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Math Help - [SOLVED] Normal Distribution

  1. #1
    Member looi76's Avatar
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    [SOLVED] Normal Distribution

    Question:
    The random variable, X, has a binomial distribution with parameters n = 40 and p = 0.3. Use a suitable approximation, which you should show in valid, to calculate the following probability: P(X = 15)


    (I should answer this question using Normal Distribution Table and not by integration!)


    Attempt:

    P(X = 15)████ n = 40████ p = 0.3████ q = 1 - 0.3 = 0.7

    np = 40 \times 0.3 = 12████ nq = 40 \times 0.7 = 28

    np > 5████ nq > 5████ So, the equation is VALID

    \mu = np = 12
    \sigma^2 = npq = 40 \times 0.3 \times 0.7 = 8.4
    \sigma = \sqrt{8.4}

    X \sim B(40 , 0.3)
    X \sim V(np, npq)
    X \sim N(\mu , \sigma^2)
    X \sim N(12 , 8.4)

    P(X = 15)

    P(14.5 \leq V \leq 15.5)████ 15 - 0.5 = 14.5████ 15 + 0.5 = 15.5

    P\left( \frac{(14.5 - 12)}{\sqrt{8.4}} \leq \frac{V - \mu}{\sigma} \leq \frac{(15.5 - 12)}{\sqrt{8.4}} \right)

    P( 0.863 \leq Z \leq 1.208 )
    = \phi(1.208) - \phi(0.863)
    = (0.8849 + 0.0015) - (0.8023 + 0.0008)
    = 0.0833

    Answer in textbook is 0.0808, where did I go wrong?
    Attached Thumbnails Attached Thumbnails [SOLVED] Normal Distribution-normal_distrubtion_table.gif  
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    The random variable, X, has a binomial distribution with parameters n = 40 and p = 0.3. Use a suitable approximation, which you should show in valid, to calculate the following probability: P(X = 15)

    (I should answer this question using Normal Distribution Table and not by integration!)

    Attempt:

    P(X = 15)████ n = 40████ p = 0.3████ q = 1 - 0.3 = 0.7

    np = 40 \times 0.3 = 12████ nq = 40 \times 0.7 = 28

    np > 5████ nq > 5████ So, the equation is VALID

    \mu = np = 12
    \sigma^2 = npq = 40 \times 0.3 \times 0.7 = 8.4
    \sigma = \sqrt{8.4}

    X \sim B(40 , 0.3)
    X \sim V(np, npq)
    X \sim N(\mu , \sigma^2)
    X \sim N(12 , 8.4)

    P(X = 15)

    P(14.5 \leq V \leq 15.5)████ 15 - 0.5 = 14.5████ 15 + 0.5 = 15.5

    P\left( \frac{(14.5 - 12)}{\sqrt{8.4}} \leq \frac{V - \mu}{\sigma} \leq \frac{(15.5 - 12)}{\sqrt{8.4}} \right)

    P( 0.863 \leq Z \leq 1.208 )
    = \phi(1.208) - \phi(0.863)
    = (0.8849 + 0.0015) - ({\color{red}0.8023} + 0.0008)
    = 0.0833

    Answer in textbook is 0.0808, where did I go wrong?
    Your error is in red. It should be 0.8051 (you looked at the '5 column' by mistake).

    Well done on showing all your working It made it easy to find your small mistake.
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