Let's consider Poisson distribution, let's calculate its mathematical expectation, dispersion, and mode. Using the MS EXCEL function POISSON.DIST(), we will construct graphs of the distribution function and probability density. Let us estimate the distribution parameter, its mathematical expectation and standard deviation.

Let's give it dry first formal definition distributions, then we give examples of situations when Poisson distribution(English) Poissondistribution) is an adequate model for describing a random variable.

If random events occur in a given period of time (or in a certain volume of matter) with an average frequency λ( lambda), then the number of events x, occurred during this period of time will have Poisson distribution.

Application of the Poisson distribution

Examples when Poisson distribution is an adequate model:

• number of calls received on telephone exchange for a certain period of time;
• the number of particles that have undergone radioactive decay over a certain period of time;
• number of defects in a piece of fabric of a fixed length.

Poisson distribution is an adequate model if the following conditions are met:

• events occur independently of each other, i.e. the probability of a subsequent event does not depend on the previous one;
• the average event rate is constant. As a consequence, the probability of an event is proportional to the length of the observation interval;
• two events cannot happen at the same time;
• the number of events must take the value 0; 1; 2…

Note: A good clue is that the observable random variable has Poisson distribution, is the fact that it is approximately equal (see below).

Below are examples of situations where Poisson distribution can't be applied:

• the number of students who leave the university within an hour (since the average flow of students is not constant: during classes there are few students, and during the break between classes the number of students increases sharply);
• the number of earthquakes with an amplitude of 5 points per year in California (since one earthquake can cause aftershocks of similar amplitude - the events are not independent);
• the number of days that patients spend in the intensive care unit (because the number of days that patients spend in the intensive care unit is always greater than 0).

Note: Poisson distribution is an approximation of more accurate discrete distributions: And .

Note: About the relationship Poisson distribution And Binomial distribution can be read in the article. About the relationship Poisson distribution And Exponential distribution can be read in the article about.

Poisson distribution in MS EXCEL

In MS EXCEL, starting from version 2010, for Distributions Poisson there is a function POISSON.DIST() , English name- POISSON.DIST(), which allows you to calculate not only the probability of what will happen over a given period of time X events (function probability density p(x), see formula above), but also (the probability that during a given period of time at least x events).

Before MS EXCEL 2010, EXCEL had the POISSON() function, which also allows you to calculate distribution function And probability density p(x). POISSON() is left in MS EXCEL 2010 for compatibility.

The example file contains graphs probability density distribution And cumulative distribution function.

Poisson distribution has a beveled shape ( long tail on the right side of the probability function), but as the parameter increases, λ becomes more and more symmetrical.

Note: Average And dispersion(square) are equal to the parameter Poisson distribution– λ (see example sheet file Example).

Task

Typical Application Poisson distributions in quality control is a model of the number of defects that may appear in an instrument or device.

For example, with an average number of defects in a chip λ (lambda) equal to 4, the probability that a randomly selected chip will have 2 or fewer defects is: = POISSON.DIST(2,4,TRUE)=0.2381

The third parameter in the function is set = TRUE, so the function will return integral function distribution, that is, the probability that the number random events will be in the range from 0 to 4 inclusive.

Calculations in this case are made according to the formula:

The probability that a randomly selected microcircuit will have exactly 2 defects is: = POISSON.DIST(2,4,FALSE)=0.1465

The third parameter in the function is set = FALSE, so the function will return the probability density.

The probability that a randomly selected microcircuit will have more than 2 defects is equal to: =1-POISSON.DIST(2,4,TRUE) =0.8535

Note: If x is not an integer, then when calculating the formula . Formulas =POISSON.DIST( 2 ; 4; LIE) And =POISSON.DIST( 2,9 ; 4; LIE) will return the same result.

Random number generation and λ estimation

For values ​​of λ >15 , Poisson distribution well approximated Normal distribution with the following parameters: μ =λ , σ 2 =λ .

More details about the relationship between these distributions can be found in the article. There are also examples of approximation, and the conditions for when it is possible and with what accuracy are explained.

ADVICE: You can read about other MS EXCEL distributions in the article.

Brief theory

Let independent tests be carried out, in each of which the probability of an event occurring is equal to . To determine the probability of an event occurring in these tests, Bernoulli's formula is used. If it is large, then use or. However, this formula is not suitable if it is small. In these cases (great, small) they resort to asymptotic Poisson's formula.

Let us set ourselves the task of finding the probability that, for very large number tests, in each of which the probability of the event is very small, the event will occur exactly once. Let us make an important assumption: the work retains a constant value, namely . This means that the average number of occurrences of an event in different series of trials, i.e. at different meanings, remains unchanged.

Example of problem solution

Problem 1

The base received 10,000 electric lamps. The probability that the lamp will break during travel is 0.0003. Find the probability that among the lamps received, five lamps will be broken.

Solution

Condition for the applicability of the Poisson formula:

If the probability of an event occurring in an individual trial is sufficiently close to zero, then even for large values ​​of the number of trials, the probability calculated by local theorem Laplace turns out to be insufficiently accurate. In such cases, use the formula derived by Poisson.

Let the event - 5 lamps be broken

Let's use Poisson's formula:

In our case:

Answer

Problem 2

The enterprise has 1000 pieces of equipment certain type. The probability of a piece of equipment failing within an hour is 0.001. Draw up a distribution law for the number of equipment failures per hour. Find numerical characteristics.

Solution

Random variable - the number of equipment failures, can take values

Let's use Poisson's law:

Let's find these probabilities:

The mathematical expectation and variance of a random variable distributed according to Poisson’s law is equal to the parameter of this distribution:

Average solution cost test work 700 - 1200 rubles (but not less than 300 rubles for the entire order). The price is greatly influenced by the urgency of the decision (from a day to several hours). The cost of online help for an exam/test is from 1000 rubles. for solving the ticket.

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Most general case various kinds probability distributions is a binomial distribution. Let us use its versatility to determine the most common particular types of distributions encountered in practice.

Binomial distribution

Let there be some event A. The probability of occurrence of event A is equal to p, the probability of non-occurrence of event A is 1 p, sometimes it is designated as q. Let n number of tests, m frequency of occurrence of event A in these n tests.

It is known that total probability everyone possible combinations outcomes is equal to one, that is:

1 = p n + n · p n 1 (1 p) + C n n 2 · p n 2 (1 p) 2 + + C n m · p m· (1 p) n – m+ + (1 p) n .

Expectation M binomial distribution is equal to:

M = n · p ,

Where n number of tests, p probability of occurrence of event A.

Standard deviation σ :

σ = sqrt( n · p· (1 p)) .

Example 1. Calculate the probability that an event that has a probability p= 0.5, in n= 10 trials will happen m= 1 time. We have: C 10 1 = 10, and further: P 1 = 10 0.5 1 (1 0.5) 10 1 = 10 0.5 10 = 0.0098. As we can see, the probability of this event occurring is quite low. This is explained, firstly, by the fact that it is absolutely not clear whether the event will happen or not, since the probability is 0.5 and the chances here are “50 to 50”; and secondly, it is required to calculate that the event will occur exactly once (no more and no less) out of ten.

Example 2. Calculate the probability that an event that has a probability p= 0.5, in n= 10 trials will happen m= 2 times. We have: C 10 2 = 45, and further: P 2 = 45 0.5 2 (1 0.5) 10 2 = 45 0.5 10 = 0.044. The likelihood of this event occurring has increased!

Example 3. Let's increase the likelihood of the event itself occurring. Let's make it more likely. Calculate the probability that an event that has a probability p= 0.8, in n= 10 trials will happen m= 1 time. We have: C 10 1 = 10, and further: P 1 = 10 0.8 1 (1 0.8) 10 1 = 10 0.8 1 0.2 9 = 0.000004. The probability has become less than in the first example! The answer, at first glance, seems strange, but since the event has a fairly high probability, it is unlikely to happen only once. It is more likely that it will happen more than once. Indeed, counting P 0 , P 1 , P 2 , P 3, , P 10 (probability that an event in n= 10 trials will happen 0, 1, 2, 3, , 10 times), we will see:

C 10 0 = 1 , C 10 1 = 10 , C 10 2 = 45 , C 10 3 = 120 , C 10 4 = 210 , C 10 5 = 252 ,
C 10 6 = 210 , C 10 7 = 120 , C 10 8 = 45 , C 10 9 = 10 , C 10 10 = 1 ;

P 0 = 1 0.8 0 (1 0.8) 10 0 = 1 1 0.2 10 = 0.0000…;
P 1 = 10 0.8 1 (1 0.8) 10 1 = 10 0.8 1 0.2 9 = 0.0000…;
P 2 = 45 0.8 2 (1 0.8) 10 2 = 45 0.8 2 0.2 8 = 0.0000…;
P 3 = 120 0.8 3 (1 0.8) 10 3 = 120 0.8 3 0.2 7 = 0.0008…;
P 4 = 210 0.8 4 (1 0.8) 10 4 = 210 0.8 4 0.2 6 = 0.0055…;
P 5 = 252 0.8 5 (1 0.8) 10 5 = 252 0.8 5 0.2 5 = 0.0264…;
P 6 = 210 0.8 6 (1 0.8) 10 6 = 210 0.8 6 0.2 4 = 0.0881…;
P 7 = 120 0.8 7 (1 0.8) 10 7 = 120 0.8 7 0.2 3 = 0.2013…;
P 8 = 45 0.8 8 (1 0.8) 10 8 = 45 0.8 8 0.2 2 = 0.3020…(highest probability!);
P 9 = 10 0.8 9 (1 0.8) 10 9 = 10 0.8 9 0.2 1 = 0.2684…;
P 10 = 1 0.8 10 (1 0.8) 10 10 = 1 0.8 10 0.2 0 = 0.1074…

Of course P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 + P 9 + P 10 = 1 .

Normal distribution

If we depict the quantities P 0 , P 1 , P 2 , P 3, , P 10, which we calculated in example 3, on the graph, it turns out that their distribution has a form close to the normal distribution law (see Fig. 27.1) (see lecture 25. Modeling of normally distributed random variables).

The binomial law becomes normal if the probabilities of occurrence and non-occurrence of event A are approximately the same, that is, we can conditionally write: p≈ (1 p) . For example, let's take n= 10 and p= 0.5 (that is p= 1 p = 0.5 ).

In essence, we will come to such a problem if, for example, we want to theoretically calculate how many boys and how many girls there will be out of 10 children born in a maternity hospital on the same day. More precisely, we will count not boys and girls, but the probability that only boys will be born, that 1 boy and 9 girls will be born, that 2 boys and 8 girls will be born, and so on. Let us assume for simplicity that the probability of having a boy and a girl is the same and equal to 0.5 (but in fact, to be honest, this is not the case, see the course “Modeling Artificial Intelligence Systems”).

It is clear that the distribution will be symmetrical, since the probability of having 3 boys and 7 girls is equal to the probability of having 7 boys and 3 girls. The greatest likelihood of birth will be 5 boys and 5 girls. This probability is 0.25, by the way, it’s not that big absolute value. Further, the probability that 10 or 9 boys will be born at once is much less than the probability that 5 ± 1 boy will be born out of 10 children. The binomial distribution will help us make this calculation. So.

C 10 0 = 1 , C 10 1 = 10 , C 10 2 = 45 , C 10 3 = 120 , C 10 4 = 210 , C 10 5 = 252 ,
C 10 6 = 210 , C 10 7 = 120 , C 10 8 = 45 , C 10 9 = 10 , C 10 10 = 1 ;

P 0 = 1 0.5 0 (1 0.5) 10 0 = 1 1 0.5 10 = 0.000977…;
P 1 = 10 0.5 1 (1 0.5) 10 1 = 10 0.5 10 = 0.009766…;
P 2 = 45 0.5 2 (1 0.5) 10 2 = 45 0.5 10 = 0.043945…;
P 3 = 120 0.5 3 (1 0.5) 10 3 = 120 0.5 10 = 0.117188…;
P 4 = 210 0.5 4 (1 0.5) 10 4 = 210 0.5 10 = 0.205078…;
P 5 = 252 0.5 5 (1 0.5) 10 5 = 252 0.5 10 = 0.246094…;
P 6 = 210 0.5 6 (1 0.5) 10 6 = 210 0.5 10 = 0.205078…;
P 7 = 120 0.5 7 (1 0.5) 10 7 = 120 0.5 10 = 0.117188…;
P 8 = 45 0.5 8 (1 0.5) 10 8 = 45 0.5 10 = 0.043945…;
P 9 = 10 0.5 9 (1 0.5) 10 9 = 10 0.5 10 = 0.009766…;
P 10 = 1 0.5 10 (1 0.5) 10 10 = 1 0.5 10 = 0.000977…

Of course P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 + P 9 + P 10 = 1 .

Let us display the quantities on the graph P 0 , P 1 , P 2 , P 3, , P 10 (see Fig. 27.2).

So, under the conditions m ≈ n/2 and p≈ 1 p or p≈ 0.5 instead of the binomial distribution, you can use the normal one. For large values n the graph shifts to the right and becomes more and more flat, as the mathematical expectation and variance increase with increasing n : M = n · p , D = n · p· (1 p) .

By the way, binomial law tends to normal and with increasing n, which is quite natural, according to the central limit theorem(see lecture 34. Recording and processing statistical results).

Now consider how the binomial law changes in the case when p ≠ q, that is p> 0 . In this case, the hypothesis of normal distribution cannot be applied, and the binomial distribution becomes a Poisson distribution.

Poisson distribution

The Poisson distribution is special case binomial distribution (with n>> 0 and at p>0 (rare events)).

A formula is known from mathematics that allows you to approximately calculate the value of any member of the binomial distribution:

Where a = n · p Poisson parameter (mathematical expectation), and the variance is equal to the mathematical expectation. Let us present mathematical calculations that explain this transition. Binomial distribution law

P m = C n m · p m· (1 p) n – m

can be written if you put p = a/n , in the form

Because p is very small, then only the numbers should be taken into account m, small compared to n. Work

very close to unity. The same applies to the size

Magnitude

very close to e – a. From here we get the formula:

Example. The box contains n= 100 parts, both high-quality and defective. The probability of receiving a defective product is p= 0.01 . Let's say that we take out a product, determine whether it is defective or not, and put it back. By doing this, it turned out that out of 100 products that we went through, two turned out to be defective. What is the likelihood of this?

From the binomial distribution we get:

From the Poisson distribution we get:

As you can see, the values ​​turned out to be close, so in the case of rare events it is quite acceptable to apply Poisson’s law, especially since it requires less computational effort.

Let us show graphically the form of Poisson's law. Let's take the parameters as an example p = 0.05 , n= 10 . Then:

C 10 0 = 1 , C 10 1 = 10 , C 10 2 = 45 , C 10 3 = 120 , C 10 4 = 210 , C 10 5 = 252 ,
C 10 6 = 210 , C 10 7 = 120 , C 10 8 = 45 , C 10 9 = 10 , C 10 10 = 1 ;

P 0 = 1 0.05 0 (1 0.05) 10 0 = 1 1 0.95 10 = 0.5987…;
P 1 = 10 0.05 1 (1 0.05) 10 1 = 10 0.05 1 0.95 9 = 0.3151…;
P 2 = 45 0.05 2 (1 0.05) 10 2 = 45 0.05 2 0.95 8 = 0.0746…;
P 3 = 120 0.05 3 (1 0.05) 10 3 = 120 0.05 3 0.95 7 = 0.0105…;
P 4 = 210 0.05 4 (1 0.05) 10 4 = 210 0.05 4 0.95 6 = 0.00096…;
P 5 = 252 0.05 5 (1 0.05) 10 5 = 252 0.05 5 0.95 5 = 0.00006…;
P 6 = 210 0.05 6 (1 0.05) 10 6 = 210 0.05 6 0.95 4 = 0.0000…;
P 7 = 120 0.05 7 (1 0.05) 10 7 = 120 0.05 7 0.95 3 = 0.0000…;
P 8 = 45 0.05 8 (1 0.05) 10 8 = 45 0.05 8 0.95 2 = 0.0000…;
P 9 = 10 0.05 9 (1 0.05) 10 9 = 10 0.05 9 0.95 1 = 0.0000…;
P 10 = 1 0.05 10 (1 0.05) 10 10 = 1 0.05 10 0.95 0 = 0.0000…

Of course P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 + P 9 + P 10 = 1 .

At n> ∞ the Poisson distribution becomes normal law, according to the central limit theorem (see.

Poisson distribution.

Let's consider the most typical situation, in which the Poisson distribution appears. Let the event A appears a certain number of times in a fixed area of ​​space (interval, area, volume) or a period of time with constant intensity. To be specific, consider the sequential occurrence of events over time, called a stream of events. Graphically, the flow of events can be illustrated by many points located on the time axis.

This could be a flow of calls in the service sector (repair of household appliances, calling an ambulance, etc.), a flow of calls to a telephone exchange, failure of some parts of the system, radioactive decay, pieces of fabric or metal sheets and the number of defects on each of them, etc. The Poisson distribution is most useful in those problems where it is necessary to determine only the number of positive outcomes (“successes”).

Imagine a raisin bun, divided into small pieces equal size. Due to random distribution raisins cannot be expected that all pieces will contain them same number. When the average number of raisins contained in these pieces is known, then the Poisson distribution gives the probability that any given piece contains X=k(k= 0,1,2,...,)number of raisins.

In other words, the Poisson distribution determines which part of a long series of pieces will contain equal to 0, or 1, or 2, or etc. number of highlights.

Let's make the following assumptions.

1. The probability of the occurrence of a certain number of events in a given time interval depends only on the length of this interval, and not on its position on the time axis. This is the property of stationarity.

2. The occurrence of more than one event in a sufficiently short period of time is practically impossible, i.e. conditional probability occurrence of another event in the same interval tends to zero at ® 0. This is the property of ordinaryness.

3. Probability of occurrence given number events in a fixed period of time does not depend on the number of events appearing in other periods of time. This is the property of lack of aftereffect.

A flow of events that satisfies the above propositions is called the simplest.

Let's consider a fairly short period of time. Based on property 2, the event may appear once in this interval or not appear at all. Let us denote the probability of an event occurring by r, and non-appearance – through q = 1-p. Probability r is constant (property 3) and depends only on the value (property 1). The mathematical expectation of the number of occurrences of an event in the interval will be equal to 0× q+ 1× p = p. Then the average number of occurrences of events per unit time is called the flow intensity and is denoted by a, those. a = .

Let's consider final segment time t and divide it by n parts = . The occurrences of events in each of these intervals are independent (property 2). Let us determine the probability that in a period of time t at constant flow intensity A the event will appear exactly X = k won't appear again n–k. Since an event can in each of n gaps appear no more than 1 time, then for its appearance k once in a segment of duration t it should appear in any k intervals from the total n. There are total such combinations, and the probability of each is equal. Consequently, by the addition theorem of probabilities we obtain for the desired probability well-known formula Bernoulli

This equality is written as an approximate one, since the initial premise for its derivation was property 2, which is fulfilled more accurately the smaller . To obtain exact equality, let us pass to the limit at ® 0 or, what is the same, n® . We will get it after replacement

P = a= and q = 1 – .

Let's introduce new parameter = at, meaning the average number of occurrences of an event in a segment t. After simple transformations and passing to the limit in the factors, we obtain.

= 1, = ,

Finally we get

, k = 0, 1, 2, ...

e = 2.718... – base natural logarithm.

Definition. Random variable X, which only accepts integers, positive values 0, 1, 2, ... has a Poisson distribution with parameter if

For k = 0, 1, 2, ...

The Poisson distribution has been proposed French mathematician S.D. Poisson (1781-1840). It is used to solve problems of calculating probabilities of relatively rare, mutually random independent events per unit of time, length, area and volume.

For the case when a) is large and b) k= , the Stirling formula is valid:

To calculate subsequent values, a recurrent formula is used

P(k + 1) = P(k).

Example 1. What is the probability that out of 1000 people on a given day: a) none, b) one, c) two, d) three people were born?

Solution. Because p= 1/365, then q= 1 – 1/365 = 364/365 "1.

Then

A) ,

b) ,

V) ,

G) .

Therefore, if there are samples of 1000 people, then the average number of people who were born on a particular day will accordingly be 65; 178; 244; 223.

Example 2. Determine the value at which with probability R the event appeared at least once.

Solution. Event A= (appear at least once) and = (not appear even once). Hence .

From here And .

For example, for R= 0.5, for R= 0,95 .

Example 3. On looms operated by one weaver, 90 thread breaks occur within an hour. Find the probability that at least one thread break will occur in 4 minutes.

Solution. By condition t = 4 min. and the average number of breaks per minute, from where . The required probability is .

Properties. The mathematical expectation and variance of a random variable having a Poisson distribution with parameter are equal to:

M(X) = D(X) = .

These expressions are obtained by direct calculations:

This is where the replacement was made n = k– 1 and the fact that .

By performing transformations similar to those used in the output M(X), we get

The Poisson distribution is used to approximate the binomial distribution at large n

In many practically important applications, the Poisson distribution plays an important role. Many of the numbers discrete quantities are implementations of a Poisson process with the following properties:

• We are interested in how many times a certain event occurs in a given range of possible outcomes random experiment. The area of ​​possible outcomes can be a time interval, a segment, a surface, etc.
• The probability of a given event is the same for all areas of possible outcomes.
• The number of events occurring in one area of ​​possible outcomes is independent of the number of events occurring in other areas.
• The probability that in the same area of ​​possible outcomes this event occurs more than once, tends to zero as the range of possible outcomes decreases.

To further understand the meaning of the Poisson process, suppose we examine the number of customers visiting a bank branch located in a central business district, during lunch, i.e. from 12 to 13 o'clock. Suppose you want to determine the number of clients arriving in one minute. Does this situation have the features listed above? First, the event that interests us is the arrival of a client, and the range of possible outcomes is a one-minute interval. How many clients will come to the bank in a minute - none, one, two or more? Secondly, it is reasonable to assume that the probability of a customer arriving within a minute is the same for all one-minute intervals. Third, the arrival of one customer during any one-minute interval is independent of the arrival of any other customer during any other one-minute interval. And finally, the probability that more than one client will come to the bank tends to zero if the time interval tends to zero, for example, becomes less than 0.1 s. So, the number of customers coming to the bank during lunch within one minute is described by the Poisson distribution.

The Poisson distribution has one parameter, denoted by λ ( greek letter“lambda”) is the average number of successful trials in a given area of ​​possible outcomes. The variance of the Poisson distribution is also λ, and its standard deviation is . Number of successful trials X Poisson random variable varies from 0 to infinity. The Poisson distribution is described by the formula:

Where P(X)- probability X successful trials, λ - expected number of successes, e- natural logarithm base equal to 2.71828, X- number of successes per unit of time.

Let's return to our example. Let's say that during the lunch break, on average, three customers come to the bank per minute. What is the probability that two customers will come to the bank at a given moment? What is the probability that more than two clients will come to the bank?

Let us apply formula (1) with the parameter λ = 3. Then the probability that two clients will come to the bank within a given minute is equal to

The probability that more than two clients will come to the bank is equal to P(X > 2) = P(X = 3) + P(X = 4) + … + P(X = ∞). Since the sum of all probabilities must be equal to 1, the terms of the series on the right side of the formula represent the probability of addition to the event X ≤ 2. In other words, the sum of this series is equal to 1 – P(X ≤ 2). Thus, P(X>2) = 1 – P(X≤2) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)]. Now, using formula (1), we get:

Thus, the probability that no more than two clients will come to the bank within a minute is 0.423 (or 42.3%), and the probability that more than two clients will come to the bank within a minute is 0.577 (or 57.7 %).

Such calculations may seem tedious, especially if the parameter λ is large enough. To avoid complex calculations, many Poisson probabilities can be found in special tables (Fig. 1). For example, the probability that two clients will come to the bank at a given minute, if on average three clients come to the bank per minute, is at the intersection of the line X= 2 and column λ = 3. Thus, it is equal to 0.2240 or 22.4%.

Rice. 1. Poisson probability at λ = 3

Nowadays, it is unlikely that anyone will use tables if Excel with its =POISSON.DIST() function is at hand (Fig. 2). This function has three parameters: number of successful trials X, average expected number of successful trials λ, parameter Integral, taking two values: FALSE – in this case the probability of the number of successful trials is calculated X(X only), TRUE – in this case the probability of the number of successful trials from 0 to X.

Rice. 2. Calculation in Excel probabilities Poisson distribution at λ = 3

Approximation of the binomial distribution using the Poisson distribution

If the number n is large and the number r- small, the binomial distribution can be approximated using the Poisson distribution. How larger number n And less number r, the higher the approximation accuracy. The following Poisson model is used to approximate the binomial distribution.

Where P(X)- probability X success with given parameters n And r, n- sample size, r- true probability of success, e- the base of the natural logarithm, X- number of successes in the sample (X = 0, 1, 2, …, n).

Theoretically, a random variable with a Poisson distribution takes values ​​from 0 to ∞. However, in situations where the Poisson distribution is used to approximate the binomial distribution, the Poisson random variable is the number of successes among n observations - cannot exceed the number n. From formula (2) it follows that with increasing number n and a decrease in the number r probability of detection large number success rate decreases and tends to zero.

As mentioned above, the expectation µ and the variance σ 2 of the Poisson distribution are equal to λ. Therefore, when approximating the binomial distribution using the Poisson distribution, formula (3) should be used to approximate the mathematical expectation.

(3) µ = E(X) = λ =n.p.

To approximate the standard deviation, formula (4) is used.

Please note that the standard deviation calculated using formula (4) tends to standard deviation in the binomial model – when the probability of success p tends to zero, and, accordingly, the probability of failure 1 – p tends to unity.

Let's assume that 8% of the tires produced at a certain plant are defective. To illustrate the use of the Poisson distribution to approximate the binomial distribution, let's calculate the probability of finding one defective tire in a sample of 20 tires. Let us apply formula (2), we obtain

If we were to calculate the true binomial distribution rather than its approximation, we would get the following result:

However, these calculations are quite tedious. However, if you use Excel to calculate probabilities, then using the Poisson distribution approximation becomes redundant. In Fig. Figure 3 shows that the complexity of calculations in Excel is the same. However, this section, in my opinion, is useful to understand that under some conditions the binomial distribution and the Poisson distribution give similar results.

Rice. 3. Comparison of the complexity of calculations in Excel: (a) Poisson distribution; (b) binomial distribution

So, in this and two previous notes three discrete numerical distributions: , and Poisson. To better understand how these distributions relate to each other, we present a small tree of questions (Fig. 4).

Rice. 4. Classification of discrete probability distributions

Materials from the book Levin et al. Statistics for Managers are used. – M.: Williams, 2004. – p. 320–328