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hba

  • 2 years ago

Stats help required

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  1. hba
    • 2 years ago
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  2. klimenkov
    • 2 years ago
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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?

  3. hba
    • 2 years ago
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    a+b/2

  4. hba
    • 2 years ago
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    Sqrt[(Sum(Xi-(a+b/2)/2)]

  5. klimenkov
    • 2 years ago
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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.

  6. hba
    • 2 years ago
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    I Couldn't get a word.

  7. hba
    • 2 years ago
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    Step by Step would be easier.

  8. klimenkov
    • 2 years ago
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    Ask a concrete question, please.

  9. klimenkov
    • 2 years ago
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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?\[\frac1{150}\cdot\sum_{i=101}^{250}x_i\]

  10. hba
    • 2 years ago
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    Okay i got the first two equations but the last one i couldn't get.

  11. hba
    • 2 years ago
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    Are we subtracting 100 items out of it?

  12. klimenkov
    • 2 years ago
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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.

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

  14. hba
    • 2 years ago
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    Got it (y).

  15. hba
    • 2 years ago
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    What next?

  16. klimenkov
    • 2 years ago
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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.

  17. hba
    • 2 years ago
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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.

  18. klimenkov
    • 2 years ago
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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.

  19. hba
    • 2 years ago
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    I know i am just terrible. :/

  20. hba
    • 2 years ago
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    I'll do this one later.

  21. hba
    • 2 years ago
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    Thanks anyways.

  22. klimenkov
    • 2 years ago
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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.

  23. hba
    • 2 years ago
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    @klimenkov I have another question Can i post that now?

  24. klimenkov
    • 2 years ago
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    Sure.

  25. hba
    • 2 years ago
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    Gimme a sec

  26. hba
    • 2 years ago
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    This one.

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  27. klimenkov
    • 2 years ago
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    Variance and standard deviation are the same?

  28. hba
    • 2 years ago
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    No SD=Sqrt(Variance)

  29. klimenkov
    • 2 years ago
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    Tell me honestly: do you need just answers?

  30. hba
    • 2 years ago
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    No

  31. hba
    • 2 years ago
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    Now please explain

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

  33. hba
    • 2 years ago
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    \[Mean=\frac{ \sum_{i=1}^{n} X_i}{ n }\]

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

  35. hba
    • 2 years ago
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    @klimenkov Right?

  36. klimenkov
    • 2 years ago
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    Yes. That is right.

  37. klimenkov
    • 2 years ago
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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\[\frac1n\sum_{i=1}^n (2x_i-3y_i+6)\]

  38. hba
    • 2 years ago
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    -18 right?

  39. hba
    • 2 years ago
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    76 for the b part?

  40. hba
    • 2 years ago
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    6 for part c?

  41. klimenkov
    • 2 years ago
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    Yes.

  42. hba
    • 2 years ago
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    For part d its 2-(5n/6)

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

  44. klimenkov
    • 2 years ago
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    Write the answer for d), please.

  45. hba
    • 2 years ago
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    The answer for d is 2-5n/6

  46. hba
    • 2 years ago
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    No No wait.

  47. hba
    • 2 years ago
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    The answer to d is 1.167

  48. hba
    • 2 years ago
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    @klimenkov Right?

  49. klimenkov
    • 2 years ago
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    Yes.

  50. hba
    • 2 years ago
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    Now Variance?

  51. hba
    • 2 years ago
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    Did i draw the equations correctly?

  52. klimenkov
    • 2 years ago
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    Yes.

  53. hba
    • 2 years ago
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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?

  54. klimenkov
    • 2 years ago
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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.

  55. hba
    • 2 years ago
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    Where did 18 coming from now?

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

  57. hba
    • 2 years ago
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    How do we get that?

  58. klimenkov
    • 2 years ago
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    You count this by yourself. Look higher: -18 right?

  59. hba
    • 2 years ago
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    I am not getting it :/

  60. klimenkov
    • 2 years ago
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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 \(x_i\).

  61. hba
    • 2 years ago
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    How are you calculating the mean here was my question

  62. hba
    • 2 years ago
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    Because we have mean x as 12 and mean y as 16

  63. klimenkov
    • 2 years ago
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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.

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