## hba Group Title Stats help required one year ago one year ago

1. hba Group Title

2. klimenkov Group Title

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

3. hba Group Title

a+b/2

4. hba Group Title

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

5. klimenkov Group Title

$\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 Group Title

I Couldn't get a word.

7. hba Group Title

Step by Step would be easier.

8. klimenkov Group Title

9. klimenkov Group Title

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 Group Title

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

11. hba Group Title

Are we subtracting 100 items out of it?

12. klimenkov Group Title

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 Group Title

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 Group Title

Got it (y).

15. hba Group Title

What next?

16. klimenkov Group Title

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 Group Title

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 Group Title

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 Group Title

I know i am just terrible. :/

20. hba Group Title

I'll do this one later.

21. hba Group Title

Thanks anyways.

22. klimenkov Group Title

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 Group Title

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

24. klimenkov Group Title

Sure.

25. hba Group Title

Gimme a sec

26. hba Group Title

This one.

27. klimenkov Group Title

Variance and standard deviation are the same?

28. hba Group Title

No SD=Sqrt(Variance)

29. klimenkov Group Title

Tell me honestly: do you need just answers?

30. hba Group Title

No

31. hba Group Title

32. klimenkov Group Title

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 Group Title

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

34. hba Group Title

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

35. hba Group Title

@klimenkov Right?

36. klimenkov Group Title

Yes. That is right.

37. klimenkov Group Title

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 Group Title

-18 right?

39. hba Group Title

76 for the b part?

40. hba Group Title

6 for part c?

41. klimenkov Group Title

Yes.

42. hba Group Title

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

43. hba Group Title

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 Group Title

45. hba Group Title

The answer for d is 2-5n/6

46. hba Group Title

No No wait.

47. hba Group Title

The answer to d is 1.167

48. hba Group Title

@klimenkov Right?

49. klimenkov Group Title

Yes.

50. hba Group Title

Now Variance?

51. hba Group Title

Did i draw the equations correctly?

52. klimenkov Group Title

Yes.

53. hba Group Title

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

54. klimenkov Group Title

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

55. hba Group Title

Where did 18 coming from now?

56. klimenkov Group Title

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

57. hba Group Title

How do we get that?

58. klimenkov Group Title

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

59. hba Group Title

I am not getting it :/

60. klimenkov Group Title

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

61. hba Group Title

How are you calculating the mean here was my question

62. hba Group Title

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

63. klimenkov Group Title

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