Calculating Probabilities for Poisson Distribution

You have to compute P(X=1)+P(X=2).In summary, the problem involves a Poisson distribution with a lambda of 2 for phone calls received at Diane's residence. For a 10 minute interval, the probability of receiving a call once is 0.2386, and the probability of receiving a call twice is 0.0498. To find the probability of receiving a call once or twice, you must add these two probabilities together. For the second part of the problem, the required time for Diane to shower without receiving a call is 20.79 minutes. To find this, you must use the value of lambda obtained earlier and set the probability of receiving no calls to be no more than 0
  • #1
Bertrandkis
25
0

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?
 
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  • #2
Bertrandkis said:
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
Bertrandkis said:
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
Bertrandkis said:
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.
 

1. What is the Poisson distribution?

The Poisson distribution is a probability distribution that is used to model the number of occurrences of a specific event within a specific time or space interval.

2. What are the characteristics of the Poisson distribution?

The Poisson distribution is characterized by a single parameter, λ (lambda), which represents the average number of occurrences of the event within the given interval. It is a discrete distribution, meaning that the possible values are whole numbers and that the probabilities for each value are independent of each other. It is also skewed to the right, with a longer tail on the right side of the distribution.

3. How is the Poisson distribution different from other probability distributions?

The Poisson distribution is different from other probability distributions, such as the normal distribution, in that it is used to model the number of occurrences of a specific event, rather than a continuous variable. It also has only one parameter, λ, whereas other distributions may have multiple parameters.

4. What types of events can be modeled using the Poisson distribution?

The Poisson distribution is commonly used to model rare events, such as the number of customers arriving at a store, the number of car accidents on a certain road, or the number of defects in a batch of products. It can also be used to model the number of radioactive particles emitted from a substance or the number of mutations in a DNA sequence.

5. How can the Poisson distribution be applied in real-life situations?

The Poisson distribution can be applied in various real-life situations, such as predicting the number of customer complaints in a month, estimating the number of calls a call center will receive in a day, or determining the likelihood of a power outage in a given area. It can also be used in quality control processes to monitor the number of defects in a manufacturing process and identify areas for improvement.

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