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hba

  • one year ago

Stats help required

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  1. hba
    • one year ago
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  2. hba
    • one year ago
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    Actually i know. \[Mean=\sum_{}^{}fx/\sum_{}^{}f\] I know, \[\sum_{}^{}f=2000\]

  3. sami-21
    • one year ago
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    do you have the answer key is the answer 7.975 ???

  4. hba
    • one year ago
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    I also think that is the answer as i tried doing it.

  5. hba
    • one year ago
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    But here x1,x2,x3 cannot be 8.5,7.5 and 8 because it is actually the mean

  6. hba
    • one year ago
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    As mentioned in the ques

  7. wio
    • one year ago
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    Okay give me a second.

  8. hba
    • one year ago
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    Sure

  9. wio
    • one year ago
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    Now suppose that we call \[ \sum fx \]The 'total'

  10. wio
    • one year ago
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    We want the 'total' of each sub population.

  11. wio
    • one year ago
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    Then we can add them

  12. wio
    • one year ago
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    To get the 'total' of the whole population.

  13. wio
    • one year ago
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    From there we can find the mean of the whole population.

  14. ParthKohli
    • one year ago
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    We know that the sum is \(700 \times 8.5 + 800 \times 7.5 + 500 \times 8\)

  15. hba
    • one year ago
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    No No No 8.5,7.5 and 8 cannot be x

  16. hba
    • one year ago
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    They are the means

  17. wio
    • one year ago
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    Yeah, but he's not using that formula.

  18. ParthKohli
    • one year ago
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    Since mean is sum of observations divided by number of observations, we know that the sum of observations multiplied by the number of observations is the sum of observations.

  19. wio
    • one year ago
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    He using this: \[ \sum fx = \frac{\sum fx}{\sum f}\times \sum f \]

  20. ParthKohli
    • one year ago
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    Can you continue from this point?

  21. sami-21
    • one year ago
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    i don't think there is any problem with taking them as x's . what you alreadyy did is correct 7.975 .

  22. ParthKohli
    • one year ago
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    lunch... g2g

  23. hba
    • one year ago
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    But how can you say mean of x is actually x?

  24. sami-21
    • one year ago
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    (8.5*700 + 800*7.5 + 500*8)/2000

  25. sami-21
    • one year ago
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    these are different random variables for the whole population . you can just use them in the formula .

  26. hba
    • one year ago
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    @wio Please justify

  27. hba
    • one year ago
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    It says that it is the mean @sami-21

  28. wio
    • one year ago
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    Okay so if you have three means... suppose they are \(m_1, m_2, m_3\)

  29. wio
    • one year ago
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    \[ m_1 = \frac{\sum_1fx}{\sum_1f} \]

  30. sami-21
    • one year ago
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    yes it does says . and requires the mean for population . which should be taking the means of the sub populatiions .

  31. hba
    • one year ago
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    One more thing,If that is not the formula,What is it?

  32. hba
    • one year ago
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    @xoya Shu away

  33. wio
    • one year ago
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    the mean of all three will be: \[ m_4 = \frac{\sum_4 fx}{\sum_4x} = \frac{\sum_1 fx + \sum_2 fx+ \sum_3 fx}{\sum_1 f + \sum_2 f + \sum_3 f} \]

  34. wio
    • one year ago
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    This is because the total frequency \(\sum_4 f\) is the sum of the sub frequencies. The total weight \(\sum_4 fx\) is equal to the sum of all the weights.

  35. wio
    • one year ago
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    Notice how \[ m_1 = \frac{\sum_1 fx}{\sum_1 f} \implies m_1\sum_1 f = \sum_1 f x \]

  36. hba
    • one year ago
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    Okay so what would the formula basically?

  37. hba
    • one year ago
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    be*

  38. wio
    • one year ago
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    Given means... \(m_1, m_2, \dots \) with total frequencies \(f_1,f_2,...\) then the mean of the totals is: \[ \frac{\sum m_if_i}{\sum f_i} \]

  39. hba
    • one year ago
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    Thanks a lot :D :D

  40. wio
    • one year ago
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    It's not the same formula, they just happen to be the same though by coincidence.

  41. hba
    • one year ago
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    I see.

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