# Homework Help: Poisson distribution (would you please verify?)

1. Nov 18, 2009

### elmarsur

1. The problem statement, all variables and given/known data

1. Suppose that the number of telephone calls an operator receives from 9:00 to 9:05 A.M. follows a Poisson distribution with mean 3. Find the probability that the operator will receive:
a. no calls in that interval tomorrow.
b. three or more calls in that interval the day after tomorrow.

2. Find the number of chocolate chips a cookie should contain
on the average if it is desired that the probability of a cookie
containing at least one chocolate chip be .99.

Thank you very much for any help!

2. Relevant equations

3. The attempt at a solution
Problem (1):
Noting probability of Poisson distribution with p, the number of calls with x, and the mean with m, we have
a) p=[(m^x)*(e^-m)]/x! = e^-m = e^-3
b) x=3 => p=1-p(x=2) = 1- [(m^2)*(e^-m)]/2! = 1- [(3^2)*(e^-3)]/2

Problem (2):
Preserving the notation above, p=.99 , x=1 =>
.99 = [(m^x)*(e^-m)]/x! = m*(e^-m)
If so, do I take ln of both sides to find m?

Thank you very much for any help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 18, 2009

### Staff: Mentor

1a looks fine, but your notation could use some work. Using m = 3, P(X = 0) = m0e-m/0! = e-3
For 1b, it asks for the probability that X is >= 3, not just P(X = 3). P(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

Last edited: Nov 18, 2009
3. Nov 18, 2009

### elmarsur

Thank you very much, Mark!
1. I don't understand how I would arrive at e^3 instead of e^-3
2. Would you have any suggestions for Problem 2 (the cookie)?

Thank you to anyone who would help!

4. Nov 18, 2009

### Staff: Mentor

It is e^(-3). I didn't carry the minus sign through from the previous step. (It's fixed, now.)
P(X >= 1) = .99, so P(X = 0) = 1 - P(X >= 1) = 1 - .99 = .01
But P(X = 0) = .01
==> m0e-m/0! = .01
Can you solve this for m?

5. Nov 18, 2009

### elmarsur

Thanks Mark!

p = .99 + p(x=0) = .99 + [(m^0)*(e^-m)]/0! = .99 + e^-m = 1 (1 is total probability = certainty)

If so, then m = -ln(0.01)

How wrong am I?

6. Nov 18, 2009

### Staff: Mentor

You're not wrong - that's correct. According to this problem, a cookie should have between 4 and 5 chocolate chips.

7. Nov 18, 2009

### elmarsur

Thank you very much, Mark, for all your help!

8. Nov 18, 2009

### elmarsur

Problems resolved.

9. Nov 18, 2009

### Staff: Mentor

I hope you take heed of the guidance I was providing on writing notation so that it makes sense.

10. Nov 18, 2009

### elmarsur

Thank you again, Mark!
Good thing you refreshed the point; I had almost forgotten.
Of course, I see that it wouldn't make sense without specifying the P(x=0) how I arrive at the end form.

I hope you are around if I need help with other things in the future.

Thanks again, and best to you.

11. Nov 18, 2009

### Staff: Mentor

You're welcome. I'm around here quite a bit, and when I'm not, there are lots of other very capable people who enjoy helping out.
Mark