As is known, random variable is called a variable that can take on certain values depending on the case. Random variables are denoted by capital letters of the Latin alphabet (X, Y, Z), and their values are denoted by the corresponding lowercase letters (x, y, z). 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 non-zero probabilities.
The distribution law of a discrete random variable is a function that connects the values of a random variable with their corresponding probabilities. 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, … .
in) by using distribution function F(x) , which determines for each value x the probability that the random variable X takes 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 – distribution polygon (polygon) (see problem 3).
Note that in order to solve some problems, it is not necessary to know the distribution law. In some cases, it is enough to know one or more 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 size of the 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 :
- Mathematical expectation
(mean value) of a discrete random variable M(X)=Σ x i p i.
For binomial distribution M(X)=np, for Poisson distribution M(X)=λ - Dispersion
discrete random variable D(X)=M2 or D(X) = M(X 2) − 2. The difference X–M(X) is called the deviation of a random variable from its mathematical expectation.
For binomial distribution D(X)=npq, for Poisson distribution D(X)=λ - Standard deviation (standard deviation) σ(X)=√D(X).
Examples of solving problems on the topic "The law of distribution of a discrete random variable"
Task 1.
1000 lottery tickets have been issued: 5 of them win 500 rubles, 10 - 100 rubles, 20 - 50 rubles, 50 - 10 rubles. Determine the law of probability distribution of the random variable X - winnings 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 winning 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 present the resulting law in the form of a table:
Find the mathematical expectation of 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
Task 3.
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 it. Find the mathematical expectation, variance and standard deviation of a discrete random variable.
Solution. 1. Discrete random variable X=(number of failed elements in one experiment) has the following possible values: x 1 =0 (none of the elements of the device failed), x 2 =1 (one element failed), x 3 =2 (two elements failed ) and x 4 \u003d 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's formula
. Given that, by condition, n=3, p=0.1, q=1-p=0.9, we determine the probabilities of the values:
P 3 (0) \u003d C 3 0 p 0 q 3-0 \u003d q 3 \u003d 0.9 3 \u003d 0.729;
P 3 (1) \u003d C 3 1 p 1 q 3-1 \u003d 3 * 0.1 * 0.9 2 \u003d 0.243;
P 3 (2) \u003d C 3 2 p 2 q 3-2 \u003d 3 * 0.1 2 * 0.9 \u003d 0.027;
P 3 (3) \u003d C 3 3 p 3 q 3-3 \u003d p 3 \u003d 0.1 3 \u003d 0.001;
Check: ∑p i = 0.729+0.243+0.027+0.001=1.
Thus, the desired binomial distribution law X has the form:
On the abscissa axis, we plot the possible values x i, and on the ordinate axis, the corresponding probabilities р i . Let's construct points M 1 (0; 0.729), M 2 (1; 0.243), M 3 (2; 0.027), M 4 (3; 0.001). Connecting these points with line segments, we obtain the desired distribution polygon.
3. Find the distribution function F(x) = P(X
For x ≤ 0 we have F(x) = P(X<0) = 0;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 it will be F(x) = 1, because the event is certain.
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Graph of the function F(x)
4.
For the binomial distribution X:
- mathematical expectation М(X) = np = 3*0.1 = 0.3;
- dispersion D(X) = npq = 3*0.1*0.9 = 0.27;
- standard deviation σ(X) = √D(X) = √0.27 ≈ 0.52.
Distribution density probabilities X call the function f(x) is the first derivative of the distribution function F(x):
The concept of the probability distribution density of a random variable X for a discrete quantity is not applicable.
Probability density f(x) is called the differential distribution function:
Property 1. The distribution density is a non-negative value:
Property 2. The improper integral of the distribution density in the range from to is equal to one:
Example 1.25. Given the distribution function of a continuous random variable X:
f(x).
Solution: The distribution density is equal to the first derivative of the distribution function:
1. Given the distribution function of a continuous random variable X:
Find the distribution density.
2. The distribution function of a continuous random variable is given X:
Find the distribution density f(x).
1.3. Numerical characteristics of continuous random
quantities
Expected value continuous random variable X, the possible values of which belong to the entire axis Oh, is determined by the equality:
It is assumed that the integral converges absolutely.
a,b), then:
f(x) is the distribution density of the random variable.
Dispersion continuous random variable X, the possible values of which belong to the entire axis, is determined by the equality:
Special case. If the values of the random variable belong to the interval ( a,b), then:
The probability that X will take values belonging to the interval ( a,b), is determined by the equality:
.
Example 1.26. Continuous random variable X
Find the mathematical expectation, variance and probability of hitting a random variable X in the interval (0; 0.7).
Solution: The random variable is distributed over the interval (0,1). Let us define the distribution density of a continuous random variable X:
a) Mathematical expectation 
:
b) Dispersion 
in) 
Tasks for independent work:
1. Random variable X given by the distribution function:
M(x);
b) dispersion D(x);
X into the interval (2,3).
2. Random value X
Find: a) mathematical expectation M(x);
b) dispersion D(x);
c) determine the probability of hitting a random variable X in the interval (1; 1.5).
3. Random value X is given by the integral distribution function:
Find: a) mathematical expectation M(x);
b) dispersion D(x);
c) determine the probability of hitting a random variable X in the interval.
1.4. Laws of distribution of a continuous random variable
1.4.1. Uniform distribution
Continuous random variable X has a uniform distribution on the interval [ a,b], if on this segment the density of the probability distribution of a random variable is constant, and outside it is equal to zero, i.e.:
Rice. four.
; 
; 
.
Example 1.27. A bus of some route moves uniformly with an interval of 5 minutes. Find the probability that a uniformly distributed random variable X– the waiting time for the bus will be less than 3 minutes.
Solution: Random value X- uniformly distributed over the interval .
Probability Density: 
.
In order for the waiting time not to exceed 3 minutes, the passenger must arrive at the bus stop within 2 to 5 minutes after the departure of the previous bus, i.e. random value X must fall within the interval (2;5). That. desired probability:
Tasks for independent work:
1. a) find the mathematical expectation of a random variable X distributed uniformly in the interval (2; 8);
b) find the variance and standard deviation of a random variable X, distributed uniformly in the interval (2;8).
2. The minute hand of an electric clock jumps at the end of each minute. Find the probability that at a given moment the clock will show the time that differs from the true time by no more than 20 seconds.
1.4.2. The exponential (exponential) distribution
Continuous random variable X is exponentially distributed if its probability density has the form:
where is the parameter of the exponential distribution.
In this way 
Rice. 5.
Numerical characteristics:
Example 1.28. Random value X- the operating time of the light bulb - has an exponential distribution. Determine the probability that the lamp will last at least 600 hours if the average lamp life is 400 hours.
Solution: According to the condition of the problem, the mathematical expectation of a random variable X equals 400 hours, so:
; 
The desired probability , where
Finally:
Tasks for independent work:
1. Write the density and distribution function of the exponential law, if the parameter .
2. Random value X
Find the mathematical expectation and variance of a quantity X.
3. Random value X given by the probability distribution function:
Find the mathematical expectation and standard deviation of a random variable.
1.4.3. Normal distribution
normal is called the probability distribution of a continuous random variable X, whose density has the form:
where a– mathematical expectation, – standard deviation X.
The probability that X will take a value belonging to the interval:
, where
is the Laplace function.
A distribution that has ; , i.e. with a probability density 
called standard.
Rice. 6.
The probability that the absolute value of the deviation is less than a positive number:
.
In particular, when a= 0 equality is true:
Example 1.29. Random value X distributed normally. Standard deviation . Find the probability that the deviation of a random variable from its mathematical expectation in absolute value will be less than 0.3.
Solution: .
Tasks for independent work:
1. Write the probability density of the normal distribution of a random variable X, knowing that M(x)= 3, D(x)= 16.
2. Mathematical expectation and standard deviation of a normally distributed random variable X are 20 and 5, respectively. Find the probability that as a result of the test X will take the value contained in the interval (15;20).
3. Random measurement errors are subject to the normal law with standard deviation mm and mathematical expectation a= 0. Find the probability that the error of at least one of 3 independent measurements does not exceed 4 mm in absolute value.
4. Some substance is weighed without systematic errors. Random weighing errors are subject to the normal law with a standard deviation r. Find the probability that the weighing will be carried out with an error not exceeding 10 g in absolute value.
Expected value
Dispersion continuous random variable X, the possible values of which belong to the entire axis Ox, is determined by the equality: 
Service assignment. The online calculator is designed to solve problems in which either distribution density f(x) , or distribution function F(x) (see example). Usually in such tasks it is required to find mathematical expectation, standard deviation, plot the functions f(x) and F(x).
Instruction. Select the type of input data: distribution density f(x) or distribution function F(x) .
The distribution density f(x) is given: 
The distribution function F(x) is given: 
A continuous random variable is defined by a probability density 
(Rayleigh distribution law - used in radio engineering). Find M(x) , D(x) .
The random variable X is called continuous
, if its distribution function F(X)=P(X< x) непрерывна и имеет производную.
The distribution function of a continuous random variable is used to calculate the probabilities of a random variable falling into a given interval:
P(α< X < β)=F(β) - F(α)
moreover, for a continuous random variable, it does not matter whether its boundaries are included in this interval or not:
P(α< X < β) = P(α ≤ X < β) = P(α ≤ X ≤ β)
Distribution density
continuous random variable is called the function
f(x)=F'(x) , derivative of the distribution function.
Distribution Density Properties
1. The distribution density of a random variable is non-negative (f(x) ≥ 0) for all values of x.2. Normalization condition:
The geometric meaning of the normalization condition: the area under the distribution density curve is equal to one.
3. The probability of hitting a random variable X in the interval from α to β can be calculated by the formula
Geometrically, the probability that a continuous random variable X falls into the interval (α, β) is equal to the area of the curvilinear trapezoid under the distribution density curve based on this interval.
4. The distribution function is expressed in terms of density as follows:
The distribution density value at the point x is not equal to the probability of taking this value; for a continuous random variable, we can only talk about the probability of falling into a given interval. Let )