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

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

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

  5. hba
    • 2 years 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
    • 2 years ago
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    As mentioned in the ques

  7. wio
    • 2 years ago
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    Okay give me a second.

  8. hba
    • 2 years ago
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    Sure

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

  10. wio
    • 2 years ago
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    We want the 'total' of each sub population.

  11. wio
    • 2 years ago
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    Then we can add them

  12. wio
    • 2 years ago
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    To get the 'total' of the whole population.

  13. wio
    • 2 years ago
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    From there we can find the mean of the whole population.

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

  15. hba
    • 2 years ago
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    No No No 8.5,7.5 and 8 cannot be x

  16. hba
    • 2 years ago
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    They are the means

  17. wio
    • 2 years ago
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    Yeah, but he's not using that formula.

  18. ParthKohli
    • 2 years 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
    • 2 years ago
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    He using this: \[ \sum fx = \frac{\sum fx}{\sum f}\times \sum f \]

  20. ParthKohli
    • 2 years ago
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    Can you continue from this point?

  21. sami-21
    • 2 years 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
    • 2 years ago
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    lunch... g2g

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

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

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

  26. hba
    • 2 years ago
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    @wio Please justify

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

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

  29. wio
    • 2 years ago
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    \[ m_1 = \frac{\sum_1fx}{\sum_1f} \]

  30. sami-21
    • 2 years 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
    • 2 years ago
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    One more thing,If that is not the formula,What is it?

  32. hba
    • 2 years ago
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    @xoya Shu away

  33. wio
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    Okay so what would the formula basically?

  37. hba
    • 2 years ago
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    be*

  38. wio
    • 2 years 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
    • 2 years ago
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    Thanks a lot :D :D

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

  41. hba
    • 2 years ago
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    I see.

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