Poisson distribution (would you please verify?)

[elmarsur]

Homework Statement

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!

Homework Equations

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!

 

[Mark44]

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)]

 

[elmarsur]

Thank you very much, Mark!
Would you please extend your help?
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!

 

[Mark44]

It is e^(-3). I didn't carry the minus sign through from the previous step. (It's fixed, now.)

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.

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?

 

[elmarsur]

Thanks Mark!
Before I saw your reply, this is what I did:

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?

 

[Mark44]

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

 

[elmarsur]

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

 

[elmarsur]

Problems resolved.

 

[Mark44]

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

 

[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.

 

[Mark44]

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