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  1. #1
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    Poisson distribution

    Breakdowns of the lifts in an office building at any given time are independent of one another and can be modelled using a Poisson Distribution with a mean 0.2 per day.

    1. Determine the probability that there will be exactly 4 breakdowns during the month of June.

    2. The probability that there are more than 3 breakdowns during the month of June.

    3. The probability that there are no breakdowns during the first 5 days of June.

    4. The probability that the first breakdown in June occures on June 3rd.

    5. It costs 1850 Euros to service the lifts when they breakdown. Find the expected cost of service for June.

    6. Determine the probability that there will be no breakdown in exactly 4 out of the first 5 days in June.

    Thanks!
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  2. #2
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    Quote Originally Posted by pantera View Post
    Breakdowns of the lifts in an office building at any given time are independent of one another and can be modelled using a Poisson Distribution with a mean 0.2 per day.

    1. Determine the probability that there will be exactly 4 breakdowns during the month of June.

    2. The probability that there are more than 3 breakdowns during the month of June.

    3. The probability that there are no breakdowns during the first 5 days of June.

    4. The probability that the first breakdown in June occures on June 3rd.

    5. It costs 1850 Euros to service the lifts when they breakdown. Find the expected cost of service for June.

    6. Determine the probability that there will be no breakdown in exactly 4 out of the first 5 days in June.

    Thanks!
    I have no time right now but will reply later if no-one else does. In the meantime, where exactly are you stuck?
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  3. #3
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    Quote Originally Posted by pantera View Post
    Breakdowns of the lifts in an office building at any given time are independent of one another and can be modelled using a Poisson Distribution with a mean 0.2 per day.

    1. Determine the probability that there will be exactly 4 breakdowns during the month of June.

    2. The probability that there are more than 3 breakdowns during the month of June.
    Hello pantera

    I'll help you with the first two to get your started.


    I think the main difficult with question 1 is know the number of days in June! there are 30 days in June so there is a mean of 0.2 \times 30 = 6 breakdowns in the month.

    Let X be the number of breakdowns in the month of June. then X \sim Po(6). You require P ( X = 4 | X \sim Po(6)) which can be calculate directly using the definition P ( X = 4) = \frac{6^4e^{-6}}{4!}.


    For question 2. X is defined in the same way as in part 1. you require P (X > 3) which is the same as 1 - P(X \leq 3). you can either use the definition or a table to calculate this.

    Bobak
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  4. #4
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    Quote Originally Posted by pantera View Post
    Breakdowns of the lifts in an office building at any given time are independent of one another and can be modelled using a Poisson Distribution with a mean 0.2 per day.

    1. Determine the probability that there will be exactly 4 breakdowns during the month of June.

    2. The probability that there are more than 3 breakdowns during the month of June.

    3. The probability that there are no breakdowns during the first 5 days of June.

    Mr F says: I think that you need to first calculate the probability of no breakdown on a single day: {\color{red}p = \frac{0.2^0 e^{-0.2}}{0!}}. Then probability of no breakdown in first five days is p^5.

    4. The probability that the first breakdown in June occures on June 3rd.

    Mr F says: p^2 (1 - p). See also geometric distribution.

    5. It costs 1850 Euros to service the lifts when they breakdown. Find the expected cost of service for June.

    Mr F says: The expected number of breakdowns in June is 6. Therefore ......

    6. Determine the probability that there will be no breakdown in exactly 4 out of the first 5 days in June.

    Mr F says: Let Y be the random variable number of breakdowns in first five days of June. Then Y ~ Binomial(n = 5, p = whatever you calculated from #3 above). Calculate Pr(Y = 4) ......

    Thanks!
    ..
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