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Thread: Normal Approximation confusion

  1. #1
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    Normal Approximation confusion

    There is an example on the normal approximation for binomial distribution.
    Given n =10, p = 0.2.
    We calculate binomial distribution from k=0 to k=6.

    $\displaystyle k=0: {10 \choose 0}p^0 q^10 = 0.1074$

    $\displaystyle k=1: {10 \choose 1}p^1 q^9 = 0.2684$

    $\displaystyle k=2: {10 \choose 2}p^2 q^8 = 0.3020$
    ... and so on.

    The normal approximation is given as:

    $\displaystyle K=0: 0.0904, 0.2307, 0.3154$...... and so on.

    Ok, I've been trying to figure out how to arrive at normal approximations given above, but to no avail.

    Do I use:$\displaystyle a_k = h*{ 1 \over {\sqrt{2 \pi}}} e^{{-0.5(kh)}^2}$

    where h=$\displaystyle {1 \over {\sqrt{2 \pi npq}}} $


    For example,
    $\displaystyle a_0 = { 1 \over {\sqrt{2 \pi}}} {1 \over {\sqrt{2 \pi 10*0.2*0.8}}} = 0.1258$

    which is different from the answer given.
    I thank you for your comments. This is a particularly vexing problem, as I want to thoroughly understand the inner workigs of Normal Approximations.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chopet View Post
    There is an example on the normal approximation for binomial distribution.
    Given n =10, p = 0.2.
    We calculate binomial distribution from k=0 to k=6.

    $\displaystyle k=0: {10 \choose 0}p^0 q^10 = 0.1074$

    $\displaystyle k=1: {10 \choose 1}p^1 q^9 = 0.2684$

    $\displaystyle k=2: {10 \choose 2}p^2 q^8 = 0.3020$
    ... and so on.

    The normal approximation is given as:

    $\displaystyle K=0: 0.0904, 0.2307, 0.3154$...... and so on.

    Ok, I've been trying to figure out how to arrive at normal approximations given above, but to no avail.

    Do I use:$\displaystyle a_k = h*{ 1 \over {\sqrt{2 \pi}}} e^{{-0.5(kh)}^2}$

    where h=$\displaystyle {1 \over {\sqrt{2 \pi npq}}} $


    For example,
    $\displaystyle a_0 = { 1 \over {\sqrt{2 \pi}}} {1 \over {\sqrt{2 \pi 10*0.2*0.8}}} = 0.1258$

    which is different from the answer given.
    I thank you for your comments. This is a particularly vexing problem, as I want to thoroughly understand the inner workigs of Normal Approximations.
    Doing it this way you should have:

    $\displaystyle
    a_0=\frac{1}{\sqrt{2\pi}\sqrt{10\times 0.2 \times 0.8}}e^{-\frac{2}{1.6}}\approx 0.0904
    $

    But to do this correctly you really need:

    $\displaystyle
    p(0)\approx P(0.5;np,npq)-P(-0.5;np,npq)
    $

    where $\displaystyle P(x;\mu,\sigma^2)$ is the cumulative normal distribution with mean $\displaystyle \mu$ and variance $\displaystyle \sigma^2$.

    RonL
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  3. #3
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    much appreciated.

    But what is the formula you applied to get:

    $\displaystyle e^{-2 \over {1.6}}$

    How does K=0 come in?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chopet View Post
    much appreciated.

    But what is the formula you applied to get:

    $\displaystyle e^{-2 \over {1.6}}$

    How does K=0 come in?
    The exponetial is of -(k-np)^2/(2npq), put k=0, n=10, p=0.2 and q=0.8

    RonL
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