Question

Estion

Answer 1

Using the Poisson distribution, it is found that there is a 0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.

What is the Poisson distribution?

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

• $\mu$ is the mean in the given interval.

Considering the average of 15 birds every 2 hours during daylight hours, the mean for a 45-minute period is given by:

$\mu = 15 \times \frac{45}{120} = 5.625$

The probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours is given by:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5.625}(5.625)^{0}}{(0)!} = 0.004$

$P(X = 1) = \frac{e^{-5.625}(5.625)^{1}}{(1)!} = 0.02$

$P(X = 2) = \frac{e^{-5.625}(5.625)^{2}}{(2)!} = 0.057$

$P(X = 3) = \frac{e^{-5.625}(5.625)^{3}}{(3)!} = 0.107$

$P(X = 4) = \frac{e^{-5.625}(5.625)^{4}}{(4)!} = 0.150$

$P(X = 5) = \frac{e^{-5.625}(5.625)^{5}}{(5)!} = 0.169$

Then:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.004 + 0.02 + 0.057 + 0.107 + 0.15 + 0.169 = 0.507$

0.507 = 50.7% probability that the bird feeder will be visited by at most 5 birds in a 45 minute period during daylight hours.

More can be learned about the Poisson distribution at brainly.com/question/13971530

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