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.

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%

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.

is there supposed to be a diagram or a graph for this one? ?

step-by-step explanation: