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Math Help - Using probability distributions - please help!

  1. #1
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    Question Using probability distributions - please help!

    1. A garden shop claims that 80% of all iris bulbs germinate and flower. A customer buys 8 bulbs.
    a) What probability distribution is most likely to model the number of bulbs which germinate and flower? Explain why you have made your choice and give any parameter(s) for the distribution. (This will be the binomial, poisson or normal distribution)

    Use your chosen probability distribution to find the probability that:
    b) all of the bulbs germinate and flower
    c) only half of the bulbs germinate and flower
    d) more than 6 bulbs germinate and flower
    e) at least one bulb fails to germinate and flower.

    2. In a warehouse, accidents have been taking place at the rate of 2 every 3 months.
    a) What probability distribution is most likely to model the number of accidents in a 3 month period? Explain why you have made your choice and give any parameter(s) for the distribution.
    b) Find the mean and standard deviation of the number of accidents per year
    c) What is the percentage of months with no accidents?
    d) If a month is chosen at random, find the probability of at least one accident.
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  2. #2
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    Quote Originally Posted by Kiwigirl
    1. A garden shop claims that 80% of all iris bulbs germinate and flower. A customer buys 8 bulbs.
    a) What probability distribution is most likely to model the number of bulbs which germinate and flower? Explain why you have made your choice and give any parameter(s) for the distribution. (This will be the binomial, poisson or normal distribution)

    Use your chosen probability distribution to find the probability that:
    b) all of the bulbs germinate and flower
    c) only half of the bulbs germinate and flower
    d) more than 6 bulbs germinate and flower
    e) at least one bulb fails to germinate and flower.
    You need to use something call binomial probability. Which states that if the probability of an event is p. Then the probability of that even happening m out of n times is,
    P(m)=_nC_mp^m(1-p)^{n-m}.

    Over here, p=.8 and n=8, the only thing changing is m.

    Now we calculate the exact probability of each event.
    P(m)= _8C_m(.8)^m(.2)^{8-m}

    Here is a table,
    \left\{\begin{array}{cc}m&P(m)\\0&.0000\\1&.0001\\  2&.0011\\3&.0092\\4&.0459\\5&.1468\\6&.2936\\7&.33  55\\8&.1678\end{array}\right

    Just as, a check we can see that they add up to one. Actually 1.001 because I approximated. Next the probabilty of exactly 0 is not impossible it is almost impossible hence the significant digits. Now, you can answer your questions. For example, the most likely is when m=7 thus, by probability theory you should chose 7 good ones and 1 bad iris bulb. You should we able to answer everything else from here. For example, at most 5 means you need to add up all from m=0 and m=5 to get your answer.
    Last edited by ThePerfectHacker; March 27th 2006 at 03:29 PM.
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    Reply

    Thank you for your help ThePerfectHacker, I don't understand much at all about this topic, so I don't really know how to go on answering the rest of questions in 1), and I don't know what to do in questions 2). Could you please help me further?
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    Quote Originally Posted by Kiwigirl
    Thank you for your help ThePerfectHacker, I don't understand much at all about this topic, so I don't really know how to go on answering the rest of questions in 1), and I don't know what to do in questions 2). Could you please help me further?
    m represents the number of bulbs that will germinate. And P(m) represent the probability that that will happen. Before I gave you a chart with the values and provided a formula for "binomial probability" of how these values where derived.

    b)That means, m=8 thus, the probability is,
    P(8)=_8C_8(.8)^8(.2)^0\approx .1678

    c)That means, m=4 thus, the probability is,
    P(4)=_8C_4(.8)^4(.2)^4\approx .0459

    d)That means, m=6\mbox{ or }7\mbox{ or}8 thus, the probability is,
    P(6)+P(7)+P(8)\approx .2936+.3355+.1678\approx .9646

    e)That means, m=0,1,2,...,7 thus, the probability is,
    P(0)+P(1)+...+P(7)=1-P(8)\approx 1-.1678\approx .8322
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