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anonymous
 4 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)
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Have you started solving the problem?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know the unbroken eggs. I know how to get the expectation and variance once I have the distribution of unbroken eggs..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, the expected value of unbroken eggs is just the difference of the total number of eggs and the expected value of broken eggs or \[ E(unbroken) = 6E(broken) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@dape what of the variance?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Calculate the deviations with the respect to the mean of the number of unbroken eggs instead of the mean of broken eggs.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is about discrete type random variable.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You use the expected value, I said mean but meant EV. Calculate the variance from the EV of the number of unbroken eggs.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Variance is calculated by summing the square of the distances of the possibilities from the mean or expectation and dividing by the number of possibilities.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, so how do you do that with this probability distribution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ Var[unbroken] = P[X=0]*(0E[X=0])^2 + P[X=1]*(1E[X=1]) ...\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you place some values in that equation? I'm not sure what E[X=0] stands for?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, you're right, that doesn't make sense, it should be \[ Var = P[X=0]*(0E[X])^2+P[X=1]*(1E[X])^2+...+P[X=6]*(6E[X])^2 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But what does EX stand for?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Expected value of the random variable.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yea, but what would be the expectd value when X= 0 for instance? Or are they all just the expectation value?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, they are all the expectation. My first formula doesn't really make sense.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So just 5.7? All of them?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, you calculate how far they are from the average.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, now it got confusing. It should be \[ P[X=0]*(6E[X])^2 \] and not \[ P[X=0]*(0E[X])^2 \] 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.

phi
 4 years ago
Best ResponseYou've already chosen the best response.0There 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

phi
 4 years ago
Best ResponseYou've already chosen the best response.0E(whole)= 5.7 var(whole)= 0.51

phi
 4 years ago
Best ResponseYou've already chosen the best response.00.8 probability of 0 eggs broken means 0.8 probability of 6 whole eggs....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It says the variance is 1.7? Yes, I'm lost.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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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