Answer:
a) 9.93% probability that exactly ten requests are received during a particular 2-hour period
b) 13.53% probability that they do not miss any calls for assistance
c) 2
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.
Poisson process with rate θ = 4 per hour.
This means that
, in which n is the number of hours.
a. Compute the probability that exactly ten requests are received during a particular 2-hour period.
n = 2, so ![\mu = 4*2 = 8](https://tex.z-dn.net/?f=%5Cmu%20%3D%204%2A2%20%3D%208)
This is P(X = 10). So
![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 = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20%5Cfrac%7Be%5E%7B-8%7D%2A8%5E%7B10%7D%7D%7B%2810%29%21%7D%20%3D%200.0993)
9.93% probability that exactly ten requests are received during a particular 2-hour period
b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?
n = 0.5, so ![\mu = 4*0.5 = 2](https://tex.z-dn.net/?f=%5Cmu%20%3D%204%2A0.5%20%3D%202)
This is P(X = 0). So
![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^{-2}*2^{0}}{(0)!} = 0.1353](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-2%7D%2A2%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.1353)
13.53% probability that they do not miss any calls for assistance
c. How many calls would you expect during their break?
n = 0.5, so ![\mu = 4*0.5 = 2](https://tex.z-dn.net/?f=%5Cmu%20%3D%204%2A0.5%20%3D%202)