# Statistics: Poisson Distribution

• whitehorsey
In summary: The relationship between Poisson and "arrival" processes is that if you model an arrival process as a collection of independent binary events, with success probability ##p## and you let the interval between events shrink as you take the limit ##p \to 0##, ##n \to \infty##, keeping ##\lambda = np## fixed, you get the same distribution as if you model it as a Poisson process with rate ##\lambda##.
whitehorsey
1. A Poisson random variable is such that it assumes the values 0 and 1 with equal probability. Find the value of the Poisson parameter, ρ ,for this variable.
2. Poisson equation: f(x) = e-λs(λs)/x!
3. I assumed the probability would be 0.5 because it can be either 0 or 1.
0.5 = e-λs(λs)/x! But now I'm wondering where in the equation would ρ be?

whitehorsey said:
1. A Poisson random variable is such that it assumes the values 0 and 1 with equal probability. Find the value of the Poisson parameter, ρ ,for this variable.

2. Poisson equation: f(x) = e-λs(λs)/x!

3. I assumed the probability would be 0.5 because it can be either 0 or 1.
0.5 = e-λs(λs)/x! But now I'm wondering where in the equation would ρ be?

No, no, no. A Poisson random variables has nonzero probabilities at 0, 1, 2, 3, ..., and all these probabilities together sum to 1. Therefore, it is impossible to have probability 1/2 at both 0 and 1; all you have is that P(0) = P(1), and these must both be < 1/2.

Anyway, your formula for the Poisson f(x) is incorrect.

Ray Vickson said:
No, no, no. A Poisson random variables has nonzero probabilities at 0, 1, 2, 3, ..., and all these probabilities together sum to 1. Therefore, it is impossible to have probability 1/2 at both 0 and 1; all you have is that P(0) = P(1), and these must both be < 1/2.

Anyway, your formula for the Poisson f(x) is incorrect.

Ah why do they both have to be < 1/2?

Oh I forgot the x! f(x) = (e-λs(λs)x)/(x!)

whitehorsey said:
Ah why do they both have to be < 1/2?
Could they add up to 1? What would that mean for the prob of 2?

whitehorsey said:
Ah why do they both have to be < 1/2?

Oh I forgot the x! f(x) = (e-λs(λs)x)/(x!)

You were worried about "where ρ would be". Well, what do you think ρ stands for? How is it related to λ and s? BTW: λ and s have nothing to do with the problem!

What's ##s## and what's ##\rho##? Never heard of these symbols used in the context of the Poisson distribution.

The probability ##P(k)## of observing a value ##k## in a Poisson process with parameter ##\lambda## is given by:

$$P(k) = \frac{\lambda^k e^{-\lambda}}{k!}$$

(##k## is the usual symbol, and it's equivalent to the way ##x## has been used in this thread by the thread starter).

It's as simple as determining ##P(0)## and ##P(1)##, both very simple expressions, then setting them equal to each other and solving for ##\lambda##. Even the solution of this equation is trivial, giving a very simple value for the parameter. Yes, you will find that the (equal) probabilities are less than half, but you don't even need to recognise that before you solve the equation.

Hmm my book has the formula that way. I think ρ =λs.
It says that:
λ = the average number of occurrences of the event per unit.
s = the length or size of the observation period.

So I got,
P(0) = P(1)
e-λsλs = e-λs
Substitute λs to ρ
ρ = 1?

Yes.

haruspex said:
Yes.

Thank you everyone!

Could you also help me on this problem?

An IT department uses the following probability distribution for the number X of computer system crashes occurring during a week:

Number of crashes, x Probability, p(x)
0 0.60
1 0.30
2 0.07
3 0.03

a. What is the probability that there would be at least 2 crashes in a given week?
b. Find the expected number of crashes in a week, E(X).
c. Find the variance of X, V(X).

This is what I got but I'm not sure if it's correct.

a. 1 - P[X=0] - P[X=1] = 0.1
b. Using the E(X) formula, I got 0.53.
c. V(x) formula = 0.5691

whitehorsey said:
Thank you everyone!

Could you also help me on this problem?

An IT department uses the following probability distribution for the number X of computer system crashes occurring during a week:

Number of crashes, x Probability, p(x)
0 0.60
1 0.30
2 0.07
3 0.03

a. What is the probability that there would be at least 2 crashes in a given week?
b. Find the expected number of crashes in a week, E(X).
c. Find the variance of X, V(X).

This is what I got but I'm not sure if it's correct.

a. 1 - P[X=0] - P[X=1] = 0.1
b. Using the E(X) formula, I got 0.53.
c. V(x) formula = 0.5691

All look right. For the first, you could also have added P(X=2) and P(X=3).

Curious3141 said:
All look right. For the first, you could also have added P(X=2) and P(X=3).

Thank You!

whitehorsey said:
Hmm my book has the formula that way. I think ρ =λs.
It says that:
λ = the average number of occurrences of the event per unit.
s = the length or size of the observation period.

So I got,
P(0) = P(1)
e-λsλs = e-λs
Substitute λs to ρ
ρ = 1?

In principle, a Poisson random variable need not have anything to do with counting "occurrences" over time; that is, you can have a ρ without having a λ and an s. A Poisson random variable with mean ρ has probability mass function $$\Pr(k) = \frac{\rho^k e^{-\rho}}{k!}, \: k = 0, 1, 2, \ldots$$
period. Whether or not ρ happens to be related to some "arrival" process has no bearing on the Poisson distribution itself. I would have hoped your textbook made that clear.

## 1. What is a Poisson distribution?

A Poisson distribution is a probability distribution used to model the number of times an event occurs in a given time period or space when the probability of the event happening is low and the events occur independently of each other. It is often used in statistics to analyze count data, such as the number of customers in a store or the number of accidents in a day.

## 2. How is a Poisson distribution different from a normal distribution?

A Poisson distribution is different from a normal distribution in several ways. Firstly, a Poisson distribution is discrete, meaning the possible outcomes are whole numbers, while a normal distribution is continuous, meaning the possible outcomes can be any real number. Additionally, the shape of a Poisson distribution is skewed to the right, while a normal distribution is symmetric. Lastly, the mean and variance of a Poisson distribution are equal, while in a normal distribution, the variance is always greater than the mean.

## 3. What is the mean and variance of a Poisson distribution?

The mean and variance of a Poisson distribution are both equal to the parameter lambda (λ), which represents the rate of occurrence of the event. This means that the mean and variance can be calculated using the formula: mean = variance = λ.

## 4. When should a Poisson distribution be used?

A Poisson distribution should be used when analyzing count data that meets the following criteria: (1) the number of events in a given time or space is independent of the number of events in any other time or space, (2) the probability of an event occurring is small, and (3) the events occur at a constant rate. Common examples of data that can be modeled using a Poisson distribution include the number of accidents in a day, the number of phone calls received in an hour, or the number of customers in a store at a given time.

## 5. How do you calculate the probability of a certain number of events occurring in a Poisson distribution?

The probability of a certain number of events (k) occurring in a Poisson distribution can be calculated using the formula: P(k;λ) = (e^-λ * λ^k) / k!, where λ is the rate of occurrence and k is the number of events. Alternatively, you can use a Poisson distribution table or a statistical software program to calculate the probability.

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