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  1. #1
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    Stats Help

    1. Quality control reports that are not available to consumers indicate that 20% of
    Hot Shot ovens have faulty control panels. The owner of an apartment complex
    has just purchased 120 of these ovens. Assume that the ovens purchased
    constitute a random sample.
    a. Can you use the normal approximation to the binomial distribution to estimate
    the probability that between 22 and 35 of them have faulty control panels?
    Explain.
    b. Estimate the probability that between 22 and 35 of them (including 22 and 35)
    have faulty control panels.


    Any ideas?
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  2. #2
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    a. Can you use the normal approximation to the binomial distribution to estimate
    the probability that between 22 and 35 of them have faulty control panels?
    Explain.
    The normal approximation is appropriate when np and n(1-p) are both large. How large depends on what textbook/lecturer you have. If you are feeling keen, calculate both the exact value and the approximation and compare them.

    b. Estimate the probability that between 22 and 35 of them (including 22 and 35)
    have faulty control panels.
    If np and n(1-p) are large, P(22 \leq X \leq 35) \approx P(21.5<Y<35.5),
    Where Y has distribution N(np, np(1-p))
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  3. #3
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    haha wow... that just confused me more. Apparently I have no idea what to do.

    So...
    p = .2
    q = .8
    n = 120

    np = 120(.2) = 24 = standard dev
    npq = 19.2

    and I have the equation Z = (x - (np))/sqrt(npq)

    no idea what to do from there...
    Last edited by iliketurtles; December 16th 2008 at 05:37 PM.
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  4. #4
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    Quote Originally Posted by iliketurtles View Post
    haha wow... that just confused me more. Apparently I have no idea what to do.
    What part of the reply you got confuses you? Note that a minimum level of knowledge on your part is assumed.
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  5. #5
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    this is what I know :

    So...
    p = .2
    q = .8
    n = 120

    np = 120(.2) = 24 = standard dev
    npq = 19.2

    and I have the equation Z = (x - (np))/sqrt(npq)

    no idea what to do from there...
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  6. #6
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    Quote Originally Posted by iliketurtles View Post
    this is what I know :

    So...
    p = .2
    q = .8
    n = 120

    np = 120(.2) = 24 = standard dev Mr F says: NO. This is the mean.
    npq = 19.2 Mr F says: This is the variance. The square root is the standard deviation.

    and I have the equation Z = (x - (np))/sqrt(npq)

    no idea what to do from there...
    You have Y ~ Normal (\mu = 24, \, \sigma^2 = 19.2).

    Calculate \Pr(21.5 < Y < 35.5).

    You must have been taught how to do calculations using a normal distribution ....?
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