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Statistics: Average values

  1. Jan 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    I thought about this earlier today. Lets say I toss a dice three times, and the outcome is 2, 4 and 5. Is the mean (2+4+5)/3 or (2+4+5)/6?

    The reason why I am asking is that we can look at the mean as a weighted average of the probability of getting each outcome. So in our case, the probability for getting {1, 2, 3, 4, 5, 6} is the same, so that is why one might be inclined to say that the mean is (2+4+5)/6.

    But then again, only 3 tosses has been made. So my question is:

    1) Am I looking at two different things here?

    2) If yes (which is probably the correct answer to #1), then is (2+4+5)/6 the expectation value (i.e. the average when tossing the dice infinitely many times), and (2+4+5)/3 the average of that sample?

    Thanks in advance. I really appreciate your help.

    Best regards,
    Niles.
     
  2. jcsd
  3. Jan 17, 2009 #2

    HallsofIvy

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    There is no need to "weight" here since the probabilities of rolling 2, 4, 5 with a single die ("dice" is the plural) are the same. If you did that, you have to take into account the fact that those are not the only possibilities. The probability of rolling one of 2, 4, or 5, as opposed to 1, 3, or 6, is 3/6= 1/2, You would have to find the "weighted" sum (1/6)(2)+ (1/6)(4)+ (1/6)(5) divided by 1/2 (the overall probability). That gives (2+4+5)/6 times 2= (2+4+5)/3.

    Not really, but you forgot that 1/6+ 1/6+ 1/6 is not 1: the total probability here is 1/2 not 1.

    Done correctly either way, the answer is (2+4+5)/3.

     
  4. Jan 18, 2009 #3
    Thanks for helping.
     
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