car security alarms go off at a mean rate of 3.8/hr. What is probability that in an hour there will be fewer than 4 alams

Question

Using poisson distribution  - Statistics Class

 

car security alarms go off at a mean rate of 3.8 per hour. What is the probability that in an hour there will be fewer than 4 alarms

 

 

Answer 1

Hello, thank you for taking the time to post your question!

The underlying formula that you want to use here is that

P(X = k) = lambda^k * e^(- lambda) / k! , where

lambda = mean rate = 3.8

k = number of events

so since it’s saying fewer than 4, I would compute the probability for

k = 0, k = 1, k = 2, k = 3 individually then sum them together. You could also use technology like Excel or a poisson distribution calculator!

either way, it should end up being 0.4735 for the end value on this one

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

The Poisson Distribution expresses the probability of x number of events happening in a fixed interval of time (or space, but in this case time), when we know the average rate of occurrence of these events and that these events are occurring randomly. 

 

The Poisson Distribution is given by:

 

P(x) = e^-λ *(λ^x/x!)              for x = 0,1,2,...,∞

 

where λ is the average rate of occurrence for an interval of time, and P(x) is the probability that x number of events will occur in any one interval of time.

 

So, here, λ = 3.8, and we are looking for probability that in an hour there will be either 0 alarms, or 1 alarm, or 2 alarms, or 3 alarms:

 

P = P(0) + P(1) + P(2) + P(3) 

 

So, you can plug in the numbers and find the answer.