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Statistics question

  1. Dec 3, 2006 #1
    30) A special roulette wheel has seven equally likely outcomes: 0, 1, 2, 3, 4, 5, 6. If you bet that an odd number comes up, you win or lose $10 according to whether or not that event occurs. If X denotes your ‘net’ gain, X=10 if we see 1, 3, or 5, and X = -10 otherwise. Suppose that you play this game 400 times. Let Y be your net gain after these 400 plays. The mean (expectation) and standard deviation of Y are, respectively (accurate to the number of figures shown):
    A) -270; 198
    B) -571; 198
    C) -270; 988
    D) -571; 988
    E) none of the other answers displayed




    I can get the mean (-571), but for the standard deviation, I only get the answer (apparently 198) when I use the formula:

    sigma² _ new = b * sigma²_old

    When the formula SHOULD use b², instead of just b.

    (Im using sigma² = Sum((x_i-u)²*p) to get the original variance.)
     
  2. jcsd
  3. Dec 4, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    On any one spin your probability of winning is 3/7 and of losing is 4/7. This is a binomial distribution with p= 3/7, q= 4/7 and N= 400. The mean value, and standard deviation for a binomial distribution with p, q, n are np and [itex]\sqrt{npq}[/itex] respectively. With your values, yes, -571 is the expectation and 988, not 198, is the is the standard deviation. I can't say anything about your formula, sigma² _ new = b * sigma²_old, since you haven said what "b" and "sigma_old" are.
     
  4. Dec 4, 2006 #3
    What I did was the mean is:

    [tex]400*[(3/7)(10)+(4/7)(-10)][/tex]
    = -571

    Now, for standard deviation, it should be:

    (For a single turn) variation =

    [tex]sigma^2=[10-(-1.428)]^2*(3/7)+[-10-(-1.428)]^2*(4/7)[/tex]

    So, sigma_new² (after 400 turns) SHOULD =

    b^2*sigma^2
    =400²*([10-(-1.428)]²*(3/7)+[-10-(-1.428)]²*(4/7))²

    However, I only get the right answer (which is 198), when I simply use 400 instead of 400².
     
    Last edited: Dec 4, 2006
  5. Dec 4, 2006 #4
    Nevermind, I got it using binomial distribution. The formula I was using before was for when you linearly transform by multiplying by 400, not when you repeat 400 times.

    Thanks
     
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