Question

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)

Answer 1

Have you started solving the problem?

Answer 2

I don't know the unbroken eggs. I know how to get the expectation and variance once I have the distribution of unbroken eggs..

Answer 3

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
\[ E(unbroken) = 6-E(broken) \]

Answer 4

@dape what of the variance?

Answer 5

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

Answer 6

How do you do that?

Answer 7

@dape

Answer 8

it is about discrete type random variable.

Answer 9

You use the expected value, I said mean but meant EV. Calculate the variance from the EV of the number of unbroken eggs.

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

But what does E|X| stand for?

Answer 16

Expected value of the random variable.

Answer 17

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

Answer 18

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

Answer 19

So just 5.7? All of them?

Answer 20

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

Answer 21

Ok. Thankyou! :)

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

Did I lose you?

Answer 26

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

Answer 27

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

Answer 28

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