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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)
Have you started solving the problem?
I don't know the unbroken eggs. I know how to get the expectation and variance once I have the distribution of unbroken eggs..
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 \[ E(unbroken) = 6-E(broken) \]
@dape what of the variance?
Calculate the deviations with the respect to the mean of the number of unbroken eggs instead of the mean of broken eggs.
it is about discrete type random variable.
You use the expected value, I said mean but meant EV. Calculate the variance from the EV of the number of unbroken eggs.
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.
Yes, so how do you do that with this probability distribution?
\[ Var[unbroken] = P[X=0]*(0-E[X=0])^2 + P[X=1]*(1-E[X=1]) ...\]
Can you place some values in that equation? I'm not sure what E[X=0] stands for?
Sorry, you're right, that doesn't make sense, it should be \[ Var = P[X=0]*(0-E[X])^2+P[X=1]*(1-E[X])^2+...+P[X=6]*(6-E[X])^2 \]
But what does E|X| stand for?
Expected value of the random variable.
Yea, but what would be the expectd value when X= 0 for instance? Or are they all just the expectation value?
Yes, they are all the expectation. My first formula doesn't really make sense.
So just 5.7? All of them?
Yes, you calculate how far they are from the average.
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.
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
E(whole)= 5.7 var(whole)= 0.51
0.8 probability of 0 eggs broken means 0.8 probability of 6 whole eggs....
It says the variance is 1.7? Yes, I'm lost.
I got calculated the variance to 0.51, which agrees with phi's calculation. Where does it say the variance is 1.7?