The distribution of a discrete random variable x is given. Discrete random variables (DSV)
Definition 2.3. A random variable, denoted by X, is called discrete if it takes a finite or countable set of values, i.e. set - a finite or countable set.
Let's consider examples of discrete random variables.
1.
Two coins are thrown once. The number of coat of arms drops in this experiment is random value NS... Its possible values are 0,1,2, i.e. 
Is a finite set.
2. The number of ambulance calls is recorded for a given period of time. Random value NS- number of calls. Its possible values are 0, 1, 2, 3, ..., i.e. = (0,1,2,3, ...) is a countable set.
3. There are 25 students in a group. On some day, the number of students who came to classes is registered - a random variable NS... Its possible values are: 0, 1, 2, 3, ..., 25 i.e. = (0, 1, 2, 3, ..., 25).
Although all 25 people in example 3 cannot skip classes, the random variable NS can take this value. This means that the values of a random variable have different probabilities.
Consider a mathematical model of a discrete random variable.
Let a random experiment be carried out, which corresponds to a finite or countable space of elementary events. Consider the mapping of this space to the set of real numbers, that is, to each elementary event we associate some real number,. In this case, the set of numbers can be finite or countable, i.e. 
or
A system of subsets, which includes any subset, including a one-point one, forms the -algebra of a numerical set (- of course or countable).
Since any elementary event is assigned certain probabilities p i(in the case of finite all), moreover, then each value of the random variable can be associated with a certain probability p i, such that.
Let be NS- an arbitrary real number. We denote P x (x) probability that a random variable NS took a value equal to NS, i.e. P X (x) = P (X = x)... Then the function P x (x) can take positive values only for those values NS that belong to a finite or countable set 
, and for all other values the probability of this value P X (x) = 0.
So, we have defined a set of values, -algebra as a system of any subsets and each event ( X = x) compared the probability for any, i.e. built a probabilistic space.
For example, the space of elementary events of an experiment consisting of a double toss of a symmetric coin consists of four elementary events:, where
When the coin was tossed twice, two lattices fell out; when the coin was tossed twice, two emblems fell out;
On the first toss of the coin, the grate fell out, and on the second, the coat of arms;
On the first toss of the coin, the coat of arms fell out, and on the second - the lattice.
Let the random variable NS- the number of fallouts of the lattice. It is defined on and many of its meanings 
... All possible subsets, including one-point ones, form an algebra, i.e. = (Ø, (1), (2), (0,1), (0,2), (1,2), (0,1,2)).
The probability of an event ( X = x i}, і = 1,2,3, we define as the probability of occurrence of an event that is its prototype:
Thus, on elementary events ( X = x i) set a numeric function P X, so 
.
Definition 2.4. The distribution law of a discrete random variable is a set of pairs of numbers (x i, p i), where x i are the possible values of the random variable, and p i are the probabilities with which it takes these values, and.
The simplest form setting the law of distribution of a discrete random variable is a table that lists the possible values of the random variable and the corresponding probabilities:
| … | … | ||||
| … | … |
Such a table is called a distribution series. To make the distribution series more visual, it is depicted graphically: on the axis Oh dot x i and draw perpendiculars of length from them p i... The resulting points are connected and get a polygon, which is one of the forms of the distribution law (Fig. 2.1).
Thus, to set a discrete random variable, you need to set its values and the corresponding probabilities.
Example 2.2. The money receiver of the machine is triggered every time a coin is dropped with a probability R... Once it is triggered, the coins are not lowered. Let be NS- the number of coins that must be lowered before the cash drawer of the machine is triggered. Construct a series of distribution of a discrete random variable NS.
Solution. Possible values of a random variable NS: x 1 = 1, x 2 = 2, ..., x k = k, ... Let's find the probabilities of these values: p 1- the probability that the money receiver will work on the first lowering, and p 1 = p; p 2 - the probability that two attempts will be made. To do this, it is necessary that: 1) the money receiver does not work on the first attempt; 2) on the second attempt - it worked. The probability of this event is (1 – p) p... Likewise 
etc, 
... Distribution series NS will take the form
| 1 | 2 | 3 | … | To | … | |
| R | qp | q 2 p | q r -1 p |
Note that the probabilities p to form a geometric progression with the denominator: 1 – p = q, q<1, therefore, such a probability distribution is called geometric.
ІІ assume further that a mathematical model has been built 
experiment described by a discrete random variable NS, and consider the calculation of the probabilities of occurrence of arbitrary events.
Let an arbitrary event contain a finite or countable set of values x i: A = {x 1, x 2, ..., x i, ...) .Event A can be represented as a combination of incompatible events of the form:. Then, applying Kolmogorov's axiom 3 , we get
since the probabilities of the occurrence of events we determined equal to the probabilities of the occurrence of events that are their prototypes. This means that the likelihood of any event 
,, can be calculated by the formula, since this event can be represented as a combination of events, where .
Then the distribution function F (x) = P (-<Х<х) is found by the formula. Hence it follows that the distribution function of a discrete random variable NS is discontinuous and increases in jumps, that is, it is a step function (Fig.2.2):
If the set is finite, then the number of terms in the formula is finite; if it is countable, then the number of terms is also countable.
Example 2.3. The technical device consists of two elements that work independently of each other. The probability of failure of the first element in time T is 0.2, and the probability of failure of the second element is 0.1. Random value NS- the number of failed elements in time T. Find the distribution function of a random variable and build its graph.
Solution. The space of elementary events of an experiment consisting in investigating the reliability of two elements of a technical device is determined by four elementary events,,,: - both elements are in good working order; - the first element is operational, the second is faulty; - the first element is faulty, the second is operational; - both elements are defective. Each of the elementary events can be expressed in terms of elementary events of spaces 
and 
, where - the first element is operational; - the first element is out of order; - the second element is serviceable; - the second element is out of order. Then, and since the elements of a technical device work independently of each other, then
8. What is the probability that the values of a discrete random variable belong to the interval?
A series of distributions of a discrete random variable is given. Find the missing probability and plot the distribution function. Calculate expected value and the variance of this quantity.
The random variable X takes only four values: -4, -3, 1 and 2. It takes each of these values with a certain probability. Since the sum of all probabilities must be equal to 1, the missing probability is:
0,3 + ? + 0,1 + 0,4 = 1,
Let's compose the distribution function of the random variable X. It is known that the distribution function, then:
Hence,
Let's plot the function F(x) .
The mathematical expectation of a discrete random variable is equal to the sum of the products of the value of the random variable by the corresponding probability, i.e.
We find the variance of a discrete random variable by the formula:
APPLICATION
Combinatorial elements Here: is the factorial of a number |
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Actions on eventsAn event is any fact that may or may not happen as a result of experience. Combining events A and V- this event WITH which consists of an appearance or an event A, or events V, or both events at the same time. Designation: Intersection of events A and V- this event WITH, which consists in the simultaneous appearance of both events. Designation:  |
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Classical definition of probabilityEvent probability A Is the ratio of the number of experiments
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Probability multiplication formulaEvent probability
If events A and B are independent (the appearance of one does not affect the appearance of the other), then the probability of an event is: |
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The formula for adding probabilitiesEvent probability Event probability A, Event probability V,
If events A and B are inconsistent (they cannot appear at the same time), then the probability of the event is equal to: |
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Total Probability FormulaLet the event A can occur simultaneously with one of the events |
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Bernoulli schemeLet n independent tests be carried out. The probability of occurrence (success) of the event A in each of them is constant and equal to p, the probability of failure (i.e. not the occurrence of an event A) q = 1 - p... Then the probability of occurrence k successes in n tests can be found using the Bernoulli formula:
Most likely number of successes |
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Random Variablesdiscrete continuous (e.g., number of girls in a family with 5 children) (e.g., working hours of the kettle) Numerical characteristics of discrete random variablesLet a discrete quantity be given by a distribution series:
,, ..., - values of a random variable NS; ,,…, Are the corresponding values of probabilities. Distribution functionThe distribution function of a random variable NS is a function defined on the whole number line and equal to the probability that NS will be less NS: |
;
;
favorable for the occurrence of the event A, to the total number of experiments 
:
can be found by the formula:
- probability of an event A,
- probability of an event V,
- probability of an event V provided that the event A has already happened.
can be found by the formula:
- probability of joint occurrence of events A and V.
,
,
…,
- let's call them hypotheses. Also known 
- probability of fulfillment i-th hypothesis and 
- the probability of occurrence of event A when performing i-th hypothesis. Then the probability of the event A can be found by the formula:
in the Bernoulli scheme, this is the number of occurrences of a certain event, which corresponds to the highest probability. Can be found by the formula:
Exam questions
Event. Operations on random events.
The concept of the probability of an event.
Rules for addition and multiplication of probabilities. Conditional probabilities.
Formula of total probability. Bayes' formula.
Bernoulli's scheme.
A random variable, its distribution function and distribution series.
Basic properties of the distribution function.
Expected value. Mathematical expectation properties.
Dispersion. Dispersion properties.
The density of the probability distribution of a one-dimensional random variable.
Distribution types: uniform, exponential, normal, binomial and Poisson distribution.
Local and integral theorems of Moivre-Laplace.
The law and distribution function of a system of two random variables.
Density of distribution of a system of two random variables.
Conditional distribution laws, conditional mathematical expectation.
Dependent and independent random variables. Correlation coefficient.
Sample. Sample processing. Polygon and histogram of frequencies. Empirical distribution function.
The concept of estimating distribution parameters. Evaluation requirements. Confidence interval. Plotting intervals for evaluating the mathematical expectation and standard deviation.
Statistical hypotheses. Consent criteria.
Chapter 1. Discrete random variable
§ 1. Concepts of a random variable.
Distribution law of a discrete random variable.
Definition : A random value is a quantity that, as a result of a test, takes only one value from a possible set of its values, unknown in advance and depending on random reasons.
There are two types of random variables: discrete and continuous.
Definition : The random variable X is called discrete (discontinuous), if the set of its values is finite or infinite, but countable.
In other words, the possible values of a discrete random variable can be renumbered.
You can describe a random variable using its distribution law.
Definition : The law of distribution of a discrete random variable is the correspondence between the possible values of a random variable and their probabilities.
The distribution law of a discrete random variable X can be specified in the form of a table, in the first line of which all possible values of the random variable are indicated in ascending order, and in the second line the corresponding probabilities of these values, i.e.
where p1 + p2 + ... + pn = 1
Such a table is called a distribution series of a discrete random variable.
If the set of possible values of a random variable is infinite, then the series p1 + p2 +… + pn +… converges and its sum is equal to 1.
The distribution law of a discrete random variable X can be depicted graphically, for which a polyline is constructed in a rectangular coordinate system, connecting successively points with coordinates (xi; pi), i = 1,2, ... n. The resulting line is called distribution polygon (fig. 1).
Organic chemistry "href =" / text / category / organicheskaya_hiimya / "rel =" bookmark "> of organic chemistry are 0.7 and 0.8, respectively. Draw up the law of distribution of a random variable X - the number of exams that a student will pass.
Solution. The considered random variable X as a result of the exam can take one of the following values: x1 = 0, x2 = 1, x3 = 2.
Let's find the probability of these values. Let's denote the events:
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 |
So, the distribution law of a random variable X is given by the table:
Control: 0.6 + 0.38 + 0.56 = 1.
§ 2. Distribution function
The distribution function also gives a complete description of the random variable.
Definition: The distribution function of a discrete random variable X the function F (x) is called, which determines for each value of x the probability that a random variable X takes a value less than x:
F (x) = P (X<х)
Geometrically, the distribution function is interpreted as the probability that the random variable X will take the value that is depicted on the number line by a point lying to the left of the point x.
1) 0≤ F (x) ≤1;
2) F (x) is a non-decreasing function on (-∞; + ∞);
3) F (x) - is continuous on the left at the points x = xi (i = 1,2, ... n) and is continuous at all other points;
4) F (-∞) = P (X<-∞)=0 как вероятность невозможного события Х<-∞,
F (+ ∞) = P (X<+∞)=1 как вероятность достоверного события Х<-∞.
If the distribution law of a discrete random variable X is given in the form of a table:
then the distribution function F (x) is determined by the formula:
https://pandia.ru/text/78/455/images/image007_76.gif "height =" 110 ">
0 for х≤ x1,
p1 at x1< х≤ x2,
F (x) = p1 + p2 at x2< х≤ х3
1 for x> xn.
Its graph is shown in Fig. 2:
§ 3. Numerical characteristics of a discrete random variable.
The mathematical expectation is one of the important numerical characteristics.
Definition: The mathematical expectation M (X) a discrete random variable X is the sum of the products of all its values by the corresponding probabilities:
M (X) = ∑ xiрi = x1р1 + x2р2 + ... + xnрn
The mathematical expectation serves as a characteristic of the average value of a random variable.
Mathematical expectation properties:
1) M (C) = C, where C is a constant value;
2) M (C X) = C M (X),
3) M (X ± Y) = M (X) ± M (Y);
4) M (X Y) = M (X) M (Y), where X, Y are independent random variables;
5) M (X ± C) = M (X) ± C, where C is a constant;
The dispersion is used to characterize the degree of dispersion of possible values of a discrete random variable around its mean value.
Definition: Dispersion D ( X ) of a random variable X is called the mathematical expectation of the square of the deviation of the random variable from its mathematical expectation:
Dispersion properties:
1) D (C) = 0, where C is a constant;
2) D (X)> 0, where X is a random variable;
3) D (C X) = C2 D (X), where C is a constant;
4) D (X + Y) = D (X) + D (Y), where X, Y are independent random variables;
To calculate the variance, it is often convenient to use the formula:
D (X) = M (X2) - (M (X)) 2,
where М (Х) = ∑ xi2рi = x12р1 + x22р2 + ... + xn2рn
The variance D (X) has the dimension of the square of a random variable, which is not always convenient. Therefore, the quantity √D (X) is also used as an indicator of the scattering of possible values of a random variable.
Definition: Mean square deviation σ (X) a random variable X is called the square root of the variance:
Problem number 2. Discrete random variable X is given by the distribution law:
Find P2, the distribution function F (x) and plot its graph, as well as M (X), D (X), σ (X).
Solution: Since the sum of the probabilities of possible values of the random variable X is equal to 1, then
P2 = 1- (0.1 + 0.3 + 0.2 + 0.3) = 0.1
Let us find the distribution function F (x) = P (X Geometrically, this equality can be interpreted as follows: F (x) is the probability that a random variable will take a value that is depicted on the numerical axis by a point lying to the left of the point x. If x≤-1, then F (x) = 0, because on (-∞; x) there is not a single value of this random variable; If -1<х≤0, то F(х)=Р(Х=-1)=0,1, т. к. в промежуток (-∞;х) попадает только одно значение x1=-1; If 0<х≤1, то F(х)=Р(Х=-1)+ Р(Х=0)=0,1+0,1=0,2, т. к. в промежуток (-∞; x) two values x1 = -1 and x2 = 0; If 1<х≤2, то F(х)=Р(Х=-1) + Р(Х=0)+ Р(Х=1)= 0,1+0,1+0,3=0,5, т. к. в промежуток (-∞;х) попадают три значения x1=-1, x2=0 и x3=1; If 2<х≤3, то F(х)=Р(Х=-1) + Р(Х=0)+ Р(Х=1)+ Р(Х=2)= 0,1+0,1+0,3+0,2=0,7, т. к. в промежуток (-∞;х) попадают четыре значения x1=-1, x2=0,x3=1 и х4=2; If x> 3, then F (x) = P (X = -1) + P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 0.1 + 0.1 + 0.3 + 0.2 + 0.3 = 1, since four values x1 = -1, x2 = 0, x3 = 1, x4 = 2 fall into the interval (-∞; x) and x5 = 3. https://pandia.ru/text/78/455/images/image006_89.gif "width =" 14 height = 2 "height =" 2 "> 0 at x≤-1, 0.1 at -1<х≤0, 0.2 at 0<х≤1, F (x) = 0.5 at 1<х≤2, 0.7 at 2<х≤3, 1 for x> 3 Let us represent the function F (x) graphically (Fig. 3): https://pandia.ru/text/78/455/images/image014_24.jpg "width =" 158 height = 29 "height =" 29 "> ≈1.2845. §
4. Binomial distribution law discrete random variable, Poisson's law. Definition: Binomial
is the law of distribution of a discrete random variable X - the number of occurrences of event A in n independent repeated tests, in each of which events A may occur with probability p or not occur with probability q = 1-p. Then P (X = m) -probability of occurrence of event A exactly m times in n tests is calculated by the Bernoulli formula: P (X = m) = Сmnpmqn-m Mathematical expectation, variance and mean standard deviation a random variable X, distributed according to a binary law, is found, respectively, by the formulas: https://pandia.ru/text/78/455/images/image016_31.gif "width =" 26 "> The probability of event A -" getting a five "in each test is the same and is equal to 1/6, that is . P (A) = p = 1/6, then P (A) = 1-p = q = 5/6, where - "not a five". The random variable X can take on the values: 0; 1; 2; 3. The probability of each of the possible values of X is found by the Bernoulli formula: P (X = 0) = P3 (0) = C03p0q3 = 1 (1/6) 0 (5/6) 3 = 125/216; P (X = 1) = P3 (1) = C13p1q2 = 3 (1/6) 1 (5/6) 2 = 75/216; P (X = 2) = P3 (2) = C23p2q = 3 (1/6) 2 (5/6) 1 = 15/216; P (X = 3) = P3 (3) = C33p3q0 = 1 (1/6) 3 (5/6) 0 = 1/216. That. the distribution law of the random variable X has the form: Control: 125/216 + 75/216 + 15/216 + 1/216 = 1. Let's find the numerical characteristics of the random variable X: M (X) = np = 3 (1/6) = 1/2, D (X) = npq = 3 (1/6) (5/6) = 5/12, Problem number 4. The automatic machine stamps parts. The probability that a manufactured part will be defective is 0.002. Find the probability that among 1000 selected parts there will be: a) 5 defective ones; b) at least one defective. Solution:
The number n = 1000 is large, the probability of manufacturing a defective part p = 0.002 is small, and the events under consideration (the part turns out to be defective) are independent, therefore, the Poisson formula takes place: Рn (m) = e-
λ
λm Find λ = np = 1000 0.002 = 2. a) Find the probability that there will be 5 defective parts (m = 5): P1000 (5) = e-2
25
= 32 0,13534
= 0,0361 b) Find the probability that there will be at least one defective part. Event A - “at least one of the selected parts is defective” is the opposite of the event - “all selected parts are not defective.” Therefore, P (A) = 1-P (). Hence the desired probability is equal to: P (A) = 1-P1000 (0) = 1- e-2
20
= 1- e-2 = 1-0.13534≈0.865. Tasks for independent work.
1.1
1.2.
Dispersed random variable X is given by the distribution law: Find p4, the distribution function F (X) and plot its graph, as well as M (X), D (X), σ (X). 1.3.
There are 9 markers in the box, of which 2 markers no longer write. Take 3 felt-tip pens at random. Random variable X is the number of writing markers among those taken. Draw up the law of distribution of a random variable. 1.4.
6 textbooks are placed in random order on the library shelf, 4 of them are bound. The librarian takes 4 textbooks at random. Random variable X is the number of bound textbooks among those taken. Draw up the law of distribution of a random variable. 1.5.
There are two tasks in the ticket. Probability correct decision the first task is 0.9, the second is 0.7. Random variable X is the number of correctly solved problems in the ticket. Draw up the distribution law, calculate the mathematical expectation and variance of this random variable, as well as find the distribution function F (x) and build its graph. 1.6.
Three arrows shoot at the target. The probability of hitting the target with one shot for the first shooter is 0.5, for the second -0.8, for the third -0.7. Random variable X is the number of hits on the target if the shooters make one shot at a time. Find the distribution law, M (X), D (X). 1.7.
The basketball player throws the ball into the basket with a hit probability of 0.8 on each shot. For each hit, he receives 10 points, and in case of a miss, he is not awarded points. Draw up the distribution law of a random variable X-number of points received by a basketball player for 3 throws. Find M (X), D (X), and the probability that he will get more than 10 points. 1.8.
The cards are written letters, only 5 vowels and 3 consonants. 3 cards are chosen at random, and each time the taken card is returned back. Random variable X is the number of vowels among those taken. Draw up the distribution law and find M (X), D (X), σ (X). 1.9.
On average, 60% of contracts are paid by the insurance company in connection with the occurrence of an insured event. Draw up the law of distribution of a random variable X - the number of contracts for which the sum insured was paid among the randomly selected four contracts. Find the numerical characteristics of this quantity. 1.10.
The radio station sends callsigns at regular intervals (no more than four) until two-way communication is established. The probability of receiving a response to the callsign is 0.3. Random X is the number of callsigns sent. Draw up the distribution law and find F (x). 1.11.
There are 3 keys, of which only one fits the lock. Draw up the distribution law of a random variable X-number of attempts to open the lock, if the tried key does not participate in subsequent attempts. Find M (X), D (X). 1.12.
Three instruments are independently tested for reliability in succession. Each subsequent device is tested only if the previous one proved to be reliable. The probability of passing the test for each device is 0.9. Draw up the distribution law of a random variable X-number of tested devices. 1.13
A discrete random variable X has three possible values: x1 = 1, x2, x3, and x1<х2<х3. Вероятность того, что Х примет значения х1 и х2, соответственно равны 0,3 и 0,2. Известно, что М(Х)=2,2, D(X)=0,76. Составить закон распределения случайной величины. 1.14.
The electronic device block contains 100 identical elements. The probability of failure of each element during the time T is 0.002. The elements work independently. Find the probability that no more than two elements will fail in time T. 1.15.
The textbook has been published with a circulation of 50,000 copies. The probability that the textbook is not stitched correctly is 0.0002. Find the probability that the circulation contains: a) four defective books, b) less than two defective books. 1
.16.
The number of calls arriving at the PBX every minute is distributed according to Poisson's law with the parameter λ = 1.5. Find the probability that in a minute you will enter: a) two calls; b) at least one call. 1.17.
Find M (Z), D (Z) if Z = 3X + Y. 1.18.
The laws of distribution of two independent random variables are given: Find M (Z), D (Z) if Z = X + 2Y. Answers:
https://pandia.ru/text/78/455/images/image007_76.gif "height =" 110 "> 1.1.
p3 = 0.4; 0 at x≤-2, 0.3 at -2<х≤0, F (x) = 0.5 at 0<х≤2, 0.9 at 2<х≤5, 1 for x> 5 0.3 at -1<х≤0, 0.4 at 0<х≤1, F (x) = 0.6 at 1<х≤2, 0.7 at 2<х≤3, 1 for x> 3 M (X) = 1; D (X) = 2.6; σ (X) ≈ 1.612. https://pandia.ru/text/78/455/images/image025_24.gif "width =" 2 height = 98 "height =" 98 "> 0 at x≤0, 0.03 at 0<х≤1, F (x) = 0.37 at 1<х≤2, 1 for x> 2 M (X) = 2; D (X) = 0.62 M (X) = 2.4; D (X) = 0.48, P (X> 10) = 0.896 1.
8
.
M (X) = 15/8; D (X) = 45/64; σ (X) ≈ M (X) = 2.4; D (X) = 0.96 https://pandia.ru/text/78/455/images/image008_71.gif "width =" 14 "> 1.11.
M (X) = 2; D (X) = 2/3 1.14.
1.22 e-0.2≈0.999 1.15.
a) 0.0189; b) 0.00049 1.16.
a) 0.0702; b) 0.77687 1.17.
3,8; 14,2 1.18.
11,2; 4. Chapter 2. Continuous random variable
Definition: Continuous
called a quantity, all possible values of which completely fill a finite or infinite interval of the numerical axis. Obviously, the number of possible values of a continuous random variable is infinite. A continuous random variable can be specified using a distribution function. Definition: F distribution function
continuous random variable X is called the function F (x), which determines for each value of xhttps: //pandia.ru/text/78/455/images/image028_11.jpg "width =" 14 "height =" 13 "> R The distribution function is sometimes called the cumulative distribution function. Distribution function properties:
1) 1≤ F (x) ≤1 2) For a continuous random variable, the distribution function is continuous at any point and differentiable everywhere, except, perhaps, at individual points. 3) The probability of a random variable X hitting one of the intervals (a; b), [a; b), [a; b] is equal to the difference between the values of the function F (x) at points a and b, i.e. P (a<Х 4) The probability that a continuous random variable X will take one separate value is equal to 0. 5) F (-∞) = 0, F (+ ∞) = 1 Specifying a continuous random variable using a distribution function is not the only one. Let us introduce the concept of probability distribution density (distribution density). Definition
:
The density of the probability distribution
f
(
x
)
continuous random variable X is called the derivative of its distribution function, i.e.: The density of the probability distribution is sometimes called the differential distribution function or differential distribution law. The plot of the probability distribution density f (x) is called probability distribution curve
.
Probability density properties:
1) f (x) ≥0, for хhttps: //pandia.ru/text/78/455/images/image029_10.jpg "width =" 285 "height =" 141 ">. Gif" width = "14" height = "62 src ="> 0 for x≤2, f (x) = c (x-2) at 2<х≤6, 0 for x> 6. Find: a) the value of c; b) the distribution function F (x) and build its graph; c) P (3≤x<5) Solution:
+
∞ a) We find the value of c from the normalization condition: ∫ f (x) dx = 1. Therefore, -∞ https://pandia.ru/text/78/455/images/image032_23.gif "height =" 38 src = "> -∞ 2 2 x if 2<х≤6, то F(x)= ∫ 0dx+∫ 1/8(х-2)dx=1/8(х2/2-2х) = 1/8(х2/2-2х - (4/2-4))= 1/8 (x2 / 2-2x + 2) = 1/16 (x-2) 2; Gif "width =" 14 "height =" 62 "> 0 at x≤2, F (x) = (x-2) 2/16 at 2<х≤6, 1 for x> 6. The graph of the function F (x) is shown in Fig. 3 https://pandia.ru/text/78/455/images/image034_23.gif "width =" 14 "height =" 62 src = "> 0 at x≤0, F (x) = (3 arctan x) / π at 0<х≤√3, 1 for x> √3. Find the Differential Distribution Function f (x) Solution:
Since f (x) = F ’(x), then https://pandia.ru/text/78/455/images/image011_36.jpg "width =" 118 "height =" 24 "> All properties of mathematical expectation and variance, considered earlier for dispersed random variables, are also valid for continuous ones. Problem number 3. The random variable X is given by the differential function f (x): https://pandia.ru/text/78/455/images/image036_19.gif "height =" 38 "> -∞ 2 X3 / 9 + x2 / 6 = 8 / 9-0 + 9 / 6-4 / 6 = 31/18, https://pandia.ru/text/78/455/images/image032_23.gif "height =" 38 "> + ∞ D (X) = ∫ x2 f (x) dx- (M (x)) 2 = ∫ x2 x / 3 dx + ∫1 / 3x2 dx = (31/18) 2 = x4 / 12 + x3 / 9 - - (31/18)2=16/12-0+27/9-8/9-(31/18)2=31/9- (31/18)2==31/9(1-31/36)=155/324, https://pandia.ru/text/78/455/images/image032_23.gif "height =" 38 "> P (1<х<5)= ∫ f(x)dx=∫ х/3 dx+∫ 1/3 dx+∫ 0 dx= х2/6 +1/3х = 4/6-1/6+1-2/3=5/6. Tasks for an independent solution.
2.1.
A continuous random variable X is given by the distribution function: 0 at x≤0, F (x) = https://pandia.ru/text/78/455/images/image038_17.gif "width =" 14 "height =" 86 "> 0 at x≤ π / 6, F (x) = - cos 3x at π / 6<х≤ π/3, 1 for x> π / 3. Find the differential distribution function f (x), and also P (2π / 9<Х< π /2). 2.3.
0 at x≤2, f (x) = c x at 2<х≤4, 0 for x> 4. 2.4.
A continuous random variable X is given by the distribution density: 0 at x≤0, f (х) = с √х at 0<х≤1, 0 for x> 1. Find: a) the number c; b) M (X), D (X). 2.5.
https://pandia.ru/text/78/455/images/image041_3.jpg "width =" 36 "height =" 39 "> at x, 0 at x. Find: a) F (x) and plot its graph; b) M (X), D (X), σ (X); c) the probability that in four independent tests the value of X will take exactly 2 times the value belonging to the interval (1; 4). 2.6.
The density of the probability distribution of a continuous random variable X is given: f (x) = 2 (x-2) at x, 0 at x. Find: a) F (x) and plot its graph; b) M (X), D (X), σ (X); c) the probability that in three independent tests the value of X will take exactly 2 times the value belonging to the segment. 2.7.
The function f (x) is given in the form: https://pandia.ru/text/78/455/images/image045_4.jpg "width =" 43 "height =" 38 src = ">. jpg" width = "16" height = "15"> [- √ 3/2; √3 / 2]. 2.8.
The function f (x) is given as: https://pandia.ru/text/78/455/images/image046_5.jpg "width =" 45 "height =" 36 src = "> .jpg" width = "16" height = "15"> [- π /4 ; π / 4]. Find: a) the value of the constant c, at which the function will be the probability density of some random variable X; b) the distribution function F (x). 2.9.
The random variable X, concentrated on the interval (3; 7), is given by the distribution function F (x) =. Find the probability that the random variable X will take on the value: a) less than 5, b) not less than 7. 2.10.
Random variable X, concentrated on the interval (-1; 4), given by the distribution function F (x) =. Find the probability that random variable X will take on the value: a) less than 2, b) not less than 4. 2.11.
https://pandia.ru/text/78/455/images/image049_6.jpg "width =" 43 "height =" 44 src = "> .jpg" width = "16" height = "15">. Find: a) the number c; b) M (X); c) the probability P (X> M (X)). 2.12.
The random variable is given by the differential distribution function: https://pandia.ru/text/78/455/images/image050_3.jpg "width =" 60 "height =" 38 src = ">. jpg" width = "16 height = 15" height = "15"> ... Find: a) M (X); b) probability P (X≤M (X)) 2.13.
The Remy distribution is given by the probability density: https://pandia.ru/text/78/455/images/image052_5.jpg "width =" 46 "height =" 37 "> at x ≥0. Prove that f (x) is indeed the density of the probability distribution. 2.14.
The density of the probability distribution of a continuous random variable X is given: https://pandia.ru/text/78/455/images/image054_3.jpg "width =" 174 "height =" 136 src = "> (Fig. 4) 2.16.
The random variable X is distributed according to the law “ right triangle»In the interval (0; 4) (Fig. 5). Find an analytical expression for the probability density f (x) on the entire number axis. Answers
0 at x≤0, f (x) = https://pandia.ru/text/78/455/images/image038_17.gif "width =" 14 "height =" 86 "> 0 at x≤ π / 6, F (x) = 3sin 3x at π / 6<х≤ π/3, 0 for x> π / 3. A continuous random variable X has a uniform distribution law on some interval (a; b), to which all possible values of X belong, if the probability distribution density f (x) is constant on this interval and is equal to 0 outside it, i.e. 0 for х≤а, f (x) = for a<х 0 for x≥b. The graph of the function f (x) is shown in Fig. 1 F (x) = https://pandia.ru/text/78/455/images/image077_3.jpg "width =" 30 "height =" 37 ">, D (X) =, σ (X) =. Problem number 1. The random variable X is uniformly distributed over the segment. Find: a) the probability distribution density f (x) and build its graph; b) the distribution function F (x) and plot its graph; c) M (X), D (X), σ (X). Solution:
Using the formulas considered above, for a = 3, b = 7, we find: https://pandia.ru/text/78/455/images/image081_2.jpg "width =" 22 "height =" 39 "> at 3≤x≤7, 0 for x> 7 Let's build its graph (Fig. 3): https://pandia.ru/text/78/455/images/image038_17.gif "width =" 14 "height =" 86 src = "> 0 at x≤3, F (x) = https://pandia.ru/text/78/455/images/image084_3.jpg "width =" 203 "height =" 119 src = "> fig. 4 D (X) = == https: //pandia.ru/text/78/455/images/image089_1.jpg "width =" 37 "height =" 43 "> == https: //pandia.ru/text/ 78/455 / images / image092_10.gif "width =" 14 "height =" 49 src = "> 0 at x<0, f (х) = λе-λх for х≥0. The distribution function of a random variable X, distributed according to the exponential law, is given by the formula: https://pandia.ru/text/78/455/images/image094_4.jpg "width =" 191 "height =" 126 src = "> pic..jpg" width = "22" height = "30">, D (X) =, σ (X) = Thus, the mathematical expectation and the standard deviation of the exponential distribution are equal to each other. The probability of hitting X in the interval (a; b) is calculated by the formula: P (a<Х Problem number 2. The mean time of failure-free operation of the device is 100 h. Assuming that the time of failure-free operation of the device has an exponential distribution law, find: a) the density of the probability distribution; b) distribution function; c) the probability that the device uptime will exceed 120 hours. Solution:
By condition, the mathematical distribution M (X) = https: //pandia.ru/text/78/455/images/image098_10.gif "height =" 43 src = "> 0 at x<0, a) f (x) = 0.01e -0.01x at x≥0. b) F (x) = 0 for x<0, 1- e -0.01x at x≥0. c) We find the desired probability using the distribution function: P (X> 120) = 1-F (120) = 1- (1- e -1.2) = e -1.2≈0.3. §
3.Normal distribution law Definition:
A continuous random variable X has normal distribution law (Gauss's law),
if its distribution density has the form: where m = M (X), σ2 = D (X), σ> 0. The curve of the normal distribution law is called normal or gaussian curve
(fig. 7) The distribution function of a random variable X, distributed according to the normal law, is expressed in terms of the Laplace function Ф (х) by the formula: where is the Laplace function. Comment:
The function Φ (x) is odd (Φ (-x) = - Φ (x)), in addition, for x> 5 we can assume Φ (x) ≈1 / 2. The graph of the distribution function F (x) is shown in Fig. eight https://pandia.ru/text/78/455/images/image106_4.jpg "width =" 218 "height =" 33 "> The probability that the absolute value of the deviation is less positive numberδ is calculated by the formula: In particular, for m = 0, the following equality is true: The Three Sigma Rule
If a random variable X has a normal distribution law with parameters m and σ, then it is practically certain that its value is contained in the interval (a-3σ; a + 3σ), since https://pandia.ru/text/78/455/images/image110_2.jpg "width =" 157 "height =" 57 src = "> a) b) Let's use the formula: https://pandia.ru/text/78/455/images/image112_2.jpg "width =" 369 "height =" 38 src = "> According to the table of values of the function Ф (х), we find Ф (1.5) = 0.4332, Ф (1) = 0.3413. So, the desired probability: P (28 Self-study tasks
3.1.
Random variable X is evenly distributed in the interval (-3; 5). Find: b) distribution functions F (x); c) numerical characteristics; d) the probability P (4<х<6). 3.2.
The random variable X is uniformly distributed over the segment. Find: a) distribution density f (x); b) distribution functions F (x); c) numerical characteristics; d) probability P (3≤x≤6). 3.3.
An automatic traffic light is installed on the highway, in which the green light is on for 2 minutes, yellow for 3 seconds and red for 30 seconds, etc. A car drives along the highway at a random time. Find the probability that the car will pass the traffic light without stopping. 3.4.
Metro trains run regularly every 2 minutes. The passenger enters the platform at a random time. What is the probability that the passenger will have to wait for the train more than 50 seconds. Find the mathematical expectation of a random variable X - the waiting time of the train. 3.5.
Find the variance and standard deviation of the exponential distribution given by the distribution function: F (x) = 0 for x<0, 1-e-8x at x≥0. 3.6.
A continuous random variable X is given by the density of the probability distribution: f (x) = 0 at x<0, 0.7 e-0.7x at x≥0. a) What is the distribution law of the considered random variable. b) Find the distribution function F (X) and the numerical characteristics of the random variable X. 3.7.
The random variable X is distributed according to the exponential law, given by the density of the probability distribution: f (x) = 0 at x<0, 0.4 e-0.4 x at x≥0. Find the probability that, as a result of the test, X will take a value from the interval (2.5; 5). 3.8.
A continuous random variable X is distributed according to the exponential law given by the distribution function: F (x) = 0 for x<0, 1-e-0.6x at x≥0 Find the probability that, as a result of the test, X will take a value from the segment. 3.9.
The mathematical expectation and standard deviation of the normally distributed random variable are, respectively, 8 and 2. Find: a) distribution density f (x); b) the probability that, as a result of the test, X will take a value from the interval (10; 14). 3.10.
The random variable X is normally distributed with a mathematical expectation of 3.5 and a variance of 0.04. Find: a) distribution density f (x); b) the probability that, as a result of the test, X will take the value from the segment. 3.11.
Random variable X is normally distributed with M (X) = 0 and D (X) = 1. Which of the events: | X | ≤0.6 or | X | ≥0.6 has the highest probability? 3.12.
Random variable X is distributed normally with M (X) = 0 and D (X) = 1 From what interval (-0.5; -0.1) or (1; 2) in one test it will take a value with a higher probability? 3.13.
The current price per share can be modeled using the normal distribution law with M (X) = 10den. units and σ (X) = 0.3 den. units Find: a) the probability that the current share price will be from 9.8 den. units up to 10.4 den. units; b) using the "three sigma rule" to find the boundaries in which the current stock price will be. 3.14.
The substance is weighed without systematic errors. Random weighing errors are subject to the normal law with the mean square ratio σ = 5g. Find the probability that in four independent experiments the error in three weighings will not occur in the absolute value of 3d. 3.15.
Random variable X is normally distributed with M (X) = 12.6. The probability of hitting a random variable in the interval (11.4; 13.8) is 0.6826. Find the standard deviation σ. 3.16.
Random variable X is normally distributed with M (X) = 12 and D (X) = 36. Find an interval in which random variable X will fall as a result of testing with a probability of 0.9973. 3.17.
A part made by an automatic machine is considered defective if the deviation X of its controlled parameter from the nominal exceeds the unit of measurement in modulus 2. It is assumed that the random variable X is distributed normally with M (X) = 0 and σ (X) = 0.7. What percentage of defective parts does the machine give out? 3.18.
The parameter X of the part is distributed normally with a mathematical expectation of 2 equal to the nominal value and a standard deviation of 0.014. Find the probability that the deviation of X from the nominal in absolute value does not exceed 1% of the nominal. Answers
https://pandia.ru/text/78/455/images/image116_9.gif "width =" 14 "height =" 110 src = "> b) 0 at x≤-3, F (x) = left "> 3.10.
a) f (x) =, b) P (3.1≤X≤3.7) ≈0.8185. 3.11.
| x | ≥0.6. 3.12.
(-0,5;-0,1). 3.13.
a) P (9.8≤X≤10.4) ≈0.6562. 3.14.
0,111. 3.15.
σ = 1.2. 3.16.
(-6;30). 3.17.
0,4%. As is known, random variable
a variable is called, which can take certain values, depending on the case. Random variables are designated by capital letters of the Latin alphabet (X, Y, Z), and their values - by the corresponding lowercase letters (x, y, z). Random variables are divided into discontinuous (discrete) and continuous. Discrete random variable
is a random variable that takes only a finite or infinite (countable) set of values with certain nonzero probabilities. The law of distribution of a discrete random variable
a function that connects the values of a random variable with the corresponding probabilities is called. The distribution law can be specified in one of the following ways. 1
. The distribution law can be given by the table:
where λ> 0, k = 0, 1, 2,…. v) by using distribution function F (x)
, which determines for each value of x the probability that the random variable X will take a value less than x, i.e. F (x) = P (X< x). Properties of the function F (x) 3
. The distribution law can be set graphically
- polygon (polygon) distribution (see task 3). Note that in order to solve some problems it is not necessary to know the distribution law. In some cases, it is enough to know one or several numbers that reflect the most important features of the distribution law. It can be a number that has the meaning of the "average value" of a random variable, or a number that shows the average deviation of a random variable from its average value. Numbers of this kind are called numerical characteristics of a random variable. Basic numerical characteristics of a discrete random variable
: 1000 lottery tickets were issued: 5 of them win 500 rubles, 10 - a win of 100 rubles, 20 - a win of 50 rubles, 50 - a win of 10 rubles. Determine the law of probability distribution of a random variable X - a payoff per ticket. Solution.
According to the condition of the problem, the following values of the random variable X are possible: 0, 10, 50, 100, and 500. The number of tickets without a win is 1000 - (5 + 10 + 20 + 50) = 915, then P (X = 0) = 915/1000 = 0.915. Similarly, we find all other probabilities: P (X = 0) = 50/1000 = 0.05, P (X = 50) = 20/1000 = 0.02, P (X = 100) = 10/1000 = 0.01 , P (X = 500) = 5/1000 = 0.005. We represent the resulting law in the form of a table: Let's find the mathematical expectation of the value X: M (X) = 1 * 1/6 + 2 * 1/6 + 3 * 1/6 + 4 * 1/6 + 5 * 1/6 + 6 * 1/6 = (1+ 2 + 3 + 4 + 5 + 6) / 6 = 21/6 = 3.5 The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Draw up a distribution law for the number of failed elements in one experiment, build a distribution polygon. Find the distribution function F (x) and plot its graph. Find the mathematical expectation, variance, and standard deviation of a discrete random variable. Solution.
1.
A discrete random variable X = (the number of failed elements in one experiment) has the following possible values: x 1 = 0 (none of the device's elements failed), x 2 = 1 (one element failed), x 3 = 2 (two elements failed ) and x 4 = 3 (three elements failed). Failures of elements are independent of each other, the probabilities of failure of each element are equal to each other, therefore it is applicable Bernoulli formula
... Considering that, by condition, n = 3, p = 0.1, q = 1-p = 0.9, we determine the probabilities of the values: Thus, the sought binomial distribution law for X has the form: On the abscissa, we plot the possible values of x i, and on the ordinate - the corresponding probabilities p i. Let's build the points M 1 (0; 0.729), M 2 (1; 0.243), M 3 (2; 0.027), M 4 (3; 0.001). Connecting these points by line segments, we obtain the desired distribution polygon. 3.
Let us find the distribution function F (x) = P (X Function graph F (x) 4.
For binomial distribution X:
1.2.
p4 = 0.1; 0 at x≤-1,
(fig. 5)
https://pandia.ru/text/78/455/images/image038_17.gif "width =" 14 "height =" 86 "> 0 at x≤a,
,
The normal curve is symmetric about the straight line x = m, has a maximum at m. X = a, equal to.
,
For the binomial distribution M (X) = np, for the Poisson distribution M (X) = λ
For the binomial distribution D (X) = npq, for the Poisson distribution D (X) = λExamples of solving problems on the topic "The law of distribution of a discrete random variable"
Objective 1.
Objective 3.
P 3 (0) = C 3 0 p 0 q 3-0 = q 3 = 0.9 3 = 0.729;
P 3 (1) = C 3 1 p 1 q 3-1 = 3 * 0.1 * 0.9 2 = 0.243;
P 3 (2) = C 3 2 p 2 q 3-2 = 3 * 0.1 2 * 0.9 = 0.027;
P 3 (3) = C 3 3 p 3 q 3-3 = p 3 = 0.1 3 = 0.001;
Check: ∑p i = 0.729 + 0.243 + 0.027 + 0.001 = 1.
for 0< x ≤1 имеем F(x) = Р(Х<1) = Р(Х = 0) = 0,729;
for 1< x ≤ 2 F(x) = Р(Х<2) = Р(Х=0) + Р(Х=1) =0,729+ 0,243 = 0,972;
for 2< x ≤ 3 F(x) = Р(Х<3) = Р(Х = 0) + Р(Х = 1) + Р(Х = 2) = 0,972+0,027 = 0,999;
for x> 3 will be F (x) = 1, since the event is valid. 
- mathematical expectation M (X) = np = 3 * 0.1 = 0.3;
- variance D (X) = npq = 3 * 0.1 * 0.9 = 0.27;
- standard deviation σ (X) = √D (X) = √0.27 ≈ 0.52.
