order 3 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)

1. dape

Have you started solving the problem?

2. order

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

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

@dape what of the variance?

5. dape

Calculate the deviations with the respect to the mean of the number of unbroken eggs instead of the mean of broken eggs.

6. order

How do you do that?

7. order

@dape

8. Frank_Yalisanda

it is about discrete type random variable.

9. dape

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

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

Yes, so how do you do that with this probability distribution?

12. dape

\[ Var[unbroken] = P[X=0]*(0-E[X=0])^2 + P[X=1]*(1-E[X=1]) ...\]

13. order

Can you place some values in that equation? I'm not sure what E[X=0] stands for?

14. dape

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

But what does E|X| stand for?

16. dape

Expected value of the random variable.

17. order

Yea, but what would be the expectd value when X= 0 for instance? Or are they all just the expectation value?

18. dape

Yes, they are all the expectation. My first formula doesn't really make sense.

19. order

So just 5.7? All of them?

20. dape

Yes, you calculate how far they are from the average.

21. order

Ok. Thankyou! :)

22. dape

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

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

E(whole)= 5.7 var(whole)= 0.51

25. phi

Did I lose you?

26. phi

0.8 probability of 0 eggs broken means 0.8 probability of 6 whole eggs....

27. order

It says the variance is 1.7? Yes, I'm lost.

28. dape

I got calculated the variance to 0.51, which agrees with phi's calculation. Where does it say the variance is 1.7?