Two random variables $X$ and $Y$ are called independent if the distribution law of one random variable does not change depending on what possible values ​​the other random variable takes. That is, for any $x$ and $y$ the events $X=x$ and $Y=y$ are independent. Since the events $X=x$ and $Y=y$ are independent, then by the theorem of the product of probabilities of independent events $P\left(\left(X=x\right)\left(Y=y\right)\right)=P \left(X=x\right)P\left(Y=y\right)$.

Example 1 . Let the random variable $X$ express the cash winnings from tickets of one lottery “Russian Lotto”, and the random variable $Y$ express the cash winnings from tickets of another lottery “Golden Key”. It is obvious that the random variables $X,\Y$ will be independent, since the winnings from tickets of one lottery do not depend on the law of distribution of winnings from tickets of another lottery. In the case where the random variables $X,\Y$ would express the winnings of the same lottery, then, obviously, these random variables would be dependent.

Example 2 . Two workers work in different workshops and produce various products that are unrelated to each other by manufacturing technologies and the raw materials used. The distribution law for the number of defective products manufactured by the first worker per shift has the following form:

$\begin(array)(|c|c|)
\hline
Number of \ defective \ products \ x & 0 & 1 \\
\hline
Probability & 0.8 & 0.2 \\
\hline
\end(array)$

The number of defective products produced by the second worker per shift obeys the following distribution law.

$\begin(array)(|c|c|)
\hline
Number of \ defective \ products \ y & 0 & 1 \\
\hline
Probability & 0.7 & 0.3 \\
\hline
\end(array)$

Let us find the distribution law for the number of defective products produced by two workers per shift.

Let the random variable $X$ be the number of defective products produced by the first worker per shift, and $Y$ the number of defective products produced by the second worker per shift. By condition, the random variables $X,\Y$ are independent.

The number of defective products produced by two workers per shift is a random variable $X+Y$. Its possible values ​​are $0,\ 1$ and $2$. Let us find the probabilities with which the random variable $X+Y$ takes its values.

$P\left(X+Y=0\right)=P\left(X=0,\ Y=0\right)=P\left(X=0\right)P\left(Y=0\right) =0.8\cdot 0.7=0.56.$

$P\left(X+Y=1\right)=P\left(X=0,\ Y=1\ or\ X=1,\ Y=0\right)=P\left(X=0\right )P\left(Y=1\right)+P\left(X=1\right)P\left(Y=0\right)=0.8\cdot 0.3+0.2\cdot 0.7 =0.38.$

$P\left(X+Y=2\right)=P\left(X=1,\ Y=1\right)=P\left(X=1\right)P\left(Y=1\right) =0.2\cdot 0.3=0.06.$

Then the law of distribution of the number of defective products manufactured by two workers per shift:

$\begin(array)(|c|c|)
\hline
Number of \ defective \ products & 0 & 1 & 2 \\
\hline
Probability & 0.56 & 0.38 & 0.06\\
\hline
\end(array)$

In the previous example, we performed an operation on random variables $X,\Y$, namely, we found their sum $X+Y$. Let us now give a more rigorous definition of operations (addition, difference, multiplication) over random variables and give examples of solutions.

Definition 1. The product $kX$ of a random variable $X$ by a constant variable $k$ is a random variable that takes values ​​$kx_i$ with the same probabilities $p_i$ $\left(i=1,\ 2,\ \dots ,\ n\ right)$.

Definition 2. The sum (difference or product) of random variables $X$ and $Y$ is a random variable that takes all possible values ​​of the form $x_i+y_j$ ($x_i-y_i$ or $x_i\cdot y_i$), where $i=1 ,\ 2,\dots ,\ n$, with probabilities $p_(ij)$ that the random variable $X$ will take the value $x_i$, and $Y$ the value $y_j$:

$$p_(ij)=P\left[\left(X=x_i\right)\left(Y=y_j\right)\right].$$

Since the random variables $X,\Y$ are independent, then according to the probability multiplication theorem for independent events: $p_(ij)=P\left(X=x_i\right)\cdot P\left(Y=y_j\right)= p_i\cdot p_j$.

Example 3 . Independent random variables $X,\ Y$ are specified by their probability distribution laws.

$\begin(array)(|c|c|)
\hline
x_i & -8 & 2 & 3 \\
\hline
p_i & 0.4 & 0.1 & 0.5 \\
\hline
\end(array)$

$\begin(array)(|c|c|)
\hline
y_i & 2 & 8 \\
\hline
p_i & 0.3 & 0.7 \\
\hline
\end(array)$

Let's formulate the law of distribution of the random variable $Z=2X+Y$. The sum of random variables $X$ and $Y$, that is, $X+Y$, is a random variable that takes all possible values ​​of the form $x_i+y_j$, where $i=1,\ 2,\dots ,\ n$ , with probabilities $p_(ij)$ that the random variable $X$ will take the value $x_i$, and $Y$ the value $y_j$: $p_(ij)=P\left[\left(X=x_i\right )\left(Y=y_j\right)\right]$. Since the random variables $X,\Y$ are independent, then according to the probability multiplication theorem for independent events: $p_(ij)=P\left(X=x_i\right)\cdot P\left(Y=y_j\right)= p_i\cdot p_j$.

So, it has distribution laws for the random variables $2X$ and $Y$, respectively.

$\begin(array)(|c|c|)
\hline
x_i & -16 & 4 & 6 \\
\hline
p_i & 0.4 & 0.1 & 0.5 \\
\hline
\end(array)$

$\begin(array)(|c|c|)
\hline
y_i & 2 & 8 \\
\hline
p_i & 0.3 & 0.7 \\
\hline
\end(array)$

For the convenience of finding all values ​​of the sum $Z=2X+Y$ and their probabilities, we will compose an auxiliary table, in each cell of which we will place in the left corner the values ​​of the sum $Z=2X+Y$, and in the right corner - the probabilities of these values ​​obtained as a result multiplying the probabilities of the corresponding values ​​of random variables $2X$ and $Y$.

As a result, we obtain the distribution $Z=2X+Y$:

$\begin(array)(|c|c|)
\hline
z_i & -14 & -8 & 6 & 12 & 10 & 16 \\
\hline
p_i & 0.12 & 0.28 & 0.03 & 0.07 & 0.15 & 0.35 \\
\hline
\end(array)$

Draw up a distribution law for the number of defective parts produced during a shift on both machines, and calculate the mathematical expectation and standard deviation of this random variable.

192. The probability that the watch needs additional adjustment is 0.2. Draw up a law for the distribution of the number of watches that need additional adjustment among three randomly selected watches. Using the resulting distribution law, find the mathematical expectation and variance of this random variable. Check the result using the appropriate formulas for the mathematical expectation and dispersion of a random variable distributed according to the binomial law.

193. From the available six lottery tickets, of which four are non-winning, one ticket is drawn at random until a winning ticket is encountered. Draw up a distribution law for the random variable X - the number of tickets taken out, if each ticket taken out does not come back. Find the mathematical expectation and standard deviation of this random variable.

194. A student can take the exam no more than four times. Draw up a distribution law for the random variable X - the number of attempts to pass the exam, if the probability of passing it is 0.75 and subsequently increases by 0.1 with each subsequent attempt. Find the variance of this random variable.

195. The laws of distribution of two independent random variables X and Y are given:

Draw up a distribution law for the random variable X–Y and check the dispersion property D(X–Y) = D(X) + D(Y).

196. Among the five clocks of the same type available in the workshop, only one has a misaligned pendulum. The master checks a randomly chosen watch. The review ends as soon as a clock with a displaced pendulum is detected (the checked clocks are not viewed again). Draw up a distribution law for the number of hours watched by the master and calculate the mathematical expectation and dispersion of this random variable.

197. Independent random variables X and Y are specified by distribution laws:

Draw up a distribution law for the random variable X 2 + 2Y and check the property of the mathematical expectation: M(X 2 + 2Y) = M(X 2) + 2M(Y).

198. It is known that a random variable X, taking two values ​​x 1 = 1 and x 2 = 2, has a mathematical expectation equal to 7/6. Find the probabilities with which the random variable X takes its values. Draw up a distribution law for a random variable 2 X 2 and find its variance.

199. Two independent random variables X and Y are specified by the distribution laws:

Find P(X= 3) and P(Y= 4). Draw up the law of distribution of the random variable X – 2Y and check the properties of the mathematical expectation and dispersion: M(X – 2Y) = M(X) – 2M(Y); D(X – 2Y) = D(X) + 4D(Y).

In problems 201–210, random variables are given that are distributed according to the normal law

201. The random variable ξ is normally distributed. Find P(0< ξ<10), если Мξ= 10 и Р(10< ξ<20)= 0,3.

202. The random variable ξ is normally distributed. Find P(35< ξ<40), если Мξ= 25 и Р(10< ξ<15)= 0,2.

203. The random variable ξ is normally distributed. Find P(1< ξ<3), если Мξ= 3 и Р(3< ξ<5)= 0,1915.

204. <σ).

205. For a random variable ξ distributed according to the normal law, find Р(|ξ–а|<2σ).

206. For a random variable ξ distributed according to the normal law, find Р(|ξ–а|<4σ).

207. Independent random variables ξ and η are normally distributed,

Мξ= –1; Dξ= 2; Мη= 5; Dη= 7. Write down the probability density and the distribution function of their sum. Find Р(ξ+η<5) и Р(–1< ξ+η<3).

208. Independent random variables ξ, η, ζ are distributed according to the normal law and Мξ= 3; Dξ= 4; Мη= –2; Dη= 0.04; Мζ= 1; Dζ= 0.09. Write down the probability density and distribution function for their sum. Find Р(ξ+η+ζ<5) и Р(–1< ξ+η+ζ<3).

209. Independent random variables ξ, η, ζ are normally distributed and Мξ= –1; Dξ= 9; Мη= 2; Dη= 4; Мζ= –3; Dζ= 0.64. Write down the probability density and distribution function for their sum. Find Р(ξ+η+ζ<0) и

Р(–3< ξ+η+ζ<0).

210. The automatic machine produces rollers, controlling their diameters ξ. Assuming that ξ is normally distributed and a = 10 mm, σ = 0.1 mm, find the interval in which the diameters of the manufactured rollers will be contained with a probability of 0.9973.

In problems 211–220, sample X of volume n =100 is given by the table:

where the measurement results x i = 0.2·a +(i –1)·0.3·b; n i – frequencies with which values ​​x i occur.

1) construct a polygon of relative frequencies w i =n i /n;

2) calculate the sample mean, sample variance D B and standard deviation σ B;

3) calculate theoretical frequencies. Construct a graph on the same drawing as the polygon;

4) using the χ 2 criterion, test the hypothesis about the normal distribution of the population at a significance level of α = 0.05.

211. a = 4; b = 3; 212 . a = 3; b = 2; 213. a = 5; b = 1; 214. a = 1; b = 4;

215. a = 3; b = 5; 216. a=2; b = 3; 217. a = 4; b = 1; 218. a = 2; b = 5; 219. a = 1; b = 2; 220. a = 5; b = 4.

In problems 221–230, a two-dimensional sample of results of joint measurements of characteristics X and Y with a volume of n = 100 is specified by a correlation table:

where x i = 0.2·a +(i –1)·0.3·b; y i = 0.5·a +(j – 1)·0.2·b.

1) Find and σ y. Take the values ​​of and σ x from the previous problem.

2) Calculate the correlation coefficient r B . Draw a conclusion about the nature of the relationship between characteristics X and Y.

3) Construct the equation of a straight line of regression of Y on X in the form.

4) Draw a correlation field on the graph, i.e. plot the points (xi, yi) and construct a straight line.

221. a = 4; b = 3; 222. a = 3; b = 2; 223. a = 5; b = 1;

224. a = 1; b = 4; 225. a = 3; b = 5; 226. a = 2; b = 3;

227. a = 4; b = 1; 228. a = 2; b = 5; 229. a = 1; b = 2

230. a = 5; b = 4

In problems 231–240, find the maximum value of the function

under conditions . Take the values ​​from the table

required:

1) solve the linear programming problem graphically;

2) solve the problem using the tabular simplex method;

3) show the correspondence between the support solutions and the vertices of the region of feasible solutions;

In problems 241–250, some homogeneous cargo concentrated among three suppliers A i () must be delivered to five consumers B j (). The cargo inventories of suppliers a i and the needs of consumers b j , as well as the cost of transporting a unit of cargo from the i-th supplier to the j-th consumer C ij are given in the table.

Need to determine an optimal transportation plan that allows all cargo to be removed from suppliers and satisfies the needs of all consumers in such a way that this plan has a minimum cost. Find the first support plan using the “northwest” angle method. Find the optimal plan using the potential method. Calculate the shipping costs for each plan.

In tasks 251-260, the industry makes capital investments in four objects. Taking into account the characteristics of the contribution and local conditions, the profit of the industry, depending on the amount of financing, is expressed by the elements of the payment matrix. To simplify the problem, assume that the industry loss is equal to the industry profit. Find optimal industry strategies. Required:

1) summarize the initial data in a table and find a solution to the matrix game in pure strategies, if it exists (otherwise, see the next step 2);

2) simplify the payment matrix;

3) create a pair of mutually dual problems equivalent to the given matrix game;

4) find the optimal solution to the direct problem (for industry B) using the simplex method;

5) using the correspondence of variables, write out the optimal solution to the dual problem (for industry A);

6) give a geometric interpretation of this solution (for industry A);

7) using the relationship between optimal solutions to a pair of dual problems, optimal strategies and the cost of the game, find a solution to the game in mixed strategies;

option 1 option 2 option 3

1. Analytical geometry and vector algebra……………….. 4

2. Systems of linear equations and complex numbers………….. 5

3. Plotting function graphs, calculating limits

and identifying breakpoints of functions.…………….……………. 6

4. Derivatives of functions, greatest and least values

on the segment..…………………………………………………….… 9

5. Research of functions and construction of graphs,

functions of several variables, least squares method..… 11

6. Indefinite, definite and improper integral….. 12

7. Solving differential equations and systems

differential equations…………….……….…….….…… 14

8. Multiple and curvilinear integrals …………………………… 15

9. Study of numerical and power series, approximate

solutions to differential equations………………...……… 17

10. Probability theory……………….……………………...……… 18

Petr Alekseevich Burov

Anatoly Nikolaevich Muravyov

Collection of tasks

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Page creation date: 2017-12-07

The law of distribution of the minimum (maximum) of two random variables. Law of distribution of order statistics

In this section we will consider first of all such a functional transformation c. c., which consists in choosing the maximum (minimum) of two values.

Problem 1. The law of distribution of the minimum of two random variables. A continuous system is given. V. (X and X 2) with p.r./(*!, x 2). Find the distribution function of r.v. Y:

Solution. Let us first find P ( Y> y) = P (Xi > y; X 2 > y). Region D(y), where X> y and X 2 > y shown in Fig. 9.6.1. Probability of hitting a point (X[, X 2) to the region D(y) is equal

Where F (x b x 2) - system distribution function c. V. (Хь Х 2), F x(jq), F 2 (x 2) - distribution functions c. V. X And X 2 respectively. Hence,

To determine p.r. g (y) you need to find the derivative of the right side (9.6.1):

If s. V. X x, X 2 independent and distributed identically with p.r. Fi(X) =/ 2 (x) =f(x), That

Example 1. We consider the operation of a device consisting of two blocks Bi and B 2, the joint operation of which is absolutely necessary for the operation of the device. Block B operating times! and B 2 represent independent s. V. X And X 2, distributed according to exponential laws with parameters X And X 2. It is required to find the distribution law c. V. U- operating time of the technical unit.

Solution. It's obvious that

Using formulas (9.6.4) we find:

i.e. at least two independent random variables, distributed according to exponential laws with parameters X x and X 2, also distributed according to exponential laws with parameter X x + X 2. ?

Problem 2. Law of distribution of the minimum of n independent random variables. Given the system n independent villages V. (X x, X 2, ..., X p) with p.r. .f (x x),f 2 (x 2), ...,f n (x n). Find f. r. and density c. V. Y= min (X X,.... X p).

Solution. By definition

Example 2. We consider the operation of an automated system (AS), consisting of n subsystems For the speakers to work, everyone needs to work n subsystems; uptime of the /th subsystem 7} distributed according to the exponential law with the parameter (/ = 1, 2, p) and does not depend on the operating time of other subsystems. Determine the law of time distribution D i) of failure-free operation of the AS.

Solution. It's obvious that

Using formula (9.6.6) we find the r.v. distribution function. D l)

Thus, the distribution law c. V. - the minimum of n independent villages c., distributed according to exponential laws, is also exponential; while its parameter i)S n)) is equal to the sum of the parameters of these exponential distributions. It follows from this that

It can be shown that the distribution law c. V. D i) when large enough n will converge to the exponential law, even if s. V. 7) (/= 1, 2, ..., p) are not distributed according to exponential laws. Let us demonstrate this using the example of equally uniformly distributed s. V.:

In this case

and this is f. r. demonstrative law.

Thus, we can draw a conclusion that is widely used in engineering applications: if any device consists of a sufficiently large number of elements n, the operation of which is absolutely necessary for the operation of the device, then the law of time distribution F p) of failure-free operation of the device is close to exponential with the parameter, determined by the formula

where M[ Tj- average failure-free operation time of the i-th element.

The failure flow of such a device will be close to Poisson with the parameter )S n ?

Problem 3. The law of distribution of the maximum of two random variables. A continuous system is given. V. (Хь X 2) with density/(lbs x 2). It is required to find the r.v. distribution law.

Solution. By definition,

Where F(x x, x 2) - system distribution function (X and X 2).

Differentiating this expression as we did before, we get:

If random variables X and X2 are equally distributed, then

If random variables X 2 are independent, then

If random variables X 2 independent and equally distributed, then

Example 3. The operation of a technical device cannot begin before the assembly of its two blocks Bi and B2 is completed. The assembly time of blocks Bi and B 2 is a system of independent s. V. X x And X 2, distributed according to exponential laws with parameters X x And X 2. Y- time of completion of assembly of both technical specification blocks.

Solution. It's obvious that Y= max (X ъ X 2). Distribution density c. V. ^is determined by formula (9.6.12)

This law is not indicative. ?

Problem 4. The law of distribution of the maximum of n independent random variables. A continuous system is given. V. (X x, X 2 , ..., X p) with density f(x x, x 2,

Find the distribution law of a random variable

Solution. By definition

Where F(x 1, X 2 ,..., x p) - system distribution function (X x, X 2, ..., X p). By differentiating, we find the distribution density:

Where Fj (Xj) - f. r. With. V. Xjfj(xj) - its density.

If s. V. x b ..., X p independent and equally distributed (Fi(y) = F(y);f (y) =f(y) (/"= 1,n)), That

If random variables X and ..., X p are independent, then

Example 4. The work of technical equipment cannot begin before the assembly of all n its blocks: B b Bg, ..., B„. The assembly times of blocks B b..., B l represent a system n independent villages V. (Huh..., X p), distributed according to exponential laws with parameters A.1,..., A, p.

We need to find the density c. V. U- completion time for all assembly n TU blocks.

Solution. Obviously y = max (X,..., X p). According to formula (9.6.16) we have

Problem 5. Law of distribution of order statistics. Let us consider a continuous system of identically distributed, independent s. V. (X v X 2, ..., X p) with f. r. F(x) and p.r./(x). Let us arrange the values ​​assumed by the random variables X v X 2, ..., X p, in ascending order and denote:

X (1) is a random variable that takes the smallest of values: (X (1) = min (X v X 2, ..., X p));

X(2) - second largest accepted value of the random variables X v X 2, ..., X p;

X(T) - y-i by the magnitude of the accepted value from random variables X x, X 2, ..., X p;

X(P) - the largest random variable according to the accepted value X, X 2, x„ (X (n) = Shah (X and X 2, ..., X p)).

Obviously,

Random variables X(i), X@),..., X(") are called ordinal statistics.

Formulas (9.6.8) and (9.6.17) give the laws of distribution of extreme terms X(i), And X(") systems (*).

Let's find the distribution function F^m)(x)s. V. X^t y Event (X^x) is that T With. V. from the system n With. V. (X ( , X 2 ,..., X n) will be less than x and (p - t) With. V. will be greater than x. Since s. V. Xt (/" = 1, 2,..., p) are independent and identically distributed, then P (X t x) = F(x) R (Xj > x) = 1 - F(x). We need to find the probability that n independent experiments event (Xj x) will appear exactly T once. Applying the binomial distribution, we get

Purpose of the service. Using the online service mathematical expectation, variance and standard deviation are calculated(see example). In addition, a graph of the distribution function F(X) is plotted.

• Online solution
• Video instructions

Properties of the mathematical expectation of a random variable

1. The mathematical expectation of a constant value is equal to itself: M[C]=C, C – constant;
2. M=C M[X]
3. The mathematical expectation of the sum of random variables is equal to the sum of their mathematical expectations: M=M[X]+M[Y]
4. The mathematical expectation of the product of independent random variables is equal to the product of their mathematical expectations: M=M[X] M[Y] , if X and Y are independent.

Dispersion properties

1. The variance of a constant value is zero: D(c)=0.
2. The constant factor can be taken out from under the dispersion sign by squaring it: D(k*X)= k 2 D(X).
3. If the random variables X and Y are independent, then the variance of the sum is equal to the sum of the variances: D(X+Y)=D(X)+D(Y).
4. If random variables X and Y are dependent: D(X+Y)=DX+DY+2(X-M[X])(Y-M[Y])
5. The following computational formula is valid for dispersion:
   D(X)=M(X 2)-(M(X)) 2

Example. The mathematical expectations and variances of two independent random variables X and Y are known: M(x)=8, M(Y)=7, D(X)=9, D(Y)=6. Find the mathematical expectation and variance of the random variable Z=9X-8Y+7.
Solution. Based on the properties of mathematical expectation: M(Z) = M(9X-8Y+7) = 9*M(X) - 8*M(Y) + M(7) = 9*8 - 8*7 + 7 = 23 .
Based on the properties of dispersion: D(Z) = D(9X-8Y+7) = D(9X) - D(8Y) + D(7) = 9^2D(X) - 8^2D(Y) + 0 = 81*9 – 64*6 = 345

Continuous random variables. Systems of random variables. Function of two random arguments. Convolution formula. Stability of normal distribution, page 3

Let a function of a random argument X be given. It is required to find the mathematical expectation of this function, knowing the distribution law of the argument.

1. Let the argument X be a discrete random variable with a distribution series

.

Example 3. Discrete random variable X is given by the distribution

Find the mathematical expectation of a function .

Possible Y values:

; ; .

2. Let the argument X be a continuous random variable specified by the distribution density p(x). To find the mathematical expectation of a function, you can first find the distribution density g(y) of the value Y, and then use the formula: .

If possible values , That .

Example 4. The random variable X is given by density in the interval (0, π/2); outside this interval p(x)=0. Find the mathematical expectation of a function .

, , , ; Hence,

§ 17. Function of two random arguments.

Convolution formula. Stability of normal distribution.

o If each pair of possible values ​​of the random variables X and Y corresponds to one possible value of the random variable Z, then Z is called function of two random arguments X and Y:

.

Further examples will show how to find the distribution of the function according to known distributions of terms. This problem often occurs in practice. For example, if the X-error of the readings of a measuring device is uniformly distributed, then the task arises to find the law of distribution of the sum of errors .

Case 1. Let X and Y- discrete independent random variables. In order to draw up the distribution law for the function Z=X+Y, it is necessary to find all possible values ​​of Z and their probabilities. In other words, a distribution series of the random variable Z is compiled.

Example 1. Discrete independent random variables X and Y, specified by distributions

3. RANDOM VARIABLES. THE CONCEPT OF A RANDOM VARIABLE

Random variable A quantity is called which, as a result of tests carried out under the same conditions, takes on different, generally speaking, values ​​depending on random factors not taken into account. Examples of random variables: the number of points rolled on a dice, the number of defective products in a batch, the deviation of the point of impact of a projectile from the target, the uptime of a device, etc. There are discrete and continuous random variables. Discrete A random variable is called, the possible values ​​of which form a countable set, finite or infinite (that is, a set whose elements can be numbered).

Continuous A random variable is called, the possible values ​​of which continuously fill some finite or infinite interval of the number line. The number of values ​​of a continuous random variable is always infinite.

We will denote random variables with capital letters from the end of the Latin alphabet: X, Y, . ; random variable values ​​– in lowercase letters: X, y,. . Thus, X Denotes the entire set of possible values ​​of a random variable, and X - Some of its specific meaning.

Law of distribution A discrete random variable is a correspondence specified in any form between the possible values ​​of a random variable and their probabilities.

Let the possible values ​​of the random variable X Are . As a result of the test, the random variable will take one of these values, i.e. One event from a complete group of pairwise incompatible events will occur.

Let the probabilities of these events also be known:

Distribution law of a random variable X Can be written in the form of a table called Near distribution Discrete random variable:

The law of distribution of two independent random variables x and y is given

q p

q
p

This is a geometric law of distribution.

(we get a convergent series, since
).

Task 4. In the party from 10 There are three non-standard parts. Two parts were selected at random. Write the law for the distribution of the number of non-standard parts among the two selected ones. Calculate the mathematical expectation of this random variable.

Solution. Random variable X– the number of non-standard parts among the two selected ones has the following possible values:


Let's find their probabilities

Let’s compose the required law of distribution of a random variable

Finding the mathematical expectation

.

Task 5. The probable forecast for the value of X - the percentage change in the value of shares in relation to their current rate over six months - is given in the form of a distribution law:

Find the probability that buying shares will be more profitable than placing money in a bank deposit at 36% per annum.

Solution. The increase in the amount on a bank deposit, subject to 3% per month, will be after 6 months. The probability that buying shares is more profitable than a bank deposit is determined by the sum of the probabilities corresponding to a higher increase in the stock price:

Problem 6. Let the daily expenses for servicing and advertising cars in a certain car dealership average 100 thousand rubles, and the number of sales X cars during the day obeys the following distribution law:

a) Find the mathematical expectation of daily profit at a car price of 150 thousand rubles. b) The variance of the daily sale of the number of cars.

Solution. a)Daily profit is calculated using the formula

P = (150 X– 100) thousand rubles

Required characteristic M(P) is found using the above properties of the mathematical expectation (in thousand rubles):

b) Law of distribution of a random variable X 2 looks like:

M(X 2) = 0 ∙ 0,25 + 1 ∙ 0,2 + 9 ∙ 0,1 + 16 ∙ 0,1 + 25 ∙ 0,1 + 36 ∙ ∙0,05 + 49 ∙ 0,05 + 64 ∙ 0,025 + 81 ∙ 0,025 = 13,475.

Expectation M(X) = 2.675. Consequently, we obtain the desired dispersion value:

Problem 7. Random variable X specified on the entire axis by the distribution function
. Find the probability density function and the probability that X will take the value contained in the interval ( 0,1 ).

Solution. By definition

It is useful to accompany the solution to the problem in Fig. 4.

Z problem 8. The distribution function of a random variable has the form shown in Fig. 5.

Find: a) probability density function; b) looking at the graph F(x), indicate the main features of a random variable, for example, the range of possible values, the most probable values, etc.; V) M(X), D(X) ; G) P(X 2 ) . Then the probability that the part is good is equal to

We consider the manufacture of a part as an independent experience with a probability of “success” p=0,31 . Then the required number of parts is determined from the relation

Task 1. The lottery includes: a car worth 5,000 den. units, 4 TVs costing 250 den. units, 5 video recorders worth 200 den. units A total of 1000 tickets are sold for 7 days. units Draw up a distribution law for the net winnings received by a lottery participant who bought one ticket.

Solution. Possible values ​​of the random variable X - the net winnings per ticket - are equal to 0 - 7 = -7 money. units (if the ticket did not win), 200 – 7 = 193, 250 – 7 = 243, 5000 – 7 = 4993 den. units (if the ticket has the winnings of a VCR, TV or car, respectively). Considering that out of 1000 tickets the number of non-winners is 990, and the indicated winnings are 5, 4 and 1, respectively, and using the classical definition of probability, we obtain:

those. distribution series

Task 2. The probability that a student will pass a semester exam in a session by discipline A And B, are equal to 0.7 and 0.9, respectively. Draw up a distribution law for the number of semester exams that a student will take.

Solution. Possible values ​​of a random variable X— number of exams passed – 0, 1, 2.

Let A i– an event consisting in the fact that the student will pass i th exam ( i=1,2). Then the probability that the student will pass exams 0, 1, 2 in session will be respectively equal (we count the events A 1 and A 2 independent):

So the distribution series of the random variable

Task 3. Calculate M(X) for a random variable X— net gain according to task 1.

those. the average gain is zero. The result means that all proceeds from ticket sales go towards winnings.

Task 4. The laws of distribution of random variables are known X And Y– the number of points scored by the 1st and 2nd shooters.

It is necessary to find out which of the two shooters shoots better.

Considering the distribution series of random variables X And Y, answering this question is far from easy due to the abundance of numerical values. In addition, the first shooter has quite high probabilities (for example, more than 0.1) with extreme values ​​​​of the number of points scored ( X= 0; 1 and X= 9; 10), and the second shooter has intermediate values ​​( Y = 4; 5; 6).

Obviously, of two shooters, the better shooter is the one who scores more points on average.

that is, the average number of points scored by two shooters is the same.

Task 5. In problem 4, calculate the variance and standard deviation of the number of points scored for each shooter.

So, if the average values ​​of the number of points scored are equal ( M(X)=M(Y)) its variance, i.e. scattering characteristic relative to the average value, less for the second shooter ( D(X)

We make sure that

Considering that the law of distribution of a random variable X binomial we have

Task 7. The distribution series of a discrete random variable consists of two unknown values. The probability that a random variable will take one of these values ​​is 0.8. Find the distribution function of a random variable if its mathematical expectation is 3.2 and its variance is 0.16.

Solution. The distribution series has the form

or

Solving the resulting system, we find two solutions:

And

We write down the expression of the distribution function:

or

Task 8. Given the distribution function of a random variable X:

a) Find the probability density f(x); b) build graphs f(x) And F(x); c) make sure that X– continuous random variable; d) find the probabilities P(X=1), P(X

Problem 10. The bank issued loans n to different borrowers in the amount S r. each at a loan interest rate r. Find a) the mathematical expectation and dispersion of the bank’s profit, as well as the condition for the interest rate, if the probability of repayment of the loan by the borrower is equal to p; b) mathematical expectation and standard deviation of profit at n =1000, p =0,8, S= 100 thousand rubles And r = 30%.

Solution. a) Since the borrowers are not related to each other, we can assume that we have n independent tests. The probability of losing a loan for the bank in each trial is q = = 1 – p. Let X– the number of borrowers who repaid the loan with interest, then the bank’s profit is determined by the formula

Where X is a random variable with a binomial distribution law.

Since issuing a loan makes sense only with a positive mathematical expectation of profit (positive average profit), then from the condition M( P) > 0, the condition for the interest rate follows:

b) The loan interest rate satisfies the condition that the mathematical expectation of profit is positive: 30 >100(1 – 0.8)/0.8. Mathematical expectation of profit:

100 ∙ 1000(30 ∙ 0.8/100 – 0.2) = 4 million rubles.

Standard deviation of profit:

Problem 1. In a batch of 25 leather jackets, 5 have a hidden defect. Buy 3 jackets. Find the law of distribution of the number of defective jackets among those purchased. Construct a distribution polygon.

Task 2. The probability that an error was made in preparing the balance sheet is 0.3. The auditor was presented with 3 balance sheets of the enterprise for his conclusion. Draw up a law for the distribution of the number of positive conclusions on the balances being checked.

Task 3. Two buyers independently make one purchase each. The probability that the first buyer will make a purchase is 0.8, and the probability that the second will make a purchase is 0.6. Random variable X– the number of purchases made by customers. Describe the distribution law of a random variable X.

Task 4. Two canneries supply products to the store in a 2:3 ratio. The share of top quality products at the first plant is 90%, and at the second - 80%. 3 cans of canned food were purchased at the store. Find the mathematical expectation and standard deviation of the number of cans with the highest quality products.

Task 5. Probability density of a continuous random variable X specified in the interval (–π/2; π/2) by the function
Outside this interval
Find parameter WITH and determine the probability of hitting a random variable X into the interval (0; π/4).

Task 6. Random variable X given by the probability density
at – ∞

4)M(X) = 2.519, σ( X) ≈ 0,64; 5)C = 1/2; 6)
7)M x= =1h., D x= 1/3 h 2; 8)σ x = 48.8 g.

SMOLENSK STATE UNIVERSITY

ACCORDING TO PROBABILITY THEORY

Limit theorems of probability theory.

For any random variable that has a mathematical expectation and variance, the Chebyshev inequality is valid:

P(| X— a|> ε )≤
(1)

P(| X— a|≤ ε )≥ 1-

Chebyshev's theorem : If the variance n independent random variables X 1 , X 2 . X n are limited to the same constant, then with an unlimited increase in the number n the arithmetic mean of a random variable converges in probability to the arithmetic mean of their mathematical expectations, i.e.

Consequence: If independent random variables X 1 , X 2 . X n have the same mathematical expectations, equal a, and their variances are limited to the same constant, then Chebyshev’s inequality and Chebyshev’s theorem take the form:

Bernoulli's theorem : Relative frequency of events in n repeated independent trials, in each of which it can occur with the same probability p, with an unlimited increase in the number n converges in probability to probability p this event in a separate test:

Central limit theorem for identically distributed quantities : If X 1 , X 2 . X n– independent random variables that have equal mathematical expectations M[ X i ] =a, variance D[ X i ]= a 2 and third order absolute central moments M(| X i — a i | 3 )= m i , (
) , then the law of distribution of the amount Y n = X 1 + X 2 +. + X n at
indefinitely approaches normal. In particular, if all random variables X i identically distributed, then the law of distribution of their sum indefinitely approaches the normal law when
.

Local theorem of Moivre-Laplace : If the probability p occurrence of an event A in each trial is constant and different from 0 and 1, then the probability P m , n that the event A will happen m once every n independent tests with a sufficiently large number n, approximately equal

,

.

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

–

Laplace function (or probability integral);

,
.

Purpose of the lesson : 1. Achieve mastery of the conditions for applying the central limit theorem.

2. Strengthen the skills of calculating probabilities associated with the normal distribution law.

3. Teach students to recognize the manifestation of the law of large numbers.

For a lesson on this topic, answers to the following questions should be prepared:

What is the essence of the law of large numbers?

What is the practical and theoretical significance of Chebyshev's inequality?

What practical significance does Chebyshev's theorem have?

Explain, using Bernoulli's theorem, the property of stability of relative frequencies.

What is the essence of the central limit theorem of probability theory?

Task 1. The average water consumption on a livestock farm is 1000 liters per day, and the standard deviation of this random variable does not exceed 200 liters. Estimate the probability that the farm's water flow on any selected day will not exceed 2000 L using Chebyshev's inequality.

Solution. Dispersion D(X)=σ 2 ≤200 2 . Since the boundaries of the interval 0≤X≤2000 are symmetrical with respect to the mathematical expectation M(X)=1000, then to estimate the probability of the desired event, we can apply Chebyshev’s inequality.

,

those. no less than 0.96.

Task 2. According to statistics, on average, 87% of newborns live to be 50 years old. Using Chebyshev's inequality, estimate the probability that out of 1000 newborns, the proportion of those who survive to 50 years of age will differ from the probability of this event by no more than 0.04 (in absolute value).

,

those. no less than 0.929.

Task 3. To determine the average burning time of electric lamps in a batch of 200 identical boxes, one lamp from each box was sampled. Estimate the probability that the average burning time of the selected 200 electric lamps differs from the average burning time of the lamps in the entire batch by no more than 5 hours (in absolute value), if it is known that the standard deviation of the burning time of the lamps in each box is less than 7 hours.

Finding the probability of the desired event

,

those. not less than 0.9902.

Task 4. How many measurements of a given quantity must be taken in order to guarantee, with a probability of at least 0.95, that the arithmetic mean of these measurements differs from the true value of the quantity by no more than 1 (in absolute value), if the standard deviation of each measurement does not exceed 5?

Need to find n, at which

.

Let's apply Chebyshev's inequality:

, where

and at
, i.e. at least 500 measurements will be required.

Task 5. Metro trains run at intervals 2 minutes. Each passenger, independently of the others, arrives at the platform at a random time. Got on this train 75 passengers. What is the probability that their total waiting time will be between one and two and a half hours?

Solution. Let's denote the waiting time i th passenger through X i. It is natural to assume that it is equally possible for a passenger to arrive at any time between trains. Formally this means that X i has a uniform distribution law with a probability density function

f(x) =

Then
And

Total waiting time Y=∑ X i represents the sum of a larger number of independent identically distributed random variables with bounded variances. By virtue of the central limit theorem, it can be stated that Y has a distribution law close to normal. The normal distribution law is determined by the mathematical expectation and dispersion. Let's count them.

N(75,25) . The problem requires calculating

Task 6. The shooter hits the top ten with a probability 0,4 , to nine - with probability 0,3 , to eight - with probability 0,2 , in seven - with probability 0,1 . What is the probability that when 25 shots fired from the shooter 250 will knock out points from 220 to 240 glasses?

Solution. Let at i-th shot the shooter dials X i points. Quantities X i independent and have the same distribution

Sum of points Y= being the sum of a large number of independent identically distributed terms with limited variances, it has a distribution law close to normal, the parameters of which

N(225,25) And P(220 2 ). What is the probability that in one measurement the error will not exceed 1 MK? To improve the measurement accuracy, we have done 25 measurements, the arithmetic mean of the observed values ​​is taken as the measured value. In this case, what is the probability that the error will not exceed 1 MK? (Instruction: use the fact of stability of the normal distribution law.) Determine the last probability if the distribution law of the measurement error is unknown, and only its variance is known, equal to 4 mk 2.

Solution. Let X– measurement error. Then

If the distribution law of the measurement error is unknown, then from Chebyshev’s inequality:

P(|  0 | 1 , then both the Moivre–Laplace theorems are valid.

a) By the local theorem of Moivre–Laplace

b) Random variable X makes sense of the relative frequency of successes in n experiments, and D

Since in Pearson's experiment a deviation of the relative frequency of success from the probability of success in one experiment was equal to
then according to the Moivre–Laplace integral theorem

Task 1. On average, 10% of the working population of a certain region are unemployed. Using Chebyshev's inequality, estimate the probability that the unemployment rate among the surveyed 10,000 working-age city residents will be in the range from 9 to 11% (inclusive).

Task 2. The experience of the insurance company shows that an insured event occurs in approximately every fifth contract. Using Chebyshev’s inequality, estimate the required number of contracts that should be concluded so that with a probability of 0.9 it can be stated that the share of insured events will deviate from 0.1 by no more than 0.01 (in absolute value).

Task 3. When examining the authorized capital of banks, it was found that a fifth of banks have an authorized capital of over 100 million rubles. Find the probability that among 1800 banks have an authorized capital of over 100 million rubles: a) at least 300; b) from 300 to 400 inclusive.

Task 4. The probability that a dealer selling securities will sell them is 0.7. How many securities should there be so that it can be stated with a probability of 0.996 that the share of those sold among them will deviate from 0.7 by no more than 0.04 (in absolute value)?

Task 5. An insurance company has 10,000 clients. Each of them, insuring against an accident, contributes 500 rubles. The probability of an accident is 0.0055, and the insurance amount paid to the victim is 50,000 rubles. What is the probability that: a) the insurance company will suffer a loss; b) more than half of all funds received from clients will be spent on paying insurance amounts?

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