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Probability Old Exam P Forum 

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#1




meaning of expected value
Hi, let's consider this case:
A small quiz has three multiple choices with each question three choices. so, using np formula, we get the expected value be 3*(1/3)=1 What does this "1" mean? how does it represent the "average", "expect" in this case? What I am asking is, what does this "one" mean back into the situation, not in theoretical sense. 
#2




Repeat the test n times selecting answers at random. Sum the scores and divide by n, That is your sample average. Specify an epsilon and a probability. You can determine an n such that the average will be within epsilon of 1, with the specified probability.
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#3




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#4




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think about the following case: There is a multiple choice test with 3 questions, each with 4 answer choices and you randomly guess each question. The expected value is 3/4. Is there really a practical explanation of wht 3/4 is? Surely you can't get 3/4 of a question right. imo in the continous case you can make an argument to say that a practical explanation does exist. but when you discretise events, your expected value might not even be in your support so it's harder to come up with a practial explanation. 
#6




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What's the maximum number of items you could get correct? What's the answer if you have 5 items? Work through the answers of the first two questions with 5 items and compare them to your answer to the previous question. Rework the min, max, and expected value for differing values of total number of items available; then look at connections to the data you've just collected.
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#8




Except saying on average has no precise meaning because your realized average will very rarely equal to true expected value. In fact as you increase n the probability that the average exactly equals the true expected value goes to zero (for finite n).
Last edited by Academic Actuary; 05122019 at 02:00 PM.. 
#10




In the continuous case the average will never equal the expected value. In the discrete case it can but the probability of equality decreases. Consider flipping a fair coin. With 2 flips there is a 50% chance the average number of heads equals .5. With 100,000,000,000 flips the probability of 50,000,000,000 heads is very small, but the probability distribution for the average is collapsing around .5.
What you could say is the average converges stochastically to the expected value. 
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