Answer:
a) Earthquakes are random and independent events.
b) There is an 85.71% probability of fewer than three quakes.
c) There is a 0.51% probability of more than five quakes.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
In this problem, we have that:
In Northern Yellowstone Lake, earthquakes occur at a mean rate of 1.3 quakes per year, so ![\mu = 1.3](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.3)
(a) Justify the use of the Poisson model.
Earthquakes are random and independent events.
You can't predict when a earthquake is going to happen, or how many are going to have in a year. It is an estimative.
(b) What is the probability of fewer than three quakes?
This is ![P(X < 3)](https://tex.z-dn.net/?f=P%28X%20%3C%203%29)
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.2725)
![P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.3543)
![P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2303)
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2725 + 0.3543 + 0.2303 = 0.8571](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.2725%20%2B%200.3543%20%2B%200.2303%20%3D%200.8571)
There is an 85.71% probability of fewer than three quakes.
(c) What is the probability of more than five quakes?
This is ![P(X > 5)](https://tex.z-dn.net/?f=P%28X%20%3E%205%29)
We know that either there are 5 or less earthquakes, or there are more than 5 earthquakes. The sum of the probabilities of these events is decimal 1.
So
![P(X \leq 5) + P(X > 5) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%29%20%2B%20P%28X%20%3E%205%29%20%3D%201)
![P(X > 5) = 1 - P(X \leq 5)](https://tex.z-dn.net/?f=P%28X%20%3E%205%29%20%3D%201%20-%20P%28X%20%5Cleq%205%29)
In which
![P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-1.3}*1.3^{0}}{(0)!} = 0.2725](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.2725)
![P(X = 1) = \frac{e^{-1.3}*1.3^{1}}{(1)!} = 0.3543](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.3543)
![P(X = 2) = \frac{e^{-1.3}*1.3^{2}}{(2)!} = 0.2303](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2303)
![P(X = 3) = \frac{e^{-1.3}*1.3^{3}}{(3)!} = 0.0980](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.0980)
![P(X = 4) = \frac{e^{-1.3}*1.3^{4}}{(4)!} = 0.0324](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.0324)
![P(X = 5) = \frac{e^{-1.3}*1.3^{5}}{(5)!} = 0.0084](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20%5Cfrac%7Be%5E%7B-1.3%7D%2A1.3%5E%7B5%7D%7D%7B%285%29%21%7D%20%3D%200.0084)
![P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2725 + 0.3543 + 0.2303 + 0.0980 + 0.0324 + 0.0084 = 0.9949](https://tex.z-dn.net/?f=P%28X%20%5Cleq%205%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%3D%200.2725%20%2B%200.3543%20%2B%200.2303%20%2B%200.0980%20%2B%200.0324%20%2B%200.0084%20%3D%200.9949)
Finally
![P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9949 = 0.0051](https://tex.z-dn.net/?f=P%28X%20%3E%205%29%20%3D%201%20-%20P%28X%20%5Cleq%205%29%20%3D%201%20-%200.9949%20%3D%200.0051)
There is a 0.51% probability of more than five quakes.