Response
You've already chosen the best response.
0
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
Best Response
You've already chosen the best response.
2
Write the answer for d), please.
hba
Best Response
You've already chosen the best response.
0
The answer for d is 2-5n/6
hba
Best Response
You've already chosen the best response.
0
No No wait.
hba
Best Response
You've already chosen the best response.
0
The answer to d is 1.167
hba
Best Response
You've already chosen the best response.
0
@klimenkov Right?
klimenkov
Best Response
You've already chosen the best response.
2
Yes.
hba
Best Response
You've already chosen the best response.
0
Now Variance?
hba
Best Response
You've already chosen the best response.
0
Did i draw the equations correctly?
klimenkov
Best Response
You've already chosen the best response.
2
Yes.
hba
Best Response
You'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?
klimenkov
Best Response
You've already chosen the best response.
2
You have to find the sum\[\sum_{i=1}^n (2x_i-3y_i+6-(-18))^2\]using variances that you have.
hba
Best Response
You've already chosen the best response.
0
Where did 18 coming from now?
klimenkov
Best Response
You've already chosen the best response.
2
This is the mean of \(2x_i-3y_i+6\).
hba
Best Response
You've already chosen the best response.
0
How do we get that?
klimenkov
Best Response
You've already chosen the best response.
2
You count this by yourself. Look higher:
-18 right?
hba
Best Response
You've already chosen the best response.
0
I am not getting it :/
klimenkov
Best Response
You've already chosen the best response.
2
\[\text{variance}=\frac1n\sum_{i=1}^n(x_i-\text{mean}(x_i))\]Now you put \(2x_i-3y_i+6\) instead of \(x_i\).
hba
Best Response
You've already chosen the best response.
0
How are you calculating the mean here was my question
hba
Best Response
You've already chosen the best response.
0
Because we have mean x as 12 and mean y as 16
klimenkov
Best Response
You've already chosen the best response.
2
\(\text{mean}(2x_i-3y_i+6)=-18\). Look higher you did it by yourself. There was a formula how to get it.