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Thread: Question on normal distribution(again)

  1. #1
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    Question on normal distribution(again)



    I don't know how to do part (c) and (d)
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  2. #2
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    Quote Originally Posted by mngeow View Post


    I don't know how to do part (c) and (d)
    For (c), you are looking for the probability that $\displaystyle 850<X<1350$, while eliminating the area of the graph up to $\displaystyle X=850$.

    which is $\displaystyle \displaystyle\frac{P(X<1350)-P(X<850)}{P(X>850)}$

    Convert to the Z-values and compute the probability.
    Bear in mind that the probability of the bulb failing within the first 850 hours is now zero.
    Hence your fraction has $\displaystyle P(X>850)$ in the denominator.

    For (d), use the Binomial conversions $\displaystyle \sigma=\sqrt{npq},\;\;\; \mu=np$

    where p=0.07, q=0.93, n=1200
    Last edited by Archie Meade; Sep 15th 2010 at 01:44 PM.
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  3. #3
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    Quote Originally Posted by mngeow View Post


    I don't know how to do part (c) and (d)
    Let P(t) be the CDF of the bulb life.

    For c you want [P(1350)-P(500)]/(1-P(500)) Imagin you have 1200 bulbs then 1200 P(500) have failed by 500 hours, 1200 P(1350) by 1350 hours, so out of the 1200 (1-P(500)) that survive the first 500 hours 1200 [P(1350)-P(500)] fail in the next 850 hours.

    CB
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    ooh,I get it now,what about part d?
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  5. #5
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    Quote Originally Posted by mngeow View Post
    ooh,I get it now,what about part d?
    The number that fail has a binomial distribution$\displaystyle B(0.07,1200)$, and you use the Normal approximation (and use a continuity correction) $\displaystyle N(0.07 \times 1200, 1200\times 0.07 \times 0.93)$

    CB
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