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Finding mean and variance

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose that, on average, 70% of graduating students want 2 guest tickets for a graduation ceremony, 20% want 1 guest ticket and the remaining10% don't want any guest tickets.

    (a) Let X be the number of tickets required by a randomly chosen student. Find the mean and variance of X.

    (b) If 500 guest tickets are available for a ceremony at which 300 students are graduating, what is the probability that there will not be enough tickets to satisfy demand?


    2. Relevant equations
    Variance = E(X2) - E2(X)


    3. The attempt at a solution
    I've got the mean to be ((0.7*2) + (0.2*1) + (0.1*0))/3 = 0.53and the variance to be ((0.72*2) + (0.22*1) + (0.12*0))/3 - 0.532 = 0.7391 but this seems wrong because the variance is then bigger than the mean?!?!?!
     
  2. jcsd
  3. Mar 1, 2009 #2
    Okay I have since realised that for part a) I think i was doing it wrong so now for the mean I have:

    ((0.7*2) + (0.2*1) + (0.1*0))/3 = 0.53

    But for the Variance I have:

    ((0.7 - 0.53)2+ (0.2 - 0.53)2 + (0.1 - 0.53)2)/3 = 0.1076 which makes far more sense!!

    Now I'm thinking of using the Central Limit Theorem to solve b)??!
     
  4. Mar 1, 2009 #3

    HallsofIvy

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    This is still incorrect. Suppose every student wanted one ticket. Would the average number be ((0.7*1) + (0.2*1) + (0.1*1))/3 = 1/3? Of course not. The ".7", ".2", and ".1" are already "averages". Don't divide by 3: 0.6*2+ 0.2*1+ 0.1*0= 1.4.

    There is no reason in the world why the variance can't be larger than the mean.

     
  5. Mar 1, 2009 #4
    Right okay, so now for the variance do I use the same formula but use 1.6 (I presume it was just a typo and it should have been 0.7 *2) instead of 0.53?
    Do I then use the Central Limit Theorem for part b?
     
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