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Variance problem

  1. Apr 20, 2009 #1
    X1, X2,...,X16 are independent and normally distributed, where mean value is 80 and variance is 18^2. Let Y = X1 + X2 + ... + X16. Calculate

    i) P(X1 > 90)

    ii) E(Y)

    iii) Var(Y)

    iv) P(Y>16*90)


    i) 0.288, easy

    ii) E(Y) = 16*80 = 1280

    iii) Var(Y) = Var(16*Var(X)) = 16^2 * 18^2 (WRONG!)

    Any suggestions?
     
  2. jcsd
  3. Apr 20, 2009 #2
    X1+...+Xn is not the same as n*X

    Use Var(a1*X1+...+an*Xn)=a1^2*Var(X1)+...+an^2*Var(Xn) with ai=1 for all i.
     
  4. Apr 20, 2009 #3
    Brilliant! Would any of

    P(X1>90) = 0.288

    E(Y) = 1280

    Var(Y) = 5184

    P(Y>16*90) = 0.013

    be correct without assuming normal distribution, but still assuming independence?
     
  5. Apr 20, 2009 #4
    (ii) and (iii) don't require normal. (ii) doesn't even require independence, but (iii) does.

    (i) and (iv) require normal.
     
  6. Apr 20, 2009 #5
    And which of them requires independence?
     
  7. Apr 20, 2009 #6
    (iii) and (iv)
     
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