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

Question

A study found that the airlines average about 4.5 fatalities per month. Assume that the

probability distribution for x, the number of fatalities per month, can be approximated by

a Poisson probability distribution

a. What is the probability that more than one fatalities will occur during any given month?

[3 Marks]

b. What is the probability that no fatality will occur during two weeks?

Answer 1

a) P(more than 1 fatality)=1-(P(0 fatality)+P(1 fatality)), since number of fatalities can be 0, 1, 2, 3, ..... and probability of anything that can happen happening is always 1, therefore to find out the probability of more than 1 fatality, we can subtract the probability of 0 or 1 fatalities from 1.

P(more than 1 fatality)= 1-((e^(-4.5))*4.5^0)/0!)+((e^(-4.5)*4.5^1)/1!))=0.938901

Since Poisson pmf formula is P(X=x)=e^(-a)*a^x/x! where a is the mean # of occurrences, fatalities in this case-and it is 4.5/month.

b) The average number of fatalities in 2 weeks is 4.5/2=2.25, if we apply Poisson pmf formula with a =2.25 and x=0, we have

P(X=0)=e^(-2.25)*2.25^0/0! =0.105399