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20 January, 06:59

Ruby has a bird feeder which is visited by an average of 13 birds every 2 hours during daylight hours. What is the probability that the bird feeder will be visited by more than 3 birds in a 40 minute period during daylight hours? Round your answer to three decimal places.

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  1. 20 January, 08:05
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    62.93%

    Step-by-step explanation:

    We have to solve it by a Poisson distribution, where:

    p (x = n) = e ^ ( - l) * l ^ (x) / x!

    Where he would come being the number of birds that there would be in 40 minutes, we know that in 2 hours, that is 120 minutes there are 13, therefore in 40 there would be:

    l = 13 * 40/120

    l = 4,333

    Now, we have p (x> 3) and that is equal to:

    p (x> 3) = 1 - p (x < = 3)

    So, we calculate the probability from 0 to 3:

    p (x = 0) = 2.72 ^ ( - 4.33) * 4.33 ^ (0) / 0! = 0.01313

    p (x = 1) = 2.72 ^ ( - 4.33) * 4.33 ^ (1) / 1! = 0.0568

    p (x = 2) = 2.72 ^ ( - 4.33) * 4.33 ^ (2) / 2! = 0.12310

    p (x = 3) = 2.72 ^ ( - 4.33) * 4.33 ^ (3) / 3! = 0.17767

    If we add each one:

    0.01313 + 0.0568 + 0.12310 + 0.17767 = 0.3707

    replacing:

    p (x> 3) = 1 - 0.3707

    p (x> 3) = 0.6293

    Which means that the probability is 62.93%
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