Local theorem of Moivre-Laplace(1730 Moivre and Laplace)

If the probability $p$ of occurrences of event $A$ is constant and $p\ne 0$ and $p\ne 1$, then the probability $P_n (k)$ is that event $A$ will appear $k$ times in $n $ tests, is approximately equal (the larger $n$, the more accurate) the value of the function $y=\frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \cdot \frac ( 1 ) ( \sqrt ( 2 \pi ) ) \cdot e^ ( - ( x^2 ) / 2 ) =\frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \cdot \varphi (x)$

for $x=\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) $. There are tables containing the values ​​of the function $\varphi (x)=\frac ( 1 ) ( \sqrt ( 2\cdot \pi ) ) \cdot e^ ( - ( x^2 ) / 2 ) $

so \begin(equation) \label ( eq2 ) P_n (k)\approx \frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \cdot \varphi (x)\,\,where\,x =\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) \qquad (2) \end(equation)

function $\varphi (x)=\varphi (( -x ))$ is even.

Example. Find the probability that event $A$ will occur exactly 80 times in 400 trials if the probability of this event occurring in each trial is $p=0.2$.

Solution. If $p=0.2$ then $q=1-p=1-0.2=0.8$.

$P_ ( 400 ) (( 80 ))\approx \frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \varphi (x)\,\,where\,x=\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) $

$ \begin(array) ( l ) x=\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) =\frac ( 80-400\cdot 0.2 ) ( \sqrt ( 400 \cdot 0.2\cdot 0.8 ) ) =\frac ( 80-80 ) ( \sqrt ( 400\cdot 0.16 ) ) =0 \\ \varphi (0)=0.3989\,\,P_ ( 400 ) (( 80 ))\approx \frac ( 0.3989 ) ( 20\cdot 0.4 ) =\frac ( 0.3989 ) ( 8 ) =0.0498 \\ \end(array) $

Moivre-Laplace integral theorem

The probability P of the occurrence of event $A$ in each trial is constant and $p\ne 0$ and $p\ne 1$, then the probability $P_n (( k_1 ,k_2 ))$ that the event $A$ will occur from $k_ ( 1 ) $ up to $k_ ( 2 ) $ times in $n$ trials, equals $ P_n (( k_1 ,k_2 ))\approx \frac ( 1 ) ( \sqrt ( 2\cdot \pi ) ) \int\limits_ ( x_1 ) ^ ( x_2 ) ( e^ ( - ( z^2 ) / 2 ) dz ) =\Phi (( x_2 ))-\Phi (( x_1 ))$

where $x_1 =\frac ( k_1 -n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) , x_2 =\frac ( k_2 -n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) $ ,where

$\Phi (x)=\frac ( 1 ) ( \sqrt ( 2\cdot \pi ) ) \int ( e^ ( - ( z^2 ) / 2 ) dz ) $ -found from tables

$\Phi (( -x ))=-\Phi (x)$-odd

Odd function. The values ​​in the table are given for $x=5$, for $x>5,\Phi (x)=0.5$

Example. It is known that 10% of products are rejected during inspection. 625 products were selected for control. What is the probability that among those selected there are at least 550 and at most 575 standard products?

Solution. If there are 10% defects, then there are 90% standard products. Then, by condition, $n=625, p=0.9, q=0.1, k_1 =550, k_2 =575$. $n\cdot p=625\cdot 0.9=562.5$. We get $ \begin(array) ( l ) P_ ( 625 ) (550.575)\approx \Phi (( \frac ( 575-562.5 ) ( \sqrt ( 625\cdot 0.9\cdot 0.1 ) ) ) )- \Phi (( \frac ( 550-562.5 ) ( \sqrt ( 626\cdot 0.9\cdot 0.1 ) )) \approx \Phi (1.67)- \Phi (-1, 67)=2 \Phi (1.67)=0.9052 \\ \end(array) $

Moivre-Laplace integral theorem . If the probability p of the occurrence of event A in each trial is constant and different from 0 and 1, then the probability that the number m of the occurrence of event A in n independent trials lies in the range from a to b (inclusive), with a sufficiently large number n is approximately equal to

Where
- Laplace function (or probability integral);

,
.

The formula is called the Moivre-Laplace integral formula. The larger n, the more accurate this formula is. If the condition npq ≥ 20 is met, the integral formula
, just like local, gives, as a rule, an error in calculating probabilities that is satisfactory for practice.

The function Ф(х) is tabulated (see table). To use this table you need to know function properties :

The function Ф(х) is odd, i.e. Ф(-х) = -Ф(х).

The function Ф(х) is monotonically increasing, and as x → +∞ Ф(х) → 1 (in practice we can assume that already for x > 4 Ф(х) ≈ 1).

Example . In some area, out of every 100 families, 80 have refrigerators. Calculate the probability that from 300 to 360 (inclusive) families out of 400 have refrigerators.

Solution. We apply the integral theorem of Moivre-Laplace (npq = 64 ≥ 20). First we define:

,

.

Now according to the formula
, taking into account the properties of Ф(х), we obtain

(according to table F(2.50) = 0.9876, F(5.0) ≈ 1)

1. Corollaries from the Moivre-Laplace integral theorem (with conclusion). Examples.

Let us consider a corollary of the integral theorem of Moivre-Laplace.

Consequence. If the probability p of the occurrence of event A in each trial is constant and different from 0 and 1, then with a sufficiently large number n of independent trials, the probability is that:

a) the number m of occurrences of event A differs from the product nр by no more than the value ε >
;

b) frequency event A is contained in the range from α to β (inclusive), i.e.
, Where
,
.

c) frequency event A differs from its probability p by no more than Δ > 0 (in absolute value), i.e.
.

□ 1) Inequality
is equivalent to the double inequality pr - E ~ m ~ pr + E. Therefore, according to the integral formula
:

.

2) Inequality
is equivalent to the inequality a ≤ m ≤ b for a = nα and b = nβ. Substituting in formulas
And
,
values ​​a and b using the obtained expressions, we obtain the formulas to be proved
And
,
.

3) Inequality
tantamount to inequality
. Substituting in the formula

, we obtain the formula to be proved
.

Example . According to statistics, on average 87% of newborns live to be 50 years old. Find the probability that out of 1000 newborns the proportion (frequency) of those surviving to 50 years of age will: a) be in the range from 0.9 to 0.95; b) will differ from the probability of this event by no more than 0.04 (in absolute value)?

Solution. a) The probability p that a newborn will live to be 50 years old is 0.87. Because n = 1000 is large (the condition npq = 1000·0.87·0.13 = 113.1 ≥ 20 is satisfied), then we use a corollary of the Moivre-Laplace integral theorem. First we define:

,
. Now according to the formula
:

B) According to the formula
:

Since inequality
tantamount to inequality
, the obtained result means that it is almost certain that from 0.83 to 0.91 of newborns out of 1000 will live to be 50 years old.

The concept of “random variable” and its description. Discrete random variable and its law (series) of distribution. Independent random variables. Examples.

Under random variable is understood as a variable that, as a result of testing, depending on the case, takes one of its possible set of values ​​(which one is not known in advance).

Examples of random variables : 1) the number of children born during the day in Moscow; 2) the number of defective products in a given batch; 3) the number of shots fired before the first hit; 4) flight range of an artillery shell; 5) electricity consumption at the plant per month.

The random variable is called discrete (discontinuous) , if the set of its values ​​is finite, or infinite, but countable.

Under continuous random variable We will understand a quantity whose infinite uncountable set of values ​​is a certain interval (finite or infinite) of the number axis.

So, in the above examples 1-3 we have discrete random variables (in examples 1 and 2 - with a finite set of values; in example 3 - with an infinite but countable set of values); and in examples 4 and 5 - continuous random variables.

For discrete random variable many possible values ​​of the random variable, i.e. functions
, finite or countable, for continuous- endless and uncountable.

Random variables are denoted by capital letters of the Latin alphabet X, Y, Z,..., and their values ​​are denoted by the corresponding lowercase letters x, y, z,....

A random variable is said to be “distributed” according to a given distribution law or “subject to” this distribution law.

For a discrete random variable distribution law m.b. given in the form of a table, analytically (in the form of a formula) and graphically.

The simplest form of specifying the distribution law of a discrete random variable X is a table (matrix), which lists in ascending order all possible values ​​of the random variable and their corresponding probabilities, i.e.

This table is called near the distribution of a discrete random variable .

Events X=x 1, X=x 2,...,X=x n, consisting in the fact that as a result of the test, the random variable X will take the values ​​x 1, x 2, ..., x n, respectively, are inconsistent and the only possible ones (because in the table lists all possible values ​​of the random variable), i.e. form a complete group. Therefore, the sum of their probabilities is equal to 1. Thus, for any discrete random variable
.

Distribution series m.b. is depicted graphically if the values ​​of a random variable are plotted along the abscissa axis, and their corresponding probabilities are plotted along the ordinate axis. The connection of the obtained points forms a broken line called polygon or probability distribution polygon .

Two random variables are called independent , if the distribution law of one of them does not change depending on what possible values ​​the other quantity takes. So, if a discrete random variable X can take values ​​x i (i = 1, 2, ..., n), and a random variable Y can take values ​​y j (j = 1, 2, ..., m), then the independence of discrete random quantities X and Y means the independence of the events X = x i and Y = y for any i = 1, 2, ... , n and j = 1, 2, ..., m. Otherwise, the random variables are called dependent .

For example , if there are tickets for two different money lotteries, then the random variables X and Y, respectively expressing the winnings for each ticket (in monetary units), will be independent, because for any win on a ticket of one lottery (for example, when X = x i), the law of distribution of winnings on another ticket (Y) will not change.

If the random variables X and Y express the winnings on tickets of one money lottery, then in this case X and Y are dependent, because any winning on one ticket (X = x i) leads to a change in the probabilities of winning on another ticket (Y), i.e. e. to a change in the distribution law of U.

Mathematical operations on discrete random objects personalities and examples of constructing distribution laws for KH, X" 1 , X + K, XV given distributions of independent cases numeric quantities X And U.

Let's define mathematical operations over discrete random variables.

Let two random variables be given:

The product kX of a random variable X by a constant value k is a random variable that takes values ​​kx i with the same probabilities p i (i = 1,2,...,n).

m th degree of the random variable X, i.e.
, is a random variable that takes values with the same probabilities p i (i = 1,2,...,n).

The sum (difference or product) of random variables X and Y is a random variable that takes all possible values ​​of the form xi+yj (xj-yj or xj·yj), where i = l,2,...,n; j =1,2,...,m, with probabilities pij that the random variable X will take the value xi, and y will take the value yj:

If random variables X and Y are independent, i.e. any events X=xi, Y=yj are independent, then by the theorem of multiplication of probabilities for independent events

3note . The above definitions of operations on discrete random variables need clarification: since in a number of cases the same values ,
,
can be obtained in different ways for different xi, yj with probabilities pi, pij, then the probabilities of such repeating values ​​are found by adding the resulting probabilities pi or pij.

Laplace's integral theorem

Theorem. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then the probability that the number m of the occurrence of event A in n independent trials lies in the range from a to b (inclusive), with a sufficiently large number of trials n is approximately equal to

The integral formula of Laplace, as well as the local formula of Moivre-Laplace, the more accurate the more n and the closer to 0.5 the value p And q. Calculation using this formula gives an insignificant error if the condition is met npq≥ 20, although the fulfillment of the condition can be considered acceptable npq > 10.

Function Ф( x) tabulated (see Appendix 2). To use this table you need to know the properties of the function Ф( x):

1. Function Ф( x) – odd, i.e. F(– x) = – Ф( x).

2. Function Ф( x) – monotonically increasing, and as x → +∞ Ф( x) → 0.5 (practically we can assume that already at x≥ 5 F( x) ≈ 0,5).

Example 3.4. Using the conditions of Example 3.3, calculate the probability that from 300 to 360 (inclusive) students will successfully pass the exam the first time.

Solution. We apply Laplace's integral theorem ( npq≥ 20). We calculate:

= –2,5; = 5,0;

P 400 (300 ≤ m≤ 360) = Ф(5.0) – Ф(–2.5).

Taking into account the properties of the function Ф( x) and using the table of its values, we find: Ф(5,0) = 0.5; Ф(–2.5) = – Ф(2.5) = – 0.4938.

We get P 400 (300 ≤ m ≤ 360) = 0,5 – (– 0,4938) = 0,9938. ◄

Let us write down the consequences of Laplace's integral theorem.

Corollary 1. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then with a sufficiently large number n of independent trials, the probability that the number m of the occurrence of event A differs from the product np by no more than ε > 0

.

(3.8)

Example 3.5. Under the conditions of Example 3.3, find the probability that from 280 to 360 students will successfully pass the probability theory exam the first time.

Solution. Calculate probability R 400 (280 ≤ m≤ 360) can be similar to the previous example using the basic integral formula of Laplace. But it’s easier to do this if you notice that the boundaries of the interval 280 and 360 are symmetrical with respect to the value n.p.=320. Then, based on Corollary 1, we obtain

= = ≈

≈ = 2Ф(5.0) ≈ 2·0.5 ≈ 1,

those. It is almost certain that between 280 and 360 students will successfully pass the exam the first time. ◄

Corollary 2. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then with a sufficiently large number n of independent trials, the probability that the frequency m/n of event A lies in the range from α to β (inclusive) is equal to

,

(3.9)

Example 3.6. According to statistics, on average 87% of newborns live to be 50 years old. Find the probability that out of 1000 newborns the proportion (frequency) of those surviving to 50 years will be in the range from 0.9 to 0.95.

Solution. The probability that a newborn will live to be 50 years old is r= 0.87. Because n= 1000 is large (i.e. the condition npq= 1000·0.87·0.13 = 113.1 ≥ 20 satisfied), then we use Corollary 2 of Laplace’s integral theorem. We find:

2,82, = 7,52.

= 0,5 – 0,4976 = 0,0024. ◄

Corollary 3. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then with a sufficiently large number n of independent trials, the probability that the frequency m/n of event A differs from its probability p by no more thanΔ > 0 (in absolute value) equals

.

(3.11)

Example 3.7. According to the conditions of the previous problem, find the probability that out of 1000 newborns, the proportion (frequency) of those surviving to 50 years will differ from the probability of this event by no more than 0.04 (in absolute value).

Solution. Using Corollary 3 of Laplace’s integral theorem, we find:

= 2F(3.76) = 2·0.4999 = 0.9998.

Since inequality is equivalent to inequality, this result means that it is almost certain that from 83 to 91% of newborns out of 1000 will live to be 50 years old. ◄

Previously, we established that for independent trials the probability of the number m occurrences of the event A V n test is found using Bernoulli's formula. If n is large, then use Laplace's asymptotic formula. However, this formula is unsuitable if the probability of the event is small ( r≤ 0.1). In this case ( n great, r little) apply Poisson's theorem

Poisson's formula

Theorem. If the probability p of the occurrence of event A in each trial tends to zero (p → 0) with an unlimited increase in the number n of trials (n→ ∞), and the product np tends to a constant number λ (np → λ), then the probability P n (m) that event A will appear m times in n independent trials satisfies the limit equality

Theorem 2 (Moivre-Laplace (local)). A in each n independent tests is equal to r n testing event A will occur once, is approximately equal (the more n, the more accurate) the value of the function

,

Where , . The table of function values ​​is given in the appendix. 1.

Example 6.5. The probability of finding a porcini mushroom among others is equal. What is the probability that among 300 porcini mushrooms there will be 75?

Solution. According to the conditions of the problem , . We find . From the table we find .

.

Answer: .

Theorem 3 (Moivre-Laplace (integral)). If the probability of an event occurring A in each n independent tests is equal to r and is different from zero and one, and the number of tests is large enough, then the probability that in n tests number of successes m is between and , approximately equal (the more n, the more accurate)

,

Where r- probability of success in each test, , , values ​​are given in appendix. 2.

Example 6.6. In a batch of 768 watermelons, each watermelon is unripe with probability . Find the probability that the number of ripe watermelons will be in the range from 564 to 600.

Solution. By condition By Laplace’s integral theorem

Answer:

Example 6.7. The city is visited daily by 1,000 tourists who go out for lunch during the day. Each of them chooses one of two city restaurants for lunch with equal probabilities and independently of each other. The owner of one of the restaurants wants that, with a probability of approximately 0.99, all tourists who come to his restaurant can dine there at the same time. How many seats should there be in his restaurant for this?

Solution. Let A= “tourist dined with interested owner.” Event occurrence A Let's consider it a "success" , . We are interested in this smallest number k, that the probability of occurrence is no less than k"successes" in a sequence of independent trials with a probability of success r= 0.5 is approximately equal to 1 – 0.99 = 0.01. This is precisely the probability of the restaurant being overcrowded. Thus, we are interested in this smallest number k, What . Let us apply the Moivre-Laplace integral theorem

Whence it follows that

.

Using the table for F(X) (Appendix 2), we find , Means . Therefore, the restaurant should have 537 seats.

Answer: 537 places.

From Laplace's integral theorem we can obtain the formula

.

Example 6.8. The probability of an event occurring in each of 625 independent trials is 0.8. Find the probability that the relative frequency of occurrence of an event will deviate from its probability in absolute value by no more than 0.04.

The probability that in n independent trials, in each of which the probability of an event occurring is p(0< p < 1), событие наступит ровно k раз, приближенно равна
Table of function values ​​φ(x); for negative values ​​of x, use the same table (function φ (x) is even: φ(-x) = φ(x)).

Example No. 1. In each of 700 independent trials, event A occurs with a constant probability of 0.35. Find the probability that event A occurs: a) exactly 270 times; b) less than 270 and more than 230 times; c) more than 270 times.
Solution. Since the number of experiments n = 700 is quite large, we use Laplace’s formulas.
a) Given: n = 700, p = 0.35, k = 270.
Let's find P 700 (270). We use Laplace's local theorem.
We find:

We find the value of the function φ(x) from the table:

b) Given: n = 700, p = 0.35, a = 230, b = 270.
Let's find P 700 (230< k < 270).
We use Laplace's integral theorem (23), (24). We find:

We find the value of the function Ф(x) from the table:

c) Given: n = 700, p = 0.35, a = 270, b = 700.
Let's find P 700 (k > 270).
We have:

Example No. 2. At a steady-state technological process in a weaving mill, 10 thread breaks occur per 100 spindles per hour. Determine: a) the probability that 7 thread breaks will occur on 80 spindles within an hour; b) the most likely number of thread breaks on 80 spindles within an hour.
Solution. The statistical probability of a thread breaking within an hour is p = 10/100 = 0.1 and, therefore, q = 1 – 0.1 = 0.9; n = 80; k = 7.
Since n is large, the local Laplace theorem (23) is used. We calculate:

Let's use the property φ(-x) = φ(x), find φ(0.37) ≈ 0.3726, and then calculate the desired probability:

Thus, the probability that 7 thread breaks will occur on 80 spindles within an hour is approximately 0.139.
The most probable number k 0 of occurrences of an event during repeated tests will be determined by formula (14). We find: 7.1< k 0 < 8,1. Поскольку k 0 может быть только целым числом, то k 0 = 8.

Example No. 3. The probability that a part is first grade is 0.4. 150 parts made. Find the probability that there are 68 first-class parts among them.

Example No. 4. The probability of an event occurring in each of the independent trials is p.
Find the probability that the event will occur n times if m tests are carried out.
Present your answer to three significant figures.
р=0.75, n=87, m=120