Question

Stats help required

Answer 1

If I have a set of two number \(\{a,b\}\), what is the mean and what is the standard deviation of those?

Answer 2

a+b/2

Answer 3

Sqrt[(Sum(Xi-(a+b/2)/2)]

Answer 4

\[\frac1{250}\cdot\sum_{i=1}^{250}x_i=15.6\]\[\frac1{250}\cdot\left(100\cdot\frac1{100}\sum_{i=1}^{100}x_i+\sum_{i=101}^{250}x_i\right)=15.6\]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.

Answer 5

I Couldn't get a word.

Answer 6

Step by Step would be easier.

Answer 7

Ask a concrete question, please.

Answer 8

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?\[\frac1{150}\cdot\sum_{i=101}^{250}x_i\]

Answer 9

Okay i got the first two equations but the last one i couldn't get.

Answer 10

Are we subtracting 100 items out of it?

Answer 11

To find a deviation you need to have a mean of the second group. The last sum is the mean of the second group.

Answer 12

Actually the set \(\{x_1,\ldots,x_{250}\}\) is divided into 2 groups:
\(\{x_1,\ldots,x_{100}\}\) and \(\{x_{101},\ldots,x_{250}\}\).

Answer 13

Got it (y).

Answer 14

What next?

Answer 15

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.

Answer 16

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.

Answer 17

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.

Answer 18

I know i am just terrible. :/

Answer 19

I'll do this one later.

Answer 20

Thanks anyways.

Answer 21

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.

Answer 22

@klimenkov I have another question Can i post that now?

Answer 23

Sure.

Answer 24

Gimme a sec

Answer 25

This one.

Answer 26

Variance and standard deviation are the same?

Answer 27

No

SD=Sqrt(Variance)

Answer 28

Tell me honestly: do you need just answers?

Answer 29

No

Answer 30

Now please explain

Answer 31

If you really want to get this, you have to get what is mean and variance.
What is mean for \(\{x_1,\ldots,x_n\}\) ?

Answer 32

\[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]

Answer 33

\[Variance=\frac{ \sum_{i=1}^{n}(X_i-X(mean) )^2}{n}\]

Answer 34

@klimenkov Right?

Answer 35

Yes. That is right.

Answer 36

Now you have\[\frac1n\sum_{i=1}^n x_i=12, \quad\frac1n\sum_{i=1}^n y_i=16\]Find\[\frac1n\sum_{i=1}^n (2x_i-3y_i+6)\]

Answer 37

-18 right?

Answer 38

76 for the b part?

Answer 39

6 for part c?

Answer 40

Yes.

Answer 41

For part d its 2-(5n/6)

Answer 42

Now for the variance part,

\[\ 2=\frac{ \sum_{i=1}^{n}(X_i-X(mean) ) }{ n }\]

\[\ 1=\frac{ \sum_{i=1}^{n}(Y_i-Y(mean) ) }{ n }\]

Answer 43

Write the answer for d), please.

Answer 44

The answer for d is 2-5n/6

Answer 45

No No wait.

Answer 46

The answer to d is 1.167

Answer 47

@klimenkov Right?

Answer 48

Yes.

Answer 49

Now Variance?

Answer 50

Did i draw the equations correctly?

Answer 51

Yes.

Answer 52

So here we know mean of x and y so should we replace it by 12 and 16 in the equations?

Answer 53

You have to find the sum\[\sum_{i=1}^n (2x_i-3y_i+6-(-18))^2\]using variances that you have.

Answer 54

Where did 18 coming from now?

Answer 55

This is the mean of \(2x_i-3y_i+6\).

Answer 56

How do we get that?

Answer 57

You count this by yourself. Look higher:
-18 right?

Answer 58

I am not getting it :/

Answer 59

\[\text{variance}=\frac1n\sum_{i=1}^n(x_i-\text{mean}(x_i))\]Now you put \(2x_i-3y_i+6\) instead of \(x_i\).

Answer 60

How are you calculating the mean here was my question

Answer 61

Because we have mean x as 12 and mean y as 16

Answer 62

\(\text{mean}(2x_i-3y_i+6)=-18\). Look higher you did it by yourself. There was a formula how to get it.