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    Part of the question. It seemed easy, but my answers are miles away from the book answers. Has anyone already done this question?

    ...Tomatoes are packed in boxes and sent to market. The number of bad tomatoes in a box has Poisson distribution with mean 0.44. Find the probability of there being
    a) fewer than 2
    b) more than 2 bad tomatoes in a box when it is opened.

    Use a normal approximation to find the probability that in 50 randomly chosen boxes there will be fewer than 20 bad tomatoes in total.
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    what's your answer? and the answer of the book??
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    (Original post by BCHL85)
    what's your answer? and the answer of the book??
    I've got 0.927 for part (a) for example, and the book has 0.605...

    Sorry, must go now. Back later.
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    (Original post by Aitch)
    ...Tomatoes are packed in boxes and sent to market. The number of bad tomatoes in a box has Poisson distribution with mean 0.44. Find the probability of there being
    a) fewer than 2
    b) more than 2 bad tomatoes in a box when it is opened.

    Use a normal approximation to find the probability that in 50 randomly chosen boxes there will be fewer than 20 bad tomatoes in total.
    Let X be the DRV 'the number of bad tomatoes in box', then:

    X ~ Po(0.44)
    a) P(X < 2) = P(X = 0) + P(X + 1)
    ............... = 1/e0.44 + 0.44/e0.44 ≈ 0.927
    b) P(X > 2) = 1 - (P(X < 2) + P(X = 2))
    ............... = 1 - (0.44²/(2e0.44) + 0.927) ≈ 0.011

    50 boxes -> 0.44 x 50 = 22 bad tomatoes.

    Let Y model X, then:

    Y ~ N(22, 22)
    P(Y < 20) = 1- Φ((20-22)/√22)
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    The last bit needs a 1/2 continuity correction doesn't it
    P(Y < 20) = Φ((19.5-22)/√22)

    the rest looks right, I got part
    a) 0.927
    b) 0.102 (using exact value from part a))
    c) Φ(-0.53300179)
    = 1 - Φ(0.53) since we round to 2dp for this!
    = 1 - 0.7019
    = 0.2981

    The answers in the book are either complete gibberish or I have misunderstood that whole course, gulp! Gets scared!
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    (Original post by mensandan)
    The last bit needs a 1/2 continuity correction doesn't it
    P(Y < 20) = Φ((19.5-22)/√22)

    the rest looks right, I got part
    a) 0.927
    b) 0.102 (using exact value from part a))
    c) Φ(-0.53300179)
    = 1 - Φ(0.53) since we round to 2dp for this!
    = 1 - 0.7019
    = 0.2981

    The answers in the book are either complete gibberish or I have misunderstood that whole course, gulp! Gets scared!
    I agree with all of this. Thanks for the answer. Do you have the text book too? Can you make any sense of the given answers to this question?

    Has anyone else done this question and wondered about the given answers?
    (a) 0.605
    (b) 0.135
    (c) 0.238
 
 
 
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