## hba one year ago Stats help required

1. hba

2. klimenkov

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

3. hba

a+b/2

4. hba

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

5. 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.

6. hba

I Couldn't get a word.

7. hba

Step by Step would be easier.

8. klimenkov

9. 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?$\frac1{150}\cdot\sum_{i=101}^{250}x_i$

10. hba

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

11. hba

Are we subtracting 100 items out of it?

12. 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.

13. klimenkov

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

14. hba

Got it (y).

15. hba

What next?

16. 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.

17. 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.

18. 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.

19. hba

I know i am just terrible. :/

20. hba

I'll do this one later.

21. hba

Thanks anyways.

22. 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.

23. hba

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

24. klimenkov

Sure.

25. hba

Gimme a sec

26. hba

This one.

27. klimenkov

Variance and standard deviation are the same?

28. hba

No SD=Sqrt(Variance)

29. klimenkov

Tell me honestly: do you need just answers?

30. hba

No

31. hba

32. klimenkov

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

33. hba

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

34. hba

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

35. hba

@klimenkov Right?

36. klimenkov

Yes. That is right.

37. klimenkov

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

38. hba

-18 right?

39. hba

76 for the b part?

40. hba

6 for part c?

41. klimenkov

Yes.

42. hba

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

43. hba

Now for the variance part, $\ 2=\frac{ \sum_{i=1}^{n}(X_i-X(mean) ) }{ n }$ $\ 1=\frac{ \sum_{i=1}^{n}(Y_i-Y(mean) ) }{ n }$

44. klimenkov

45. hba

The answer for d is 2-5n/6

46. hba

No No wait.

47. hba

The answer to d is 1.167

48. hba

@klimenkov Right?

49. klimenkov

Yes.

50. hba

Now Variance?

51. hba

Did i draw the equations correctly?

52. klimenkov

Yes.

53. hba

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

54. klimenkov

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

55. hba

Where did 18 coming from now?

56. klimenkov

This is the mean of $$2x_i-3y_i+6$$.

57. hba

How do we get that?

58. klimenkov

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

59. hba

I am not getting it :/

60. klimenkov

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

61. hba

How are you calculating the mean here was my question

62. hba

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

63. klimenkov

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