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Poisson distribution Question

  • #1

Homework Statement



Phone calls are received at Diane residence have a Poisson distribution with [tex]\lambda[/tex] =2.
a) If Diane takes a shower for 10 min, what is the probability that the phone rings Once or Twice.
b) How long can she shower if the probability of receiving no calls be at most 0.5

Homework Equations



The Attempt at a Solution


a) For 10 min interval [tex]\lambda[/tex] =2x(10/60)=0.33333
Probability that phone rings once-> P(X=1)= [tex] e^{-0.33333}[/tex] = 0.7166

The answer given in the book is 0.28 which it not what I am getting. Where did I go wrong?

b) P(X=0)=0.5= [tex] e^{-\Lambda}[/tex]
If I take ln of both sided I get [tex] {\Lambda}[/tex] =0.69314
The book says required time=20.79 min.
How do I get this time from the value of lambda i obtained earlier?
 

Answers and Replies

  • #2
1,357
0
Probability that phone rings once-> P(X=1)= [tex] e^{-0.33333}[/tex] = 0.7166
That is wrong. Check the def. of the mass func. of the Poisson distro.

The answer given in the book is 0.28 which it not what I am getting. Where did I go wrong?
Hint: The problem asks for the probability of receiving a call once or twice in ten minutes.

b) P(X=0)=0.5= [tex] e^{-\Lambda}[/tex]
If I take ln of both sided I get [tex] {\Lambda}[/tex] =0.69314
The book says required time=20.79 min.
How do I get this time from the value of lambda i obtained earlier?
You have that 2 calls per hour equals 0.69313 calls per x hours because they're the same ratio. Can you find x? (BTW: Diane can shower as long as she wants. The question should be: How long can Diane shower without receiving a call given that the prob. of not receiving a call is at most 0.5?)
 
  • #3
I made a mistake here I ommited the multiplication by [tex]\lambda[/tex] hence the result I am getting is

Probability that phone rings once-> P(X=1)=0.3333*[tex] e^{-0.33333}[/tex]=0.2386 which is still different from 0.28 given in the book. How do I get to this value?
 
  • #5
1,357
0
Probability that phone rings once-> P(X=1)=0.3333*[tex] e^{-0.33333}[/tex]=0.2386 which is still different from 0.28 given in the book. How do I get to this value?
Let me try this again:

HINT: The problem asks for the probability of receiving a call once or twice in ten minutes.
 
  • #6
I managed to get the answer to the second part.an I got the time as 20.79.
But for the first part I the probability of 0.28 is the one for the phone to ring once as I said earlier.
The probability of the phone ringing ONCE or TWICE is not given.
Does this mean P(X<=2)?
 
  • #7
1,357
0
The probability of the phone ringing ONCE or TWICE is not given.
That's because you're supposed to compute it.

Does this mean P(X<=2)?
No. That's the prob. of receive 0, 1, or 2 calls. The problem is asking for 1 or 2 calls.
 

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