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order

  • 2 years ago

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)

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  1. dape
    • 2 years ago
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    Have you started solving the problem?

  2. order
    • 2 years ago
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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..

  3. dape
    • 2 years ago
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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 \[ E(unbroken) = 6-E(broken) \]

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

  5. dape
    • 2 years ago
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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.

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

  7. order
    • 2 years ago
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    @dape

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

  9. dape
    • 2 years ago
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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.

  10. dape
    • 2 years ago
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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.

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

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

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

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

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

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

  17. order
    • 2 years ago
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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?

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

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

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

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

  22. dape
    • 2 years ago
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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.

  23. phi
    • 2 years ago
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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

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

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

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

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

  28. dape
    • 2 years ago
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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?

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