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Math Help - central limit theorm help

  1. #1
    Senior Member
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    Exclamation central limit theorm help

    Z1,Z2,.... are identically independently distributed with uniform distribution [-20,10]. How do i use the central limit theorem to estimate the the probability
    P(summation (900, i=1) : Zi >= -4470)

    My attempt
    E(Zi)=-5
    Var(Zi)=80
    P(summation Zi <= -4470) = P([summation Zi]/900 <= [-4470]/900)
    =P(sample Z (900) <= 80)
    =P([sample Z (900) + 5]/sqrt(80/900) <= [80+5]/sqrt(80/900))
    ~=P(Z<=285.1) where Z~Normal(0,1)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Sneaky View Post
    Z1,Z2,.... are identically independently distributed with uniform distribution [-20,10]. How do i use the central limit theorem to estimate the the probability
    P(summation (900, i=1) : Zi >= -4470)

    My attempt
    E(Zi)=-5
    Var(Zi)=80
    P(summation Zi <= -4470) = P([summation Zi]/900 <= [-4470]/900)
    =P(sample Z (900) <= 80)
    =P([sample Z (900) + 5]/sqrt(80/900) <= [80+5]/sqrt(80/900))
    ~=P(Z<=285.1) where Z~Normal(0,1)
    The distribution of the sum on $$n iid RVs with mean $$ m and variance $$v has distribution N(n \times m,n \times v) (approximatly, by the CLT)

    The mean of U(a,b) is (a+b)/2 and its variance is (b-a)^2/12

    CB
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  3. #3
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    so i redid it and now i get

    1-P(Z<=0.115)

    where Z ~ N(0,1)

    Is this right?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Sneaky View Post
    so i redid it and now i get

    1-P(Z<=0.115)

    where Z ~ N(0,1)

    Is this right?
    Yes

    CB
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