How do you do this poisson distribution question?

a telephone soliciting company obtains an average of five orders per 500 solicitations. if the company reaches 250 potential customers, find the probability of obtaining at least 2 orders.

the poisson distribution is:

(e^(- lambda) x lambda^X) / X!

I know that X = 2, but i can't seem to figure out lambda. at first i thought lambda = 5/1000, but the answer in the back of the book is .3554 so 5/1000 does not work. does anyone know how to calculate lambda?

The Poisson distribution can be derived from the binomial distribution. The Poisson is nothing more than the limiting case of the Binomial where n is large and p is small.

A good way to identify when you need to use the Poisson distribution is when the problem requires you to use a rate. This is not always true, but more often than not remembering this will help you to identify a Poisson model.

Let X be the number of orders. X has the Poisson distribution with parameter 位t = 2.5

In general you have:

X ~ Poisson( 位t )
P(X = x) = ( 位t )^x * exp( -位t ) / x! for x = 0, 1, 2, 3, 4, ...
P(X = x) = 0 otherwise

the mean of the Poisson distribution is the parameter, 位t
the variance of the Poisson distribution is the parameter, 位t

In this problem we have
位 = 5 orders per 500 customers
t = 0.5 time unit(s) (250 is 50% of 500)

this results in our random variable X ~ Poisson( 2.5 )


Find P( X 鈮?2 ) =

鈭?
鈭?P(X = i)
i= 2

This sum is an infinite sum and is not easy to solve so instead let's rewrite the sum in terms of a finite sum.


Find P( X 鈮?2 ) = 1 - P( X < 2 ) =

. . . 1
1 - 鈭?P(X = i)
. . . i=0

= 0.7127025

lambda=5 for 500customers
lambda=2.5 for 250customers

To get the answer, you need to find P(X<=2) which is
P(X=0) + P(X=1) + P(X=2)