One of the most important concepts in probability theory is the concept random variable.
Random called size, which, as a result of testing, accepts certain possible values, unknown in advance and depending on random reasons that cannot be taken into account in advance.
Random variables are designated by capital letters of the Latin alphabet X, Y, Z etc. or in capital letters of the Latin alphabet with a right lower index, and the values that random variables can take - in the corresponding small letters of the Latin alphabet x, y, z etc.
The concept of a random variable is closely related to the concept of a random event. Connection with a random event lies in the fact that accepting a random variable of some numerical value is a random event characterized by probability 
.
In practice, there are two main types of random variables:
1. Discrete random variables;
2. Continuous random variables.
A random variable is a numerical function of random events.
For example, a random variable is the number of points obtained when throwing dice, or the height of a student randomly selected from the study group.
Discrete random variables are called random variables that take only values distant from each other that can be listed in advance.
Law of distribution(distribution function and distribution series or probability density) completely describe the behavior of a random variable. But in a number of problems, it is enough to know some numerical characteristics of the quantity under study (for example, its average value and possible deviation from it) in order to answer the question posed. Let's consider the main numerical characteristics of discrete random variables.
Distribution law of a discrete random variable every relation is called , establishing a connection between possible values of a random variable and their corresponding probabilities .
The distribution law of a random variable can be represented as tables:
| … | … | ||||
| … | … |
The sum of the probabilities of all possible values of a random variable is equal to one, i.e.
The distribution law can be depicted graphically: the possible values of a random variable are plotted along the abscissa axis, and the probabilities of these values are plotted along the ordinate axis; the resulting points are connected by segments. The constructed polyline is called distribution polygon.
Example. A hunter with 4 cartridges shoots at the game until he makes the first hit or uses up all the cartridges. The probability of hitting on the first shot is 0.7, with each subsequent shot it decreases by 0.1. Draw up a distribution law for the number of cartridges spent by a hunter.
Solution. Since a hunter, having 4 cartridges, can fire four shots, then the random variable X- the number of cartridges spent by the hunter can take values 1, 2, 3, 4. To find the corresponding probabilities, we introduce the events:
- “hit with i- oh shot”, ;
- “miss when i- om shot”, and the events and are pairwise independent.
According to the problem conditions we have:
, 
Using the multiplication theorem for independent events and the addition theorem for incompatible events, we find:
(the hunter hit the target with the first shot);
(the hunter hit the target with the second shot);
(the hunter hit the target with the third shot);
(the hunter hit the target with the fourth shot or missed all four times).
Check: - true.
Thus, the law of distribution of a random variable X has the form:
| 0,7 | 0,18 | 0,06 | 0,06 |
Example. A worker operates three machines. The probability that within an hour the first machine will not require adjustment is 0.9, the second - 0.8, the third - 0.7. Draw up a distribution law for the number of machines that will require adjustment within an hour.
Solution. Random value X- the number of machines that will require adjustment within an hour can take values 0.1, 2, 3. To find the corresponding probabilities, we introduce the events:
- “i- the machine will require adjustment within an hour,” ;
- “i- the machine will not require adjustment within an hour,” .
According to the conditions of the problem we have:
, 
.
On this page we have collected a brief theory and examples of solving educational problems in which a discrete random variable is already specified by its distribution series (tabular form) and it is required to study it: find numerical characteristics, construct graphs, etc. Examples on known species distributions can be found at the following links:
Brief theory about DSV
A discrete random variable is specified by its distribution series: a list of values $x_i$ that it can take and the corresponding probabilities $p_i=P(X=x_i)$. The number of values of a random variable can be finite or countable. For definiteness, we will consider the case $i=\overline(1,n)$. Then the tabular representation of the discrete random variable has the form:
$$ \begin(array)(|c|c|) \hline X_i & x_1 & x_2 & \dots & x_n \\ \hline p_i & p_1 & p_2 & \dots & p_n \\ \hline \end(array) $ $
In this case, the normalization condition is satisfied: the sum of all probabilities must be equal to one
$$\sum_(i=1)^(n) p_i=1$$
Graphically, the distribution series can be represented distribution polygon(or distribution polygon). To do this, points with coordinates $(x_i,p_i)$ are plotted on the plane and connected in order by a broken line. You will find detailed examples.
Numerical characteristics of DSV
Expected value:
$$M(X) = \sum_(i=1)^(n) x_i \cdot p_i$$
Dispersion:
$$ D(X)=M(X^2)-(M(X))^2 = \sum_(i=1)^(n) x_i^2 \cdot p_i - (M(X))^2$ $
Standard deviation:
$$\sigma (X) = \sqrt(D(X))$$
The coefficient of variation:
$$V(X) = \frac(\sigma(X))(M(X))$$.
Mode: value $Mo=x_k$ with the highest probability $p_k=\max_i(p_i)$.
You can use online calculators to calculate the expected value, variance, and standard deviation of the DSV.
DSV distribution function
From the distribution series one can compile distribution function discrete random variable $F(x)=P(X\lt x)$. This function specifies the probability that the random variable $X$ will take a value less than a certain number $x$. Examples of construction with detailed calculations and graphs you will find in the examples below.
Examples of solved problems
Task 1. A discrete random variable is specified by a distribution series:
1 2 3 4 5 6 7
0,05 0,15 0,3 0,2 0,1 0,04 0,16
Construct a distribution polygon and distribution function $F(x)$. Calculate: $M[X], D[X], \sigma[X]$, as well as the coefficient of variation, skewness, kurtosis, mode and median.
Task 2. The law of distribution of a discrete random variable X is given. Required:
a) determine expected value M(x), variance D(x) and standard deviation (x) of the random variable X; b) construct a graph of this distribution.
xi 0 1 2 3 4 5 6
pi 0.02 0.38 0.30 0.16 0.08 0.04 0.02
Task 3. For a random variable X with a given distribution series
-1 0 1 8
0.2 0.1 $р_1$ $р_2$
A) find $p_1$ and $p_2$ so that $M(X)=0.5$
B) after this, calculate the mathematical expectation and variance of the random variable $X$ and plot its distribution function
Task 4. Discrete SV $X$ can take only two values: $x_1$ and $x_2$, and $x_1 \lt x_2$. The probability $P$ of a possible value, the mathematical expectation $M(x)$ and the variance $D(x)$ are known. Find: 1) The distribution law of this random variable; 2) SV distribution function $X$; 3) Construct a graph of $F(x)$.
$P=0.3; M(x)=6.6; D(x)=13.44.$
Task 5. The random variable X takes three values: 2, 4 and 6. Find the probabilities of these values if $M(X)=4.2$, $D(X)=1.96$.
Task 6. A series of distribution of discrete r.v. is given. $X$. Find the numerical characteristics of the position and dispersion of r.v. $X$. Find m.o. and dispersion r.v. $Y=X/2-2$, without writing down the r.v. distribution series. $Y$, check the result using the generating function.
Construct the r.v. distribution function. $X$.
¦ x¦ 8 ¦ 12 ¦ 18 ¦ 24 ¦ 30 ¦
¦ p¦ 0.3¦ 0.1¦ 0.3¦ 0.2¦ 0.1¦
Task 7. The distribution of a discrete random variable $X$ is given by the following table (distribution row):
-6 3 9 15
0,40 0,30 ? 0,10
Determine the missing value in the distribution table. Calculate the main numerical characteristics of the distribution: $M_x, D_x, \sigma_x$. Find and construct the distribution function $F(x)$. Determine the probability that the random variable $X$ will take the following values:
A) more than 6,
B) less than 12,
C) no more than 9.
Task 8. The problem requires finding: a) mathematical expectation; b) dispersion; c) the standard deviation of a discrete random variable X according to a given law of its distribution, given in a table (the first row of the table indicates possible values, the second row indicates the probabilities of possible values).
Task 9. The distribution law of a discrete random variable $X$ is given (the first line shows possible values of $x_i$, the second line shows the probabilities of possible values of $p_i$).
Find:
A) mathematical expectation $M(X)$, variance $D(X)$ and standard deviation $\sigma(X)$;
B) compose the distribution function of the random variable $F(x)$ and construct its graph;
C) calculate the probability of a random variable $X$ falling into the interval $x_2 \lt X \lt x_4$, using the compiled distribution function $F(x)$;
D) draw up a distribution law for the value $Y=100-2X$;
D) calculate the mathematical expectation and variance of the compiled random variable $Y$ in two ways, i.e. taking advantage
property of mathematical expectation and dispersion, as well as directly according to the distribution law of the random variable $Y$.
10 20 30 40 50
0,1 0,2 0,1 0,2 0,4
Problem 10. A discrete random variable is given to a table. Calculate its initial and central moments up to 4th order inclusive. Find the probabilities of the events $\xi \lt M\xi$, $\xi \ge M \xi$, $\xi \lt 1/2 M \xi$, $\xi \ge 1/2 M \xi$.
X 0 0.3 0.6 0.9 1.2
P 0.2 0.4 0.2 0.1 0.1
Educational institution "Belarusian State
agricultural Academy"
Department of Higher Mathematics
Guidelines
to study the topic “Random Variables” by students of the Faculty of Accounting for Correspondence Education (NISPO)
Gorki, 2013
Random variables
Discrete and continuous random variables
One of the main concepts in probability theory is the concept random variable . Random variable is a quantity that, as a result of testing, takes only one of its many possible values, and it is not known in advance which one.
There are random variables discrete and continuous . Discrete random variable (DRV) is a random variable that can take on a finite number of values isolated from each other, i.e. if the possible values of this quantity can be recalculated. Continuous random variable (CNV) is a random variable, all possible values of which completely fill a certain interval of the number line.
Random variables are denoted by capital letters of the Latin alphabet X, Y, Z, etc. Possible values of random variables are indicated by the corresponding small letters.
Record 
means "the probability that a random variable X will take a value of 5, equal to 0.28.”
Example 1 . The dice are thrown once. In this case, numbers from 1 to 6 may appear, indicating the number of points. Let us denote the random variable X=(number of points rolled). This random variable as a result of the test can take only one of six values: 1, 2, 3, 4, 5 or 6. Therefore, the random variable X there is DSV.
Example 2 . When a stone is thrown, it travels a certain distance. Let us denote the random variable X=(stone flight distance). This random variable can take any, but only one, value from a certain interval. Therefore, the random variable X there is NSV.
Distribution law of a discrete random variable
A discrete random variable is characterized by the values it can take and the probabilities with which these values are taken. The correspondence between possible values of a discrete random variable and their corresponding probabilities is called law of distribution of a discrete random variable .
If all possible values are known 
random variable X and probabilities 
appearance of these values, then it is believed that the law of distribution of DSV X is known and can be written in table form:
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The DSV distribution law can be depicted graphically if points are depicted in a rectangular coordinate system 
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…,
and connect them with straight line segments. The resulting figure is called a distribution polygon.
Example 3 . Grain intended for cleaning contains 10% weeds. 4 grains were selected at random. Let us denote the random variable X=(number of weeds among the four selected). Construct the DSV distribution law X and distribution polygon.
Solution . According to the example conditions. Then:
Let's write down the distribution law of DSV X in the form of a table and construct a distribution polygon:
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Expectation of a discrete random variable
The most important properties of a discrete random variable are described by its characteristics. One of these characteristics is expected value random variable.
Let the DSV distribution law be known X:
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Mathematical expectation
DSV X is the sum of the products of each value of this quantity and the corresponding probability: 
.
The mathematical expectation of a random variable is approximately equal to the arithmetic mean of all its values. Therefore, in practical problems, the average value of this random variable is often taken as the mathematical expectation.
Example 8 . The shooter scores 4, 8, 9 and 10 points with probabilities of 0.1, 0.45, 0.3 and 0.15. Find the mathematical expectation of the number of points with one shot.
Solution . Let us denote the random variable X=(number of points scored). Then . Thus, the expected average number of points scored with one shot is 8.2, and with 10 shots - 82.
Main properties mathematical expectation are:
.
.
, Where 
,
.
.
, Where X And Y are independent random variables.
Difference 
called deviation
random variable X from its mathematical expectation. This difference is a random variable and its mathematical expectation is zero, i.e. 
.
Variance of a discrete random variable
To characterize a random variable, in addition to the mathematical expectation, we also use dispersion , which makes it possible to estimate the dispersion (spread) of the values of a random variable around its mathematical expectation. When comparing two homogeneous random variables with equal mathematical expectations, the “best” value is considered to be the one that has less spread, i.e. less dispersion.
Variance random variable X is called the mathematical expectation of the squared deviation of a random variable from its mathematical expectation: .
In practical problems, an equivalent formula is used to calculate the variance.
The main properties of the dispersion are:
.
Chapter 1. Discrete random variable
§ 1. Concepts of a random variable.
Distribution law of a discrete random variable.
Definition : Random is a quantity that, as a result of testing, takes only one value out of 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.
A random variable can be described using its distribution law.
Definition : Distribution law of a discrete random variable call the correspondence between 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 row of which all possible values of the random variable are indicated in ascending order, and in the second row the corresponding probabilities of these values, i.e.
where р1+ р2+…+ рn=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 broken line is constructed in a rectangular coordinate system, connecting sequentially 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">organic chemistry are 0.7 and 0.8, respectively. Draw up a distribution law for the 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:
https://pandia.ru/text/78/455/images/image004_81.jpg" width="259" height="66 src=">
 |
So, the distribution law of the random variable X is given by the table:
Control: 0.6+0.38+0.56=1.
§ 2. Distribution function
A complete description of a random variable is also given by the distribution function.
Definition: Distribution function of a discrete random variable X is called a function F(x), which determines for each value x the probability that the 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 represented on the number line by a point lying to the left of point x.
1)0≤ F(x) ≤1;
2) F(x) is a non-decreasing function on (-∞;+∞);
3) F(x) - continuous on the left at points x= xi (i=1,2,...n) and 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 x≤ x1,
р1 at x1< х≤ x2,
F(x)= р1 + р2 at x2< х≤ х3
1 for x>xn.
Its graph is shown in Fig. 2:
§ 3. Numerical characteristics of a discrete random variable.
One of the important numerical characteristics is the mathematical expectation.
Definition: Mathematical expectation M(X) discrete random variable X is the sum of the products of all its values and their corresponding probabilities:
M(X) = ∑ xiрi= x1р1 + x2р2+…+ xnрn
The mathematical expectation serves as a characteristic of the average value of a random variable.
Properties of mathematical expectation:
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 value;
To characterize the degree of dispersion of possible values of a discrete random variable around its mean value, dispersion is used.
Definition: Variance D ( X ) random variable X is the mathematical expectation of the squared 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 value;
4)D(X+Y)=D(X)+D(Y), where X, Y are independent random variables;
To calculate variance it is often convenient to use the formula:
D(X)=M(X2)-(M(X))2,
where M(X)=∑ xi2рi= x12р1 + x22р2+…+ xn2рn
The variance D(X) has the dimension of a squared random variable, which is not always convenient. Therefore, the value √D(X) is also used as an indicator of the dispersion of possible values of a random variable.
Definition: Standard deviation σ(X) random variable X is called the square root of the variance:
Task No. 2. The discrete random variable X is specified 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
Р2=1- (0.1+0.3+0.2+0.3)=0.1
Let's find the distribution function F(x)=P(X Geometrically, this equality can be interpreted as follows: F(x) is the probability that the random variable will take the value that is represented on the number axis by the point lying to the left of the point x. If x≤-1, then F(x)=0, since there is not a single value of this random variable on (-∞;x); 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) there are 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, because 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 at x>3 Let's 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 called the law of distribution of a discrete random variable X - the number of occurrences of event A in n independent repeated trials, in each of which event A may occur with probability p or not occur with probability q = 1-p. Then P(X=m) - the probability of occurrence of event A exactly m times in n trials is calculated using the Bernoulli formula: Р(Х=m)=Сmnpmqn-m The mathematical expectation, dispersion and standard deviation of a random variable X distributed according to a binary law are found, respectively, using the formulas: https://pandia.ru/text/78/455/images/image016_31.gif" width="26"> The probability of event A - “rolling out a five” in each trial is the same and equal to 1/6, i.e. . P(A)=p=1/6, then P(A)=1-p=q=5/6, where - “failure to get an A.” The random variable X can take the following values: 0;1;2;3. We find the probability of each of the possible values of X using Bernoulli’s formula: Р(Х=0)=Р3(0)=С03р0q3=1 (1/6)0 (5/6)3=125/216; Р(Х=1)=Р3(1)=С13р1q2=3 (1/6)1 (5/6)2=75/216; Р(Х=2)=Р3(2)=С23р2q =3 (1/6)2 (5/6)1=15/216; Р(Х=3)=Р3(3)=С33р3q0=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, Task No. 4. An 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; b) at least one is defective. Solution:
The number n=1000 is large, the probability of producing 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 holds: Рn(m)= e-
λ
λm Let's find λ=np=1000 0.002=2. a) Find the probability that there will be 5 defective parts (m=5): Р1000(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.
The dispersed random variable X is specified 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, 2 of which no longer write. Take 3 markers at random. Random variable X is the number of writing markers among those taken. Draw up a law of distribution of a random variable. 1.4.
There are 6 textbooks randomly arranged on a library shelf, 4 of which are bound. The librarian takes 4 textbooks at random. Random variable X is the number of bound textbooks among those taken. Draw up a law of distribution of a random variable. 1.5.
There are two tasks on the ticket. Probability the right decision the first problem is 0.9, the second is 0.7. Random variable X is the number of correctly solved problems in the ticket. Draw up a distribution law, calculate the mathematical expectation and variance of this random variable, and also find the distribution function F(x) and build its graph. 1.6.
Three shooters are shooting at a target. The probability of hitting the target with one shot is 0.5 for the first shooter, 0.8 for the second, and 0.7 for the third. Random variable X is the number of hits on the target if the shooters fire one shot at a time. Find the distribution law, M(X),D(X). 1.7.
A basketball player throws the ball into the basket with a probability of hitting each shot of 0.8. For each hit, he receives 10 points, and if he misses, no points are awarded to him. Draw up a distribution law for the random variable X - the number of points received by a basketball player in 3 shots. Find M(X),D(X), as well as the probability that he gets more than 10 points. 1.8.
Letters are written on the cards, a total of 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 a distribution law and find M(X),D(X),σ(X). 1.9.
On average, 60% of contracts Insurance Company pays insurance amounts in connection with the occurrence of an insured event. Draw up a distribution law for the random variable X - the number of contracts for which the insurance amount was paid among four contracts selected at random. Find the numerical characteristics of this quantity. 1.10.
The radio station sends call signs (no more than four) at certain intervals until two-way communication is established. The probability of receiving a response to a call sign is 0.3. Random variable X is the number of call signs sent. Draw up a distribution law and find F(x). 1.11.
There are 3 keys, of which only one fits the lock. Draw up a law for the distribution of the 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.
Consecutive independent tests of three devices are carried out for reliability. Each subsequent device is tested only if the previous one turned out to be reliable. The probability of passing the test for each device is 0.9. Draw up a distribution law for the random variable X-number of tested devices. 1.13
.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 time T is 0.002. The elements work independently. Find the probability that no more than two elements will fail during time T. 1.15.
The textbook was published in a circulation of 50,000 copies. The probability that the textbook is bound incorrectly 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 the following will arrive: 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 at 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 at 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
is a quantity whose all possible values completely fill a finite or infinite span of the number line. 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
a continuous random variable X is called a function F(x), which determines for each value 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. Properties of the distribution function:
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 falling into 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. R(a)<Х 4) The probability that a continuous random variable X will take one separate value is 0. 5) F(-∞)=0, F(+∞)=1 Specifying a continuous random variable using a distribution function is not the only way. Let us introduce the concept of probability distribution density (distribution density). Definition
:
Probability distribution density
f
(
x
)
of a continuous random variable X is the derivative of its distribution function, i.e.: The probability density function is sometimes called the differential distribution function or differential distribution law. The graph of the probability density distribution f(x) is called probability distribution curve
.
Properties of probability density distribution:
1) f(x) ≥0, at xhttps://pandia.ru/text/78/455/images/image029_10.jpg" width="285" height="141">.gif" width="14" height ="62 src="> 0 at x≤2, f(x)= c(x-2) at 2<х≤6, 0 for x>6. Find: a) the value of c; b) distribution function F(x) and plot it; 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 dispersion, discussed earlier for dispersed random variables, are also valid for continuous ones. Task No. 3. Random variable X is given 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. Problems for independent solution.
2.1.
A continuous random variable X is specified 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 for 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 Р(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 specified by the distribution density: 0 at x≤0, f(x)= c √x at 0<х≤1, 0 for x>1. Find: a) 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 construct its graph; b) M(X),D(X), σ(X); c) the probability that in four independent trials the value of X will take exactly 2 times the value belonging to the interval (1;4). 2.6.
The probability distribution density of a continuous random variable X is given: f(x)= 2(x-2) at x, 0 at x. Find: a) F(x) and construct its graph; b) M(X),D(X), σ (X); c) the probability that in three independent trials the value of X will take exactly 2 times the value belonging to the segment . 2.7.
The function f(x) is given as: 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) distribution function F(x). 2.9.
The random variable X, concentrated on the interval (3;7), is specified by the distribution function F(x)= . Find the probability that random variable X will take the value: a) less than 5, b) not less than 7. 2.10.
Random variable X, concentrated on the interval (-1;4), is given by the distribution function F(x)= . Find the probability that random variable X will take 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) number c; b) M(X); c) probability P(X> M(X)). 2.12.
The random variable is specified 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 Rem distribution is given by the probability density: https://pandia.ru/text/78/455/images/image052_5.jpg" width="46" height="37"> for x ≥0. Prove that f(x) is indeed a probability density function. 2.14.
The probability distribution density 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 line. Answers
0 at x≤0, f(x)= https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for 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 a certain interval (a;b), which contains all possible values of X, if the probability distribution density f(x) is constant on this interval and 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)=. Task No. 1. The random variable X is uniformly distributed on the segment. Find: a) probability distribution density f(x) and plot it; b) the distribution function F(x) and plot it; c) M(X),D(X), σ(X). Solution:
Using the formulas discussed above, with a=3, b=7, we find: https://pandia.ru/text/78/455/images/image081_2.jpg" width="22" height="39"> at 3≤х≤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(x)= λе-λх for x≥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=">fig..jpg" width="22" height="30"> , D(X)=, σ (Х)= Thus, the mathematical expectation and the standard deviation of the exponential distribution are equal to each other. The probability of X falling into the interval (a;b) is calculated by the formula: P(a<Х Task No. 2. The average failure-free operation time of the device is 100 hours. Assuming that the failure-free operation time of the device has an exponential distribution law, find: a) probability distribution density; b) distribution function; c) the probability that the device’s failure-free operation time will exceed 120 hours. Solution:
According to the 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 for x≥0. b) F(x)= 0 at 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 law distributions (Gauss's law),
if its distribution density has the form: where m=M(X), σ2=D(X), σ>0. The normal distribution curve is called normal or Gaussian curve
(Fig.7) The distribution function of a random variable X, distributed according to the normal law, is expressed through the Laplace function Ф (x) according to the formula: where is the Laplace function. Comment:
The function Ф(x) is odd (Ф(-х)=-Ф(х)), in addition, for x>5 we can assume Ф(х) ≈1/2. The graph of the distribution function F(x) is shown in Fig. 8 https://pandia.ru/text/78/455/images/image106_4.jpg" width="218" height="33"> The probability that absolute value deviations less than a positive number δ are calculated by the formula: In particular, for m=0 the following equality holds: "Three Sigma Rule"
If a random variable X has a normal distribution law with parameters m and σ, then it is almost certain that its value lies in the interval (a-3σ; a+3σ), because 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="> From the table of function values Ф(х) we find Ф(1.5)=0.4332, Ф(1)=0.3413. So, the desired probability: P(28 Tasks for independent work
3.1.
The random variable X is uniformly distributed in the interval (-3;5). Find: b) distribution function F(x); c) numerical characteristics; d) probability P(4<х<6). 3.2.
The random variable X is uniformly distributed on the segment. Find: a) distribution density f(x); b) distribution function F(x); c) numerical characteristics; d) probability P(3≤х≤6). 3.3.
There is an automatic traffic light on the highway, in which the green light is on for 2 minutes, yellow for 3 seconds, red for 30 seconds, etc. A car drives along the highway at a random moment in time. Find the probability that a car will pass a traffic light without stopping. 3.4.
Subway trains run regularly at intervals of 2 minutes. A passenger enters the platform at a random time. What is the probability that a passenger will have to wait more than 50 seconds for a train? Find the mathematical expectation of the random variable X - the waiting time for the train. 3.5.
Find the variance and standard deviation of the exponential distribution given by the distribution function: F(x)= 0 at x<0, 1st-8x for x≥0. 3.6.
A continuous random variable X is specified by the probability distribution density: f(x)= 0 at x<0, 0.7 e-0.7x at x≥0. a) Name the distribution law of the random variable under consideration. 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 specified by the probability distribution density: 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 specified by the distribution function: F(x)= 0 at x<0, 1st-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 expected value and standard deviation of a 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.
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 a value from the segment . 3.11.
The 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 is more likely? 3.12.
The random variable X is distributed normally with M(X)=0 and D(X)=1. From which interval (-0.5;-0.1) or (1;2) is it more likely to take a value during one test? 3.13.
The current price per share can be modeled using the normal distribution law with M(X)=10 den. 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 days units; b) using the “three sigma rule”, find the boundaries within which the current stock price will be located. 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 an error in three weighings will not occur in absolute value 3r. 3.15.
The random variable X is normally distributed with M(X)=12.6. The probability of a random variable falling into the interval (11.4;13.8) is 0.6826. Find the standard deviation σ. 3.16.
The random variable X is distributed normally with M(X)=12 and D(X)=36. Find the interval into which the random variable X will fall as a result of the test with a probability of 0.9973. 3.17.
A part manufactured by an automatic machine is considered defective if the deviation X of its controlled parameter from the nominal value exceeds modulo 2 units of measurement. 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 produce? 3.18.
The X parameter 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 value will not exceed 1% of the nominal value. Answers
https://pandia.ru/text/78/455/images/image116_9.gif" width="14" height="110 src="> b) 0 for x≤-3, F(x)= left"> 3.10.
a)f(x)= , b) Р(3.1≤Х≤3.7) ≈0.8185. 3.11.
|x|≥0.6. 3.12.
(-0,5;-0,1). 3.13.
a) P(9.8≤Х≤10.4) ≈0.6562. 3.14.
0,111. 3.15.
σ=1.2. 3.16.
(-6;30). 3.17.
0,4%. Distribution law of random variable X: Example No. 2. The probability of one shooter hitting the target with one shot for the first shooter is 0.8, for the second shooter – 0.85. The shooters fired one shot at the target. Considering hitting the target as independent events for individual shooters, find the probability of event A – exactly one hit on the target.
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 for x≤a,
,
The normal curve is symmetrical with respect to the straight line x=m, has a maximum at x=a, equal to .
,
Solution.
Probability that no emblems were drawn: P(0) = 0.5*0.5*0.5= 0.125
P(1) = 0,5
*0,5*0,5 + 0,5*0,5
*0,5 + 0,5*0,5*0,5
= 3*0,125=0,375
P(2) = 0,5
*0,5
*0,5 + 0,5
*0,5*0,5
+ 0,5*0,5
*0,5
= 3*0,125=0,375
Probability of getting three coats of arms: P(3) = 0.5*0.5*0.5 = 0.125
Check: P = P(0) + P(1) + P(2) + P(3) = 0.125 + 0.375 + 0.375 + 0.125 = 1 X 0
1
2
3
P 0,125
0,375
0,375
0,125
Solution.
Consider event A - one hit on the target. Possible options for this event to occur are as follows:
Then the probability of event A – exactly one hit on the target – will be equal to: P(A) = 0.12+0.17+0.68 = 0.97