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Math Help - Normail approximation to binomial distribution

  1. #1
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    Normail approximation to binomial distribution

    Suppose X1+X2+...+X80 follows a binomial distribution with n = 80, p = 0.74, q = 0.26, mean = 59.2, variance = 15.392
    then we may say that X1 + X2 + ... + X80 follows a normal distribution(approximately) with mean = 59.2, variance = 15.392

    Yi is another random variable such that Yi = 7Xi - 1 for i=1, 2, ..., 80
    Y1 + Y2 + ... + Y80 = 7(X1 + X2 + ... + X80) - 80
    so Mean(Y1 + Y2 + ... Y80) = 7Mean(X1+X2+...+X80) - 80 = 7*59.2 - 80 = 334.4
    Var(Y1 + Y2 + ... + Y80) = 49Var(X1+X2+...+X80) = 49 * 15.392 = 754.208
    so we may also say that Y1+Y2+...+Y80 follows a normal distribution with mean = 334.4, variance = 754.208
    Is this correct up to this point?

    Now suppose we want to calculate:
    probability(Y1+Y2+...+Y80 >= 340)
    =P( Z >= (339.5- 334.4)/sqrt(754.208) )
    =P( Z >= 0.1857 )

    On the other hand
    probability(Y1+Y2+...+Y80 >= 340)
    =P( 7(X1+X2+...+X80)-80 >= 340 )
    =P( X1+X2+...+X80 >= 60 )
    =P( Z >= (59.5-59.2)/sqrt(15.392) )
    =P( Z >= 0.0765 )
    which is completely different.
    It seems the methods do not give consistent answers.

    Now when we want to calculate
    probability(Y1+Y2+...+Y80 >= 300)
    =P( Z >= (299.5 - 334.4)/sqrt(754.208) )
    =P( Z >= -1.2708 )

    on the other hand
    P(Y1+Y2+...+Y80 >= 300)
    =P( 7(X1+X2+...+X80)-80 >= 300 )
    =P( X1+X2+...+X80 >= 55 )
    =P( Z >= (54.5-59.2)/15.392 )
    =P( Z >= -1.1980 )
    which is quite close to the answer from method 1

    So which method is the correct one?
    I also don't understand the reasons for such a large discrepancy in the first calculation.
    Thanks in advance.
    Last edited by iamnobody917; June 5th 2011 at 06:01 AM.
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  2. #2
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    Quote Originally Posted by iamnobody917 View Post
    Suppose X follows a binomial distribution with n = 80, p = 0.74, q = 0.26, mean = 59.2, variance = 15.392
    then we may say that X1 + X2 + ... + X80 follows a normal distribution(approximately) with mean = 59.2, variance = 15.392
    No. The sum of 80 of these RVs has mean 80*59.2, and variance 80*15.392

    If X were a RV with distribution B(80,0.74) then its mean would be 59.2 and variance 15.392 and so X is approximatly ~N(59.2,15.392)

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    No. The sum of 80 of these RVs has mean 80*59.2, and variance 80*15.392

    If X were a RV with distribution B(80,0.74) then its mean would be 59.2 and variance 15.392 and so X is approximatly ~N(59.2,15.392)

    CB
    I have made a mistake in the question. It should be X1+X2+...+X80 ~ B(80, 0.74)
    X1, X2, ..., X80 are the Bernoulli variables.
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  4. #4
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    Quote Originally Posted by iamnobody917 View Post
    Suppose X1+X2+...+X80 follows a binomial distribution with n = 80, p = 0.74, q = 0.26, mean = 59.2, variance = 15.392
    then we may say that X1 + X2 + ... + X80 follows a normal distribution(approximately) with mean = 59.2, variance = 15.392

    Yi is another random variable such that Yi = 7Xi - 1 for i=1, 2, ..., 80
    Y1 + Y2 + ... + Y80 = 7(X1 + X2 + ... + X80) - 80
    so Mean(Y1 + Y2 + ... Y80) = 7Mean(X1+X2+...+X80) - 80 = 7*59.2 - 80 = 334.4
    Var(Y1 + Y2 + ... + Y80) = 49Var(X1+X2+...+X80) = 49 * 15.392 = 754.208
    so we may also say that Y1+Y2+...+Y80 follows a normal distribution with mean = 334.4, variance = 754.208
    Is this correct up to this point?

    Now suppose we want to calculate:
    probability(Y1+Y2+...+Y80 >= 340)
    =P( Z >= (339.5- 334.4)/sqrt(754.208) )
    =P( Z >= 0.1857 )

    On the other hand
    probability(Y1+Y2+...+Y80 >= 340)
    =P( 7(X1+X2+...+X80)-80 >= 340 )
    =P( X1+X2+...+X80 >= 60 )
    =P( Z >= (59.5-59.2)/sqrt(15.392) )
    =P( Z >= 0.0765 )
    which is completely different.
    It seems the methods do not give consistent answers.

    Now when we want to calculate
    probability(Y1+Y2+...+Y80 >= 300)
    =P( Z >= (299.5 - 334.4)/sqrt(754.208) )
    =P( Z >= -1.2708 )

    on the other hand
    P(Y1+Y2+...+Y80 >= 300)
    =P( 7(X1+X2+...+X80)-80 >= 300 )
    =P( X1+X2+...+X80 >= 55 )
    =P( Z >= (54.5-59.2)/15.392 )
    =P( Z >= -1.1980 )
    which is quite close to the answer from method 1

    So which method is the correct one?
    I also don't understand the reasons for such a large discrepancy in the first calculation.
    Thanks in advance.
    You are applying the wrong continuity correction when using the normal approximation for the distribution of Y. The correction should be 3.5 not 0.5.

    CB
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