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Thread: central limit theorm help

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    Angry 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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    Re: central limit theorm help


    Whenever you meet a sequence of i.i.d. random variables

    Z_1, Z_2, ..., Z_n and you are interested to use the CLT

    to approximate the CDF of the sum \sum_{i=1}^n Z_i,

    all you need to do is to find out

    1. E[Z_1], Var[Z_1]

    2. Then E\left[\sum_{i=1}^n Z_i \right] = nE[Z_1]

    SD\left[\sum_{i=1}^n Z_i \right] = \sqrt{n}SD[Z_1]

    And thus

    \Pr\left\{\sum_{i=1}^n Z_i  \leq z \right\}
= \Pr\left\{\frac {\displaystyle \sum_{i=1}^n Z_i  - nE[Z_1]}
{\sqrt{n}SD[Z_1]} \leq \frac {z - nE[Z_1]} {\sqrt{n}SD[Z_1]} \right\}

    \approx \Phi\left(\frac {z - nE[Z_1]} {\sqrt{n}SD[Z_1]} \right) by CLT, where \Phi is the standard normal CDF

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