A law of distribution of a discrete random variable x is given. Theoretical material on the modules "probability theory and mathematical statistics"
LAW OF DISTRIBUTION AND CHARACTERISTICS
RANDOM VALUES
Random variables, their classification and methods of description.
Random is a quantity that, as a result of experience, can take on a particular value, but which one is not known in advance. For a random variable, thus, you can specify only values, one of which it will necessarily take as a result of the experiment. In what follows, these values will be called the possible values of the random variable. Since a random variable quantitatively characterizes the random result of an experiment, it can be considered as a quantitative characteristic of a random event.
Random variables are usually denoted by capital letters of the Latin alphabet, for example, X..Y..Z, and their possible values are indicated by the corresponding small letters.
There are three types of random variables:
Discrete; Continuous; Mixed.
Discrete such a random variable is called, the number of possible values of which forms a countable set. In turn, a set is called countable, the elements of which can be numbered. The word "discrete" comes from the Latin discretus, which means "intermittent, consisting of separate parts» .
Example 1. A discrete random variable is the number of defective parts X in a batch of ntuk. Indeed, the possible values of this random variable are a series of integers from 0 to n.
Example 2. A discrete random variable is the number of shots before the first hit on the target. Here, as in example 1, the possible values can be numbered, although in the limiting case the possible value is an infinitely large number.
Continuous a random variable is called, the possible values of which continuously fill a certain interval of the numerical axis, sometimes called the interval of existence of this random variable. Thus, on any finite interval of existence, the number of possible values of a continuous random variable is infinitely large.
Example 3. A continuous random variable is the energy consumption at the enterprise for a month.
Example 4. A continuous random variable is the altitude measurement error using an altimeter. Let it be known from the principle of operation of the altimeter that the error lies in the range from 0 to 2 m. Therefore, the interval of existence of this random variable is the interval from 0 to 2 m.
The law of distribution of random variables.
A random variable is considered to be completely given if its possible values are indicated on the numerical axis and the distribution law is established.
The distribution law of a random variable is called the relationship that establishes the relationship between the possible values of a random variable and the corresponding probabilities.
A random variable is said to be distributed according to a given law, or is subject to a given distribution law. A number of probabilities are used as distribution laws, distribution function, probability density, characteristic function.
The distribution law gives a complete probable description of the random variable. According to the distribution law, it is possible to judge before experiment which possible values of a random variable will appear more often, and which - less often.
For a discrete random variable, the distribution law can be specified in the form of a table, analytically (in the form of a formula) and graphically.
The simplest form for setting the law of distribution of a discrete random variable is a table (matrix), which lists in ascending order all possible values of the random variable and their corresponding probabilities, i.e.
Such a table is called a distribution series of a discrete random variable. 1
Events X 1, X 2, ..., X n, consisting in the fact that as a result of the test, a random variable X will take the values x 1, x 2, ... x n, respectively, are incompatible and the only possible (because the table lists all possible values of the random variable), i.e. form a complete group. Consequently, the sum of their probabilities is 1. Thus, for any discrete random variable
(This unit is somehow distributed between the values of the random variable, hence the term "distribution").
The distribution series can be depicted graphically if the values of the random variable are plotted along the abscissa axis, and their corresponding probabilities are plotted along the ordinate axis. The connection of the obtained points forms a polyline, called a polygon or a polygon of the probability distribution (Fig. 1).
Example The lottery draws: a car worth 5,000 den. units, 4 TVs worth 250 den. units, 5 video recorders worth 200 den. units A total of 1000 tickets are sold for 7 days. units Draw up the distribution law of the net winnings received by the lottery participant who bought one ticket.
Solution... Possible values of the random variable X - the net winnings per ticket - are 0-7 = -7 den. units (if the ticket is not won), 200-7 = 193, 250-7 = 243, 5000-7 = 4993 den. units (if the ticket contains 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 get.
Examples of solving problems on the topic "Random variables".
Task 1 ... There are 100 tickets issued in the lottery. One win of 50 USD was played. and ten wins of $ 10 each. Find the distribution law of the value X - the cost of the possible gain.
Solution. Possible values of X: x 1 = 0; x 2 = 10 and x 3 = 50. Since there are 89 "empty" tickets, then p 1 = 0.89, the probability of winning is $ 10. (10 tickets) - p 2 = 0.10 and for a win of 50 USD. - p 3 = 0.01. Thus:
|
0,89 |
0,10 |
0,01 |
Easy to control:.
Task 2. The probability that the buyer has read the advertisement of the product in advance is 0.6 (p = 0.6). A selective control of the quality of advertising is carried out by interviewing buyers before the first one who has studied the advertisement in advance. Draw up a series of distributions of the number of surveyed buyers.
Solution. According to the condition of the problem, p = 0.6. Where: q = 1 -p = 0.4. Substituting these values, we get: and build a distribution series:
|
p i |
0,24 |
Task 3. A computer consists of three independently operating elements: a system unit, a monitor and a keyboard. With a single sharp increase in voltage, the probability of failure of each element is 0.1. Based on the Bernoulli distribution, draw up the distribution law for the number of failed elements during a voltage surge in the network.
Solution. Consider Bernoulli distribution(or binomial): the probability that in n trials event A will appear exactly k once: 
, or:
|
q n |
p n |
V Let's get back to the problem.
Possible values of X (number of failures):
x 0 = 0 - none of the elements failed;
x 1 = 1 - failure of one element;
x 2 = 2 - failure of two elements;
x 3 = 3 - failure of all elements.
Since, by condition, p = 0.1, then q = 1 - p = 0.9. Using Bernoulli's formula, we get
, ,
, .
Control: .
Therefore, the sought distribution law:
|
0,729 |
0,243 |
0,027 |
0,001 |
Problem 4... Produced 5,000 rounds. Probability that one cartridge is defective 
... What is the probability that there will be exactly 3 defective cartridges in the whole batch?
Solution. Applicable Poisson distribution: This distribution is used to determine the probability that for a very large
the number of trials (mass trials), in each of which the probability of event A is very small, event A will occur k times: 
, where .
Here n = 5000, p = 0.0002, k = 3. We find, then the desired probability: 
.
Problem 5... When shooting before the first hit with hit probability p = 0.6 when firing, you need to find the probability that the hit will occur on the third shot.
Solution. We apply a geometric distribution: let independent tests be carried out, in each of which event A has the probability of occurrence p (and non-occurrence q = 1 - p). Trials end as soon as event A.
Under such conditions, the probability that event A will occur on the k-th test is determined by the formula:. Here p = 0.6; q = 1 - 0.6 = 0.4; k = 3. Therefore,.
Problem 6... Let the distribution law of a random variable X be given:
Find expected value.
Solution. ...
Note that the probabilistic meaning of the mathematical expectation is the average value of a random variable.
Problem 7... Find the variance of a random variable X with the following distribution law:
Solution. Here .
The distribution law of the square of the quantity X 2 :
|
X 2 |
|||
The desired variance:.
Dispersion characterizes the measure of deviation (dispersion) of a random variable from its mathematical expectation.
Problem 8... Let the random variable be given by the distribution:
|
10m |
|||
Find its numerical characteristics.
Solution: m, m 2 ,
M 2 , m.
About a random variable X, one can say either - its mathematical expectation is 6.4 m with a variance of 13.04 m 2 , or - its mathematical expectation is 6.4 m with a deviation of m. The second formulation is obviously clearer.
Task 9.
Random value X given by the distribution function: 
.
Find the probability that, as a result of the test, the value of X will take the value enclosed in the interval 
.
Solution. The probability that X will take a value from a given interval is equal to the increment of the integral function in this interval, i.e. ... In our case and, therefore
.
Task 10. Discrete random variable X given by the distribution law:
Find the distribution function F (x ) and plot its graph.
Solution. Since the distribution function,
for 
, then
at ;
at ;
at ;
at ;
Corresponding graph:
Problem 11. Continuous random variable X given by the differential distribution function: 
.
Find the probability of hit X in the interval
Solution. Note that this is a special case of the exponential distribution law.
Let's use the formula: 
.
Task 12. Find the numerical characteristics of a discrete random variable X given by the distribution law:
|
–5 |
|||||||||
|
X 2:
|
Is the Laplace function.
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 on 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 the 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 will take 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 value;
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 felt-tip pens 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 in succession for reliability. 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 a 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 graph of the density of the probability distribution 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 plot 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.
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 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 x≤a, 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 uptime of the device 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 a green light is on for 2 minutes, yellow for 3 seconds and red for 30 seconds, etc. The 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 more than 50 seconds for the train? 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 8 and 2, respectively. 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.
Random variable X is normally distributed with mathematical expectation 3.5 and variance 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 distributed normally 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 distributed normally 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 normally distributed 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 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 axis we plot the possible values of x i, and on the ordinate axis - 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: Definition 1 A random variable $ X $ is called discrete (discontinuous) if the set of its values is infinite or finite, but countable. In other words, a quantity is called discrete if its values can be numbered. You can describe a random variable using the distribution law. The distribution law of the 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: Picture 1. where $ p1 + p2 + ... + pn = 1 $. This table is distribution of a discrete random variable. If the set of possible values of the random variable is infinite, then the series $ р1 + р2 + ... + рn + ... $ converges and its sum will be equal to $ 1 $. The distribution law of a discrete random variable $ X $ can be represented graphically, for which a polyline is constructed in a (rectangular) coordinate system, which sequentially connects points with coordinates $ (xi; pi), i = 1,2, ... n $. The line that got called distribution polygon. Figure 2. The distribution law of a discrete random variable $ X $ can also be represented analytically (using the formula): $ P (X = xi) = \ varphi (xi), i = 1,2,3 ... n $. When solving many problems of the theory of probability, it is necessary to carry out the operations of multiplying a discrete random variable by a constant, adding two random variables, multiplying them, and subtracting them to a degree. In these cases, it is necessary to adhere to the following rules over random discrete quantities: Definition 3 By multiplication a discrete random variable $ X $ on a constant $ K $ is a discrete random variable $ Y = KX, $ which is determined by the equalities: $ y_i = Kx_i, \ \ p \ left (y_i \ right) = p \ left (x_i \ right) = p_i, \ \ i = \ overline (1, \ n). $ Definition 4 Two random variables $ x $ and $ y $ are called independent, if the distribution law of one of them does not depend on what possible values the second quantity has acquired. Definition 5 The sum two independent discrete random variables $ X $ and $ Y $ are called the random variable $ Z = X + Y, $ is determined by the equalities: $ z_ (ij) = x_i + y_j $, $ P \ left (z_ (ij) \ right) = P \ left (x_i \ right) P \ left (y_j \ right) = p_ip "_j $, $ i = \ overline (1, n) $, $ j = \ overline (1, m) $, $ P \ left (x_i \ right) = p_i $, $ P \ left (y_j \ right) = p "_j $. Definition 6 By multiplication two independent discrete random variables $ X $ and $ Y $ are called the random variable $ Z = XY, $ is determined by the equalities: $ z_ (ij) = x_iy_j $, $ P \ left (z_ (ij) \ right) = P \ left ( x_i \ right) P \ left (y_j \ right) = p_ip "_j $, $ i = \ overline (1, n) $, $ j = \ overline (1, m) $, $ P \ left (x_i \ right ) = p_i $, $ P \ left (y_j \ right) = p "_j $. Let's take into account that some products $ x_ (i \ \ \ \ \) y_j $ may be equal to each other. In this case, the probability of adding the product is equal to the sum of the corresponding probabilities. For example, if $ x_2 \ \ y_3 = x_5 \ \ y_7, \ $ then the probability of $ x_2y_3 $ (or the same $ x_5y_7 $) will be $ p_2 \ cdot p "_3 + p_5 \ cdot p" _7. $ The above also applies to the amount. If $ x_1 + \ y_2 = x_4 + \ \ y_6, $ then the probability $ x_1 + \ y_2 $ (or the same $ x_4 + \ y_6 $) will be $ p_1 \ cdot p "_2 + p_4 \ cdot p" _6. $ Letnm random variables $ X $ and $ Y $ are given by distribution laws: Figure 3. Where $ p_1 + p_2 + p_3 = 1, \ \ \ p "_1 + p" _2 = 1. $ Then the distribution law of the sum $ X + Y $ will have the form Figure 4. And the distribution law of the product $ XY $ will have the form Figure 5. The distribution function also gives a complete description of the random variable. Geometrically, the distribution function is explained as the probability that the random variable $ X $ takes on a value that is represented on the number line by a point lying on the left side of the point $ 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 with respect to 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.
Actions on discrete probabilities
Distribution function
