Poisson Probability

Question

Can someone please show me how to solve this. Your help is greatly appreciated. 

 

Airline fatalities
U.S. airlines average about 4.5 fatalities per month (Statistics Abstract of the United States: 2012). Assume the probability distribution for x, the number of fatalities per month, can be approximated by a Poisson probability distribution.
1. What is the probability that no fatalities will occur during any given month?
2. What is the probability that one fatality will occur during a month?
3. Find E (x) and the standard deviation of x

 

Thank you very much for your help!

Answer 1

Call the mean μ and the number of occurrences we're interested in x. Then, for a Poisson Distribution:

 

P(x) = (e^-μμ^x) / (x!)

 

For you, μ = 4.5 fatalities. That means your Poisson Distribution's probability formula is:

 

P(x) = (e^-4.54.5^x) / (x!)

 

 

1.

 

The probability that no fatalities will occur is P(0). So, let x = 0 and solve for P(0):

 

P(0) = (e^-4.54.5⁰) / (0!)

P(0) = (0.0111(1)) / 1

P(0) = 0.0111

 

 

2.

 

The probability that one fatality occurs is P(1). So, let x = 1 and solve for P(1):

 

P(1) = (e^-4.54.5¹) / (1!)

P(1) = (0.0111(4.5)) / 1

P(1) = 0.0500

 

 

3.

 

In a Poisson Distribution, the expected value E(x) is equal to μ. That's the value we expect to see. So:

 

E(x) = 4.5

 

The variance is also equal to μ. Since standard deviation is the square root of the variance, your standard deviation is:

 

Standard deviation = √4.5 = 2.121