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    1) A car-hire firm finds that daily demand for is cars follows a Poisson distribution with mean 3.6.

    What is the probability that 10 consecutive days will include 2 or more on which the demand is zero?


    2) The number of vehicles arriving at a toll bridge during a 5-minute period can be modelled by a Poisson distribution with mean 3.6.

    a) State the value for the standard deviation of the number of vehicles arriving at the toll during a 5-minute period.

    b) Find the probability that at least 3 vehicles arrive in each of 3 successive 5-minute periods.
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    We want to help, not mindlessly do your homework for you - do you have any thoughts of your own about these questions?
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    I'm a private candidate for A level Maths, I don't have homework lol! I've attempted all 3 questions.

    The first one, I thought I was getting the answer to, but cannot seem to get the correct solution. 2a I have no idea, 2b I'm unsure about the mean to obtain.

    Basically, I would like to see how somebody else works it out and compare it to the notes which I have.
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    (Original post by pprobson92)
    I'm a private candidate for A level Maths, I don't have homework lol! I've attempted all 3 questions.

    The first one, I thought I was getting the answer to, but cannot seem to get the correct solution. 2a I have no idea, 2b I'm unsure about the mean to obtain.

    Basically, I would like to see how somebody else works it out and compare it to the notes which I have.
    Can you post a picture of your working?
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    No, my notes are literally just undecipherable scribbles.

    With question 1, I have the probability that the demand will be zero on one day, but cannot get the answer with the probability being zero on two or fewer days. I know that you can use cumulative Poisson means, but the cumulative Poisson distribution in my book only goes up to mean =15 not 36.

    The dilemma is similar with 2b. If I times 3.6 by 3, the mean comes out as 10.2, which is not a mean option in the distribution table. So I am thinking there must be an alternative way to get the answers.
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    (Original post by pprobson92)
    No, my notes are literally just undecipherable scribbles.

    With question 1, I have the probability that the demand will be zero on one day, but cannot get the answer with the probability being zero on two or fewer days. I know that you can use cumulative Poisson means, but the cumulative Poisson distribution in my book only goes up to mean =15 not 36.

    The dilemma is similar with 2b. If I times 3.6 by 3, the mean comes out as 10.2, which is not a mean option in the distribution table. So I am thinking there must be an alternative way to get the answers.
    If you quote someone they can see that you've replied to them

    (Original post by Zacken)
    Q
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    (Original post by pprobson92)
    No, my notes are literally just undecipherable scribbles.

    With question 1, I have the probability that the demand will be zero on one day, but cannot get the answer with the probability being zero on two or fewer days. I know that you can use cumulative Poisson means, but the cumulative Poisson distribution in my book only goes up to mean =15 not 36.

    The dilemma is similar with 2b. If I times 3.6 by 3, the mean comes out as 10.2, which is not a mean option in the distribution table. So I am thinking there must be an alternative way to get the answers.
    For Q1, think about using the binomial (cumulative) distribution with number of trials 10 and probability found by using the poisson distribution.
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    Thank you, I applied binomial to the answer obtained before and got the correct answer. I also attempted the other questions again after realising that mean = variance within Poisson and got the correct answers too.

    I just needed a push in the right direction. Help much appreciated :-)
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    (Original post by pprobson92)
    Thank you, I applied binomial to the answer obtained before and got the correct answer. I also attempted the other questions again after realising that mean = variance within Poisson and got the correct answers too.

    I just needed a push in the right direction. Help much appreciated :-)
    Yep! That's the best way, hence why I didn't want to answer your question at first, just wanted to push you in the right direction so that you could do most of the work and grow mathematically! Well done on getting the answer for both parts!!
    First class work. :-)
 
 
 
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