Question

A telephone soliciting company obtains an average of five orders per 1000 solicitations. If the company reaches 250 potential customers, find the probability of obtaining at least two orders.

Answer 1

How do you solve it using Poisson Distribution?

Answer 2

@danielle02 i don't know what is Poisson Distribution and how to solve this.
Only way i can hep you that i can send you best link in which this question explains very nicely.

Answer 3

*help

Answer 4

can i send you link???

Answer 5

@mayankdevnani sure

Answer 6

this is link:
http://answers.yahoo.com/question/index?qid=20080326213331AA8Qm3v

Answer 7

i hope this link will help you.

Answer 8

i'm trying to figure out what the lambda is and it doesn't state here how to get the lambda. but thanks anyway.

Answer 9

sorry.... i can't help you out....
and for thanks... welcome:)
lol

Answer 10

The average number of orders per 250 customers is given by
\[5\times \frac{250}{1000}=1.25\]
If X is a random variable with Poisson distribution, then
\[P(X=x)=\frac{e ^{- \lambda}\lambda ^{x}}{x!}\]
In this case lambda is 1.25, therefore
\[P(X=2)=\frac{e ^{-1.25}1.25^{2}}{2!}=you\ can\ calculate\]

Answer 11

I found at the back of a book that the answer is 0.3554

Answer 12

Ok. So I finally figured out how to find the answer.

Answer 13

After finding P(X=2) it is necesary to find P(X=3), P(X=4) ......... and add the probabilities to find the probability of at least 2 orders.

Answer 14

As suggested by @Callisto the quickest way to find the solution is to calculate
P(X=0) + P(X=1), and then subtract the result from 1.
P(X=at least 2 orders) = 1 - {P(X=0) + P(X=1)}

Answer 15

@kropot72 I know this question is like a year old, but... I was wondering if you happen to see this message, could you explain how you figured out that n = 5 and p = 250/1000?
I understand why x = 2, and I know how to solve for lambda, but I just don't understand how you were able to get n and p from reading the problem... I get confused with that ; - ;

Answer 16

It is necessary to find the average number of orders when 250 potential customers are reached. It is given that an average of 5 orders are obtained when 1000 customers are reached. Therefore when 1 customer is reached an average of 5/1000 orders are obtained, and when 250 customers are reached an average of
\[\large \frac{5}{1000} \times250=1.25\ orders\ are\ obtained\]
Therefore the value of lambda for the question is 1.25.
Note that this question relates to a Poisson distribution. It appears from your posting that you are looking for values of n and p which would apply to a binomial distribution.