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order Group Title

Some of the eggs at a market are sold in boxes of six. The number, X, of broken eggs in a box has the probability distribution given in the following table. x 0 1 2 3 4 5 6 P(X=x) 0.8 0.14 0.03 0.02 0.01 0 0 (Find the expectation and variance of the number of unbroken eggs in a box)

  • 2 years ago
  • 2 years ago

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  1. dape Group Title
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    Have you started solving the problem?

    • 2 years ago
  2. order Group Title
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    I don't know the unbroken eggs. I know how to get the expectation and variance once I have the distribution of unbroken eggs..

    • 2 years ago
  3. dape Group Title
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    Well, the expected value of unbroken eggs is just the difference of the total number of eggs and the expected value of broken eggs or \[ E(unbroken) = 6-E(broken) \]

    • 2 years ago
  4. order Group Title
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    @dape what of the variance?

    • 2 years ago
  5. dape Group Title
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    Calculate the deviations with the respect to the mean of the number of unbroken eggs instead of the mean of broken eggs.

    • 2 years ago
  6. order Group Title
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    How do you do that?

    • 2 years ago
  7. order Group Title
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    @dape

    • 2 years ago
  8. Frank_Yalisanda Group Title
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    it is about discrete type random variable.

    • 2 years ago
  9. dape Group Title
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    You use the expected value, I said mean but meant EV. Calculate the variance from the EV of the number of unbroken eggs.

    • 2 years ago
  10. dape Group Title
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    Variance is calculated by summing the square of the distances of the possibilities from the mean or expectation and dividing by the number of possibilities.

    • 2 years ago
  11. order Group Title
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    Yes, so how do you do that with this probability distribution?

    • 2 years ago
  12. dape Group Title
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    \[ Var[unbroken] = P[X=0]*(0-E[X=0])^2 + P[X=1]*(1-E[X=1]) ...\]

    • 2 years ago
  13. order Group Title
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    Can you place some values in that equation? I'm not sure what E[X=0] stands for?

    • 2 years ago
  14. dape Group Title
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    Sorry, you're right, that doesn't make sense, it should be \[ Var = P[X=0]*(0-E[X])^2+P[X=1]*(1-E[X])^2+...+P[X=6]*(6-E[X])^2 \]

    • 2 years ago
  15. order Group Title
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    But what does E|X| stand for?

    • 2 years ago
  16. dape Group Title
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    Expected value of the random variable.

    • 2 years ago
  17. order Group Title
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    Yea, but what would be the expectd value when X= 0 for instance? Or are they all just the expectation value?

    • 2 years ago
  18. dape Group Title
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    Yes, they are all the expectation. My first formula doesn't really make sense.

    • 2 years ago
  19. order Group Title
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    So just 5.7? All of them?

    • 2 years ago
  20. dape Group Title
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    Yes, you calculate how far they are from the average.

    • 2 years ago
  21. order Group Title
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    Ok. Thankyou! :)

    • 2 years ago
  22. dape Group Title
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    Well, now it got confusing. It should be \[ P[X=0]*(6-E[X])^2 \] and not \[ P[X=0]*(0-E[X])^2 \] since the 0 is the number of broken eggs and we want to calculate the variance of unbroken eggs, so X=0 means 6 unbroken eggs and so on.

    • 2 years ago
  23. phi Group Title
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    There might be another way to do this, but if you notice x 6 5 4 3 2 1 0 P(X=x) 0.8 0.14 0.03 0.02 0.01 0 0 is the probability distribution of whole eggs, you can make progress

    • 2 years ago
  24. phi Group Title
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    E(whole)= 5.7 var(whole)= 0.51

    • 2 years ago
  25. phi Group Title
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    Did I lose you?

    • 2 years ago
  26. phi Group Title
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    0.8 probability of 0 eggs broken means 0.8 probability of 6 whole eggs....

    • 2 years ago
  27. order Group Title
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    It says the variance is 1.7? Yes, I'm lost.

    • 2 years ago
  28. dape Group Title
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    I got calculated the variance to 0.51, which agrees with phi's calculation. Where does it say the variance is 1.7?

    • 2 years ago
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