Poisson Probability Distribution Problem

In summary: The problem is that you are not asking for the probability that X will exceed 13, but the probability that X will occur at all. To solve for the probability that X will exceed 13, you would use the binomial distribution: P(X>13) = P(X>=11+2X) = P(X>=12) = P(X>13) = 1-P(X≤13)To solve for the probability that X will occur at all, you would use the Poisson distribution: P(X>=0) = P(X>=1) = P(X>=2) = P(X>=
  • #1
Callix
106
0

Homework Statement


An article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is 0.4 year.

a). How many loads can be expected to occur during a 4-year period?

b). What is the probability that more than 13 loads occur during a 4-year period?

c). How long must a time period be so that the probability of no loads occurring during that period is at most 0.3?

Homework Equations

The Attempt at a Solution


I tried setting up the equation as a Poisson probability distribution for a). as (e^(-0.4)*0.4^(4)) / (4!) but I wasn't sure if this was correct. I couldn't move on to b or c without knowing a. If anyone could help give me some direction with good details that would be appreciated! I want to be able to learn the material and reasoning, not simply obtain the answer.

Also, I apologize if this is not in the right category. I didn't see any homework-related sub-forums for probability and stats.

Thanks in advance!
 
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  • #2
Is a). actually just 4/0.4?
 
  • #3
Callix said:

Homework Statement


An article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is 0.4 year.

a). How many loads can be expected to occur during a 4-year period?

b). What is the probability that more than 13 loads occur during a 4-year period?

c). How long must a time period be so that the probability of no loads occurring during that period is at most 0.3?

Homework Equations

The Attempt at a Solution


I tried setting up the equation as a Poisson probability distribution for a). as (e^(-0.4)*0.4^(4)) / (4!) but I wasn't sure if this was correct. I couldn't move on to b or c without knowing a. If anyone could help give me some direction with good details that would be appreciated! I want to be able to learn the material and reasoning, not simply obtain the answer.

Also, I apologize if this is not in the right category. I didn't see any homework-related sub-forums for probability and stats.

Thanks in advance!

Your ##\tau = 0.4## yr. is a time, not a rate. The parameter in the Poisson distribution is dimensionless:
[tex] \text{parameter} = \text{mean number} = \text{rate} \times \text{time}. [/tex]
In your problem, the rate is ##\lambda = 1/\tau = 1/0.4 = 2.5## events per year.

Problems in probability and/or statistics are usually posted here or in the "Calculus and Beyond" forum, depending on the level of the question and the mathematical tools needed to deal with it. Occasionally they appear in the Elementary or Advanced Physics forums, especially if they have something to do with experimental error analysis or statistical mechanics and the like.
 
  • #4
Ray Vickson said:
Your ##\tau = 0.4## yr. is a time, not a rate. The parameter in the Poisson distribution is dimensionless:
[tex] \text{parameter} = \text{mean number} = \text{rate} \times \text{time}. [/tex]
In your problem, the rate is ##\lambda = 1/\tau = 1/0.4 = 2.5## events per year.

Problems in probability and/or statistics are usually posted here or in the "Calculus and Beyond" forum, depending on the level of the question and the mathematical tools needed to deal with it. Occasionally they appear in the Elementary or Advanced Physics forums, especially if they have something to do with experimental error analysis or statistical mechanics and the like.

Ah, that makes sense, so in this case what is x? Or is x the value that I am solving for?
 
  • #5
So far my equation is [itex]\frac{e^{-10}10^x}{x!}[/itex]
 
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  • #6
Oh, so ##\mu## (aka ##2.5 \times 4##) is the expected value, so that should be the value that I'm looking for for a)?
 
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  • #7
Callix said:
Ah, that makes sense, so in this case what is x? Or is x the value that I am solving for?

There was no letter "x" in the above; there was the multiplication sign ##\times##. Sorry if that confused you.
 
  • #8
Ray Vickson said:
There was no letter "x" in the above; there was the multiplication sign ##\times##. Sorry if that confused you.

I was referring to the equation that I made in the following post. But I realized that the value that I'm looking for is ##\mu## anyway, which if I'm understanding correctly, should be 10.

I'm confused about how to go about b). It gave me a table of CDF values listed for ##\mu## and x, so and it wants P(X>13). My logic is that that would simply sum to infinity with infinite terms, so I did 1-P(X≤13). Since the table is CDF, that would mean I simple take 1-P(13) wouldn't it?
 
  • #9
Callix said:
I was referring to the equation that I made in the following post. But I realized that the value that I'm looking for is ##\mu## anyway, which if I'm understanding correctly, should be 10.

I'm confused about how to go about b). It gave me a table of CDF values listed for ##\mu## and x, so and it wants P(X>13). My logic is that that would simply sum to infinity with infinite terms, so I did 1-P(X≤13). Since the table is CDF, that would mean I simple take 1-P(13) wouldn't it?

Yes, that would be the way to do it.
 
  • #10
Ray Vickson said:
Yes, that would be the way to do it.

Alright. And for C, do I just solve for ##\mu## from the Poisson equation? And then divide by the rate?
 
  • #11
Hmm, or could I solve C using the CDF tables?
 
  • #12
Callix said:
Alright. And for C, do I just solve for ##\mu##? And then divide by the rate?

Solve for ##\mu## how? What would be the equation you want to solve?
 
  • #13
Well it tells me that the probability would be 0.3 so I was thinking I could set it up as
[itex]\frac{e^{-\mu}\mu^x}{x!}=0.3[/itex]
But then again I don't have a specified x...
 
  • #14
Oh wait x would just be 0 because it's referring to no loads
 
  • #15
Callix said:
Oh wait x would just be 0 because it's referring to no loads
Yes, exactly.
 
  • #16
I know this is a necropost, but just curious as to how the constant ##\lambda## is determined in real life.
 
  • #17
WWGD said:
I know this is a necropost, but just curious as to how the constant ##\lambda## is determined in real life.
It's the expected number of events in whatever the time interval happens to be.
 
  • #18
Mark44 said:
It's the expected number of events in whatever the time interval happens to be.
And you use long-term data to compute it, e.g., the number of patients arriving at the emergency room?
 
  • #19
WWGD said:
And you use long-term data to compute it, e.g., the number of patients arriving at the emergency room?
I would think so -- the average number of patients arriving per day or month, or whatever.
 
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  • #20
Mark44 said:
I would think so -- the average number of patients arriving per day or month, or whatever.
Yes, thanks, I meant the average.
 

FAQ: Poisson Probability Distribution Problem

What is the Poisson Probability Distribution?

The Poisson Probability Distribution is a statistical distribution used to model the probability of a certain number of events occurring within a specified time period, given a known average rate of occurrence.

How is the Poisson Probability Distribution calculated?

The Poisson Probability Distribution is calculated using the formula P(x; μ) = (e^-μ) * (μ^x) / x!, where x is the number of events, μ is the average rate of occurrence, and e is the mathematical constant approximately equal to 2.718.

What types of problems can be solved using the Poisson Probability Distribution?

The Poisson Probability Distribution can be used to solve problems involving the probability of a certain number of events occurring, such as the number of customers arriving at a store in a given time period, the number of accidents on a highway in a day, or the number of goals scored in a soccer match.

What are the assumptions made when using the Poisson Probability Distribution?

The Poisson Probability Distribution assumes that the events occur independently of each other, that the average rate of occurrence remains constant over the specified time period, and that the probability of an event occurring is proportional to the length of the time period.

How is the Poisson Probability Distribution related to the normal distribution?

The Poisson Probability Distribution is a discrete distribution, meaning it is used for counting events that can only take whole number values. The normal distribution, on the other hand, is a continuous distribution used for measuring a range of values. However, as the number of events (x) increases, the Poisson Distribution approaches the shape of the normal distribution with a mean of μ and a variance of μ.

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