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hba Group Title

Stats help required

  • one year ago
  • one year ago

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  1. hba Group Title
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    • one year ago
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  2. klimenkov Group Title
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    If I have a set of two number \(\{a,b\}\), what is the mean and what is the standard deviation of those?

    • one year ago
  3. hba Group Title
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    a+b/2

    • one year ago
  4. hba Group Title
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    Sqrt[(Sum(Xi-(a+b/2)/2)]

    • one year ago
  5. klimenkov Group Title
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    \[\frac1{250}\cdot\sum_{i=1}^{250}x_i=15.6\]\[\frac1{250}\cdot\left(100\cdot\frac1{100}\sum_{i=1}^{100}x_i+\sum_{i=101}^{250}x_i\right)=15.6\]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.

    • one year ago
  6. hba Group Title
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    I Couldn't get a word.

    • one year ago
  7. hba Group Title
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    Step by Step would be easier.

    • one year ago
  8. klimenkov Group Title
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    Ask a concrete question, please.

    • one year ago
  9. klimenkov Group Title
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    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?\[\frac1{150}\cdot\sum_{i=101}^{250}x_i\]

    • one year ago
  10. hba Group Title
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    Okay i got the first two equations but the last one i couldn't get.

    • one year ago
  11. hba Group Title
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    Are we subtracting 100 items out of it?

    • one year ago
  12. klimenkov Group Title
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    To find a deviation you need to have a mean of the second group. The last sum is the mean of the second group.

    • one year ago
  13. klimenkov Group Title
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    Actually the set \(\{x_1,\ldots,x_{250}\}\) is divided into 2 groups: \(\{x_1,\ldots,x_{100}\}\) and \(\{x_{101},\ldots,x_{250}\}\).

    • one year ago
  14. hba Group Title
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    Got it (y).

    • one year ago
  15. hba Group Title
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    What next?

    • one year ago
  16. klimenkov Group Title
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    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.

    • one year ago
  17. hba Group Title
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    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.

    • one year ago
  18. klimenkov Group Title
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    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.

    • one year ago
  19. hba Group Title
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    I know i am just terrible. :/

    • one year ago
  20. hba Group Title
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    I'll do this one later.

    • one year ago
  21. hba Group Title
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    Thanks anyways.

    • one year ago
  22. klimenkov Group Title
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    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.

    • one year ago
  23. hba Group Title
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    @klimenkov I have another question Can i post that now?

    • one year ago
  24. klimenkov Group Title
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    Sure.

    • one year ago
  25. hba Group Title
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    Gimme a sec

    • one year ago
  26. hba Group Title
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    This one.

    • one year ago
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  27. klimenkov Group Title
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    Variance and standard deviation are the same?

    • one year ago
  28. hba Group Title
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    No SD=Sqrt(Variance)

    • one year ago
  29. klimenkov Group Title
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    Tell me honestly: do you need just answers?

    • one year ago
  30. hba Group Title
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    No

    • one year ago
  31. hba Group Title
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    Now please explain

    • one year ago
  32. klimenkov Group Title
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    If you really want to get this, you have to get what is mean and variance. What is mean for \(\{x_1,\ldots,x_n\}\) ?

    • one year ago
  33. hba Group Title
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    \[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]

    • one year ago
  34. hba Group Title
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    \[Variance=\frac{ \sum_{i=1}^{n}(X_i-X(mean) )^2}{n}\]

    • one year ago
  35. hba Group Title
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    @klimenkov Right?

    • one year ago
  36. klimenkov Group Title
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    Yes. That is right.

    • one year ago
  37. klimenkov Group Title
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    Now you have\[\frac1n\sum_{i=1}^n x_i=12, \quad\frac1n\sum_{i=1}^n y_i=16\]Find\[\frac1n\sum_{i=1}^n (2x_i-3y_i+6)\]

    • one year ago
  38. hba Group Title
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    -18 right?

    • one year ago
  39. hba Group Title
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    76 for the b part?

    • one year ago
  40. hba Group Title
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    6 for part c?

    • one year ago
  41. klimenkov Group Title
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    Yes.

    • one year ago
  42. hba Group Title
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    For part d its 2-(5n/6)

    • one year ago
  43. hba Group Title
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    Now for the variance part, \[\ 2=\frac{ \sum_{i=1}^{n}(X_i-X(mean) ) }{ n }\] \[\ 1=\frac{ \sum_{i=1}^{n}(Y_i-Y(mean) ) }{ n }\]

    • one year ago
  44. klimenkov Group Title
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    Write the answer for d), please.

    • one year ago
  45. hba Group Title
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    The answer for d is 2-5n/6

    • one year ago
  46. hba Group Title
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    No No wait.

    • one year ago
  47. hba Group Title
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    The answer to d is 1.167

    • one year ago
  48. hba Group Title
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    @klimenkov Right?

    • one year ago
  49. klimenkov Group Title
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    Yes.

    • one year ago
  50. hba Group Title
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    Now Variance?

    • one year ago
  51. hba Group Title
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    Did i draw the equations correctly?

    • one year ago
  52. klimenkov Group Title
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    Yes.

    • one year ago
  53. hba Group Title
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    So here we know mean of x and y so should we replace it by 12 and 16 in the equations?

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

    • one year ago
  55. hba Group Title
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    Where did 18 coming from now?

    • one year ago
  56. klimenkov Group Title
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    This is the mean of \(2x_i-3y_i+6\).

    • one year ago
  57. hba Group Title
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    How do we get that?

    • one year ago
  58. klimenkov Group Title
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    You count this by yourself. Look higher: -18 right?

    • one year ago
  59. hba Group Title
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    I am not getting it :/

    • one year ago
  60. klimenkov Group Title
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    \[\text{variance}=\frac1n\sum_{i=1}^n(x_i-\text{mean}(x_i))\]Now you put \(2x_i-3y_i+6\) instead of \(x_i\).

    • one year ago
  61. hba Group Title
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    How are you calculating the mean here was my question

    • one year ago
  62. hba Group Title
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    Because we have mean x as 12 and mean y as 16

    • one year ago
  63. klimenkov Group Title
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    \(\text{mean}(2x_i-3y_i+6)=-18\). Look higher you did it by yourself. There was a formula how to get it.

    • one year ago
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