Consider discrete distributions that are often used in modeling service systems.

Bernoulli distribution. The Bernoulli scheme is a sequence of independent trials, in each of which only two outcomes are possible - "success" and "failure" with probabilities R and q = 1 - R. Let the random variable X can take two values ​​with corresponding probabilities:

The Bernoulli distribution function has the form

Its graph is shown in Fig. 11.1.

A random variable with such a distribution is equal to the number of successes in one trial of the Bernoulli scheme.

The generating function, according to (11.1) and (11.15), is calculated as

Rice. 11.1.

Using formula (11.6), we find the mathematical expectation of the distribution:

We calculate the second derivative of the generating function according to (11.17)

By (11.7) we obtain the distribution variance

The Bernoulli distribution plays a big role in the theory of mass service, being a model of any random experiment, the outcomes of which belong to two mutually exclusive classes.

Geometric distribution. Assume that events occur at discrete times independently of each other. The probability that an event will occur is R, and the probability that it won't happen is q = 1-p, For example, a customer who has come to place an order.

Denote by r to the probability that an event will occur 1st time at a time to, those. to-th client made an order, and the previous to- 1 no clients. Then the probability of this complex event can be determined by the theorem of multiplication of the probabilities of independent events

The probabilities of events with a geometric distribution are shown in fig. 11.2.

The sum of the probabilities of all possible events

is a geometric progression, hence the distribution is called geometric. Since (1 - R)

Random value Xs geometric distribution has the meaning of the number of the first successful trial in the Bernoulli scheme.

Rice. 11.2.

Determine the probability that an event will occur for X>k

and the geometric distribution function

Let us calculate the generating function of the geometric distribution according to (11.1) and (11.20)

mathematical expectation of the geometric distribution according to (11.6)

and the dispersion according to (11.7)

The geometric distribution is considered to be a discrete version of the continuous exponential distribution and also has a number of properties useful for modeling service systems. In particular, like the exponential distribution, the geometric distribution has no memory:

those. if / failed experiments, then the probability that for the first success it is necessary to conduct more j of new trials is the same as the probability that a new series of trials requires a first success. /" trials. In other words, previous trials have no effect on future trials and trials are independent. Often this is true. and orders are made randomly.

Let us consider an example of a system whose functioning parameters are subject to a geometric distribution.

The master has at his disposal P identical spare parts. Every detail is likely q has a defect. During repair, the part is installed in the device, which is checked for operability. If the device does not work, then the part is replaced with another one. We consider a random variable X- the number of parts to be checked.

The probabilities of the number of tested parts will have the values ​​shown in the table:

					rya"~ x

Here q = 1 - R.

The mathematical expectation of the number of checked parts is defined as

Binomial distribution. Consider a random variable

where Xj obeys the Bernoulli distribution with the parameter R and random variables Xj independent.

Random value X will be equal to the number of occurrences of units at P tests, i.e. random variable with binomial distribution has the meaning of the number of successes in P independent tests.

According to (11.9), the generating function of the sum of mutually independent random variables, each of which has a Bernoulli distribution, is equal to the product of their generating functions (11.17):

Expanding the generating function (11.26) into a series, we obtain

In accordance with the definition of the generating function (11.1), the probability that the random variable X will take on the meaning to:

where are binomial coefficients.

11since & units per P places can be arranged in C* ways, then the number of samples containing to units will obviously be the same.

The distribution function for the binomial law is calculated by the formula

The distribution is called binomial due to the fact that the probabilities in form are terms of the expansion of the binomial:

It is clear that the total probability of all possible outcomes is equal to 1:

From (11.29) one can obtain a number of useful properties of binomial coefficients. For example, when R =1, q=1 we get

If we put R =1, q= - 1 , then

For any 1k, the following relations are valid:

The probabilities that in P tests, the event will occur: 1) less than × 2) more to once; 3) at least × 4) no more than & times, find, respectively, according to the formulas:

Using (11.6), we define the expectation of the binomial distribution

and according to (11.7) - dispersion:

Let us consider several examples of systems whose operation parameters are described by the binomial distribution.

1. A batch of 10 products contains one non-standard. Let us find the probability that with a random sample of 5 products, all of them will be standard (event BUT).

Number of all random samples p - S, e 0 , and the number of samples favoring the event is P= C 9 5 . Thus, the desired probability is equal to

2. At the entrance to a new apartment, 2 to new electric lamps. Each electric lamp burns out during the year with a probability R. Let us find the probability that during the year at least half of the initially switched on lamps will have to be replaced with new ones (the event BUT):

3. A person belonging to a certain group of consumers prefers product 1 with a probability of 0.2, product 2 with a probability of 0.3, product 3 with a probability of 0.4, product 4 with a probability of 0.1. A group of 6 consumers. Find the probabilities of the following events: BUT - the group includes at least 4 consumers who prefer product 3; AT- there is at least one consumer in the group who prefers product 4.

These probabilities are:

For large /? probability calculations become cumbersome, so limit theorems are used.

Local Laplace theorem, according to which the probability R p (k) is determined by the formula

where - Gaussian function;

Laplace integral theorem used to calculate the probability that P independent tests, the event will occur at least to ( once and no more to 2 once:

Let's consider examples of using these theorems.

1. The sewing workshop produces tailor-made clothes, among which 90% are of the highest quality. Find the probability that among 200 products there will be at least 160 and at most 170 of the highest quality.

Solution:

2. An insurance company has 12,000 clients. Each of them, insuring himself against an accident, contributes 10 thousand rubles. Probability of an accident R - 0.006, and the payment to the victim of 1 million rubles. Let's find the profit of the insurance company, provided with a probability of 0.995; in other words, what profit can the insurance company expect at a risk level of 0.005.

Solution: The total contribution of all clients is 12,000-10,000 = 120 million rubles. The company's profit depends on the number to accidents and is determined by the equality R = 120,000-1000 /: thousand rubles.

Therefore, it is necessary to find such a number A/ that the probability of the event P(k > M) did not exceed 0.005. Then, with a probability of 0.995, profit will be provided R = 120,000-10,004 / thousand rubles.

Inequality P(k > M) P(k0.995. Since to > 0, then R( 0 0.995. To estimate this probability, we use the Laplace integral theorem for P- 12,000 and /?=0.006, #=0.994:

Because*! F(x]) = -0.5.

Thus, it is necessary to find A/ for which

We find (M- 72)/8.5 > 2.58. Consequently, M>12 + 22 = 94.

So, with a probability of 0.995, the company guarantees a profit

It is often necessary to determine the most probable number to 0 . Probability of an event with a number of successes to 0 exceeds or at least not less than the probability of other possible test outcomes. Most Likely Number to 0 determined from the double inequality

3. Let there be 25 samples of consumer goods. The probability that each of the samples will be acceptable to the client is 0.7. It is necessary to determine the most likely number of samples that will be acceptable to customers. By (11.39)

From here to 0 - 18.

Poisson distribution. The Poisson distribution determines the probability that, given a very large number of trials, P, in each of which the probability of an event R very small, the event will occur exactly to schz.

Let the work pr \u003d k; this means that the average number of occurrences of an event in different series of trials, i.e. at various P, remains unchanged. In this case, the Poisson distribution can be used to approximate the binomial distribution:

Since for big P

The generating function of the Poisson distribution is calculated from (11.1) as

where by the Maclaurin formula

In accordance with the property of the coefficients of the generating function, the probability of occurrence to successes with an average number of successes X is calculated as (11.40).

On fig. 11.3 shows the probability density of the Poisson distribution.

The generating function of the Poisson distribution can also be obtained by using the series expansion of the generating function of the binomial distribution for pr \u003d X at P-» oo and the Maclaurin formula (11.42):

Rice. 11.3.

We define the mathematical expectation by (11.6)

and the dispersion according to (11.7)

Consider an example of a system with a Poisson distribution of parameters.

The company sent 500 products to the store. The probability of product damage in transit is 0.002. Find the probabilities that products will be damaged in transit: exactly 3 (event R); less than 3 (event AT) more than 3 (event Q; at least one (event D).

Number P= 500 is big, probability R= 0.002 is small, the considered events (product damage) are independent, so the Poisson formula (11.40) can be used.

At x=pr= 500 0.002=1 we get:

The Poisson distribution has a number of useful properties for modeling service systems.

1. Sum of random variables X \u003d X ( + X 2 with a Poisson distribution is also distributed according to Poisson's law.

If random variables have generating functions:

then, according to (11.9), the generating function of the sum of independent random variables with a Poisson distribution will have the form:

The parameter of the resulting distribution is X x + X 2.

2. If the number of elements./V of the set obeys the Poisson distribution with the parameter X and each element is chosen independently with probability R, then the sample elements of size Y distributed according to the Poisson law with the parameter pX.

Let , where corresponds to the Bernoulli distribution, and N- Poisson distribution. The corresponding generating functions, according to (11.17), (11.41):

Generating function of a random variable Y is calculated according to (11.14)

those. generating function corresponds to the Poisson distribution with the parameter pX.

3. As a consequence of property 2, the following property holds. If the number of elements of the set is distributed according to the Poisson law with the parameter X and the set is randomly distributed with probabilities /?, and p 2 = 1 - R into two groups, then the sizes of the sets are 7V, and N 2 are independent and Poisson-distributed with parameters p(k and p(k.

For ease of use, we present the results obtained for discrete distributions in the form of a table. 11.1 and 11.2.

Table 11.1. Main characteristics of discrete distributions

Distribution	Density	Range	Options	tn |		C X--2
Bernoulli	P(X = ) = p P (X = 0} = R + Z= 1	P - 0,1
Geometric	p(-p) to - 1	k = 1,2,...			^ 1 1 |	1 -R
Binomial	with to p to (- R g to	* = 1,2,...,#"			pr( - p)	1 -r pr
Poisson	E's to!	k = 1,2,...

Table 11. 2. Generating functions of discrete distributions

TEST QUESTIONS

• 1. What probability distributions are classified as discrete?
• 2. What is a generating function and what is it used for?
• 3. How to calculate the moments of random variables using the generating function?
• 4. What is the generating function of the sum of independent random variables?
• 5. What is called a composite distribution and how are the generating functions of composite distributions calculated?
• 6. Give the main characteristics of the Bernoulli distribution, give an example of its use in service tasks.
• 7. Give the main characteristics of the geometric distribution, give an example of use in service tasks.
• 8. Give the main characteristics of the binomial distribution, give an example of use in service tasks.
• 9. Give the main characteristics of the Poisson distribution, give an example of its use in service tasks.

Examples of random variables distributed according to the normal law are the height of a person, the mass of the caught fish of the same species. Normal distribution means the following : there are values ​​​​of human height, the mass of fish of the same species, which on an intuitive level are perceived as "normal" (and in fact - averaged), and they are much more common in a sufficiently large sample than those that differ up or down.

The normal probability distribution of a continuous random variable (sometimes the Gaussian distribution) can be called bell-shaped due to the fact that the density function of this distribution, which is symmetric about the mean, is very similar to the cut of a bell (red curve in the figure above).

The probability of meeting certain values ​​in the sample is equal to the area of ​​​​the figure under the curve, and in the case of a normal distribution, we see that under the top of the "bell", which corresponds to values ​​tending to the average, the area, and hence the probability, is greater than under the edges. Thus, we get the same thing that has already been said: the probability of meeting a person of "normal" height, catching a fish of "normal" weight is higher than for values ​​that differ up or down. In very many cases of practice, measurement errors are distributed according to a law close to normal.

Let's stop again at the figure at the beginning of the lesson, which shows the density function of the normal distribution. The graph of this function was obtained by calculating some data sample in the software package STATISTICS. On it, the histogram columns represent intervals of sample values ​​whose distribution is close (or, as they say in statistics, do not differ significantly from) to the normal distribution density function graph itself, which is a red curve. The graph shows that this curve is indeed bell-shaped.

The normal distribution is valuable in many ways because knowing only the mean of a continuous random variable and the standard deviation, you can calculate any probability associated with that variable.

The normal distribution has the added benefit of being one of the easiest to use statistical criteria used to test statistical hypotheses - Student's t-test- can be used only in the case when the sample data obey the normal distribution law.

The density function of the normal distribution of a continuous random variable can be found using the formula:

,

where x- value of the variable, - mean value, - standard deviation, e\u003d 2.71828 ... - the base of the natural logarithm, \u003d 3.1416 ...

Properties of the normal distribution density function

Changes in the mean move the bell curve in the direction of the axis Ox. If it increases, the curve moves to the right, if it decreases, then to the left.

If the standard deviation changes, then the height of the curve vertex changes. When the standard deviation increases, the top of the curve is higher, when it decreases, it is lower.

The probability that the value of a normally distributed random variable will fall within a given interval

Already in this paragraph, we will begin to solve practical problems, the meaning of which is indicated in the title. Let us analyze what possibilities the theory provides for solving problems. The starting concept for calculating the probability of a normally distributed random variable falling into a given interval is the integral function of the normal distribution.

Integral normal distribution function:

.

However, it is problematic to obtain tables for every possible combination of mean and standard deviation. Therefore, one of the simple ways to calculate the probability of a normally distributed random variable falling into a given interval is to use probability tables for a standardized normal distribution.

A normal distribution is called a standardized or normalized distribution., whose mean value is , and the standard deviation is .

Density function of the standardized normal distribution:

.

Cumulative function of the standardized normal distribution:

.

The figure below shows the integral function of the standardized normal distribution, the graph of which was obtained by calculating some data sample in the software package STATISTICS. The graph itself is a red curve, and the sample values ​​are approaching it.

To enlarge the picture, you can click on it with the left mouse button.

Standardizing a random variable means moving from the original units used in the task to standardized units. Standardization is performed according to the formula

In practice, all possible values ​​of a random variable are often not known, so the values ​​of the mean and standard deviation cannot be accurately determined. They are replaced by the arithmetic mean of the observations and the standard deviation s. Value z expresses the deviations of the values ​​of a random variable from the arithmetic mean when measuring standard deviations.

Open interval

The probability table for the standardized normal distribution, which is available in almost any book on statistics, contains the probabilities that a random variable having a standard normal distribution Z takes on a value less than a certain number z. That is, it will fall into the open interval from minus infinity to z. For example, the probability that the value Z less than 1.5 is equal to 0.93319.

Example 1 The company manufactures parts that have a normally distributed lifetime with a mean of 1000 and a standard deviation of 200 hours.

For a randomly selected part, calculate the probability that its service life will be at least 900 hours.

Solution. Let's introduce the first notation:

The desired probability.

The values ​​of the random variable are in the open interval. But we can calculate the probability that a random variable will take a value less than a given value, and according to the condition of the problem, it is required to find an equal or greater value than a given one. This is the other part of the space under the bell curve. Therefore, in order to find the desired probability, it is necessary to subtract from one the mentioned probability that the random variable will take a value less than the specified 900:

Now the random variable needs to be standardized.

We continue to introduce the notation:

z = (X ≤ 900) ;

x= 900 - given value of a random variable;

μ = 1000 - average value;

σ = 200 - standard deviation.

Based on these data, we obtain the conditions of the problem:

.

According to the tables of a standardized random variable (interval boundary) z= −0.5 corresponds to the probability 0.30854. Subtract it from unity and get what is required in the condition of the problem:

So, the probability that the life of the part will be at least 900 hours is 69%.

This probability can be obtained using the MS Excel function NORM.DIST (the value of the integral value is 1):

P(X≥900) = 1 - P(X≤900) = 1 - NORM.DIST(900; 1000; 200; 1) = 1 - 0.3085 = 0.6915.

About calculations in MS Excel - in one of the subsequent paragraphs of this lesson.

Example 2 In some city, the average annual family income is a normally distributed random variable with a mean value of 300,000 and a standard deviation of 50,000. It is known that the income of 40% of families is less than the value A. Find value A.

Solution. In this problem, 40% is nothing more than the probability that a random variable will take a value from an open interval that is less than a certain value, indicated by the letter A.

To find the value A, we first compose the integral function:

According to the task

μ = 300000 - average value;

σ = 50000 - standard deviation;

x = A is the value to be found.

Making up equality

.

According to the statistical tables, we find that the probability of 0.40 corresponds to the value of the interval boundary z = −0,25 .

Therefore, we make the equality

and find its solution:

A = 287300 .

Answer: income of 40% of families is less than 287300.

Closed interval

In many problems, it is required to find the probability that a normally distributed random variable takes a value in the interval from z 1 to z 2. That is, it will fall into the closed interval. To solve such problems, it is necessary to find in the table the probabilities corresponding to the boundaries of the interval, and then find the difference between these probabilities. This requires subtracting the smaller value from the larger one. Examples for solving these common problems are as follows, and it is proposed to solve them yourself, and then you can see the correct solutions and answers.

Example 3 The profit of an enterprise for a certain period is a random variable subject to the normal distribution law with an average value of 0.5 million c.u. and a standard deviation of 0.354. Determine, with an accuracy of two decimal places, the probability that the profit of the enterprise will be from 0.4 to 0.6 c.u.

Example 4 The length of the manufactured part is a random variable distributed according to the normal law with parameters μ =10 and σ =0.071 . Find, with an accuracy of two decimal places, the probability of marriage if the allowable dimensions of the part should be 10 ± 0.05.

Hint: in this problem, in addition to finding the probability of a random variable falling into a closed interval (the probability of obtaining a non-defective part), one more action is required.

allows you to determine the probability that the standardized value Z not less -z and no more +z, where z- an arbitrarily chosen value of a standardized random variable.

An Approximate Method for Checking the Normality of a Distribution

An approximate method for checking the normality of the distribution of sample values ​​is based on the following property of a normal distribution: skewness β 1 and kurtosis coefficient β 2 zero.

Asymmetry coefficient β 1 numerically characterizes the symmetry of the empirical distribution with respect to the mean. If the skewness is equal to zero, then the arithmetric mean, median and mode are equal: and the distribution density curve is symmetrical about the mean. If the coefficient of asymmetry is less than zero (β 1 < 0 ), then the arithmetic mean is less than the median, and the median, in turn, is less than the mode () and the curve is shifted to the right (compared to the normal distribution). If the coefficient of asymmetry is greater than zero (β 1 > 0 ), then the arithmetic mean is greater than the median, and the median, in turn, is greater than the mode () and the curve is shifted to the left (compared to the normal distribution).

Kurtosis coefficient β 2 characterizes the concentration of the empirical distribution around the arithmetic mean in the direction of the axis Oy and the degree of peaking of the distribution density curve. If the kurtosis coefficient is greater than zero, then the curve is more elongated (compared to the normal distribution) along the axis Oy(the graph is more pointed). If the kurtosis coefficient is less than zero, then the curve is more flattened (compared to a normal distribution) along the axis Oy(the graph is more obtuse).

The skewness coefficient can be calculated using the MS Excel function SKRS. If you are checking one array of data, then you need to enter a range of data in one "Number" box.

The coefficient of kurtosis can be calculated using the MS Excel function kurtosis. When checking one data array, it is also enough to enter the data range in one "Number" box.

So, as we already know, with a normal distribution, the skewness and kurtosis coefficients are equal to zero. But what if we got skewness coefficients equal to -0.14, 0.22, 0.43, and kurtosis coefficients equal to 0.17, -0.31, 0.55? The question is quite fair, since in practice we deal only with approximate, selective values ​​of asymmetry and kurtosis, which are subject to some inevitable, uncontrollable scatter. Therefore, it is impossible to require strict equality of these coefficients to zero, they should only be sufficiently close to zero. But what does enough mean?

It is required to compare the received empirical values ​​with admissible values. To do this, you need to check the following inequalities (compare the values ​​of the coefficients modulo with the critical values ​​- the boundaries of the hypothesis testing area).

For the asymmetry coefficient β 1 .

Three sigma rule.

Substitute a value? into the formula (*), we get:

So, with a probability arbitrarily close to one, it can be argued that the modulus of deviation of a normally distributed random variable from its mathematical expectation does not exceed three times the standard deviation.

Central limit theorem.

The central limit theorem is a group of theorems devoted to establishing the conditions under which the normal distribution law arises. Among these theorems, the most important place belongs to Lyapunov's theorem.

If the random variable X is the sum of a large number mutually? independent random variables, that is, the influence of each of which on the entire amount is negligible, then the random variable X has a distribution indefinitely approaching a normal distribution.

Initial and central moments of a continuous random variable, asymmetry and kurtosis. Mode and median.

In applied problems, for example, in mathematical statistics, in the theoretical study of empirical distributions that differ from the normal distribution, it becomes necessary to quantify these differences. For this purpose, special dimensionless characteristics are introduced.

Definition. The mode of a continuous random variable (Mo (X)) is its most probable value, for which the probability p i or the probability density f(x) reaches its maximum.

Definition. Median of a continuous random variable X (Me(X)) is its value for which the equality is true:

Geometrically, the vertical line x = Me (X) divides the area of ​​the figure under the curve into two equal parts.

At the point X = Me(X), the distribution function F(Me(X)) =

Find the mode Mo, the median Me and the mathematical expectation M of a random variable X with a probability density f(x) = 3x 2 , for x I [ 0; one ].

The probability density f(x) is maximum at x = 1, i.e. f (1) = 3, hence Mo (X) = 1 on the interval [ 0; one ].

To find the median, denote Me (X) = b.

Since Me (X) satisfies the condition P (X 3 = .

b 3 = ; b = » 0.79

M(X) ==+ =

Note the resulting 3 values ​​Mo (x), Me (X), M (X) on the Ox axis:

Definition. asymmetry theoretical distribution is the ratio of the central moment of the third order to the cube of the standard deviation:

Definition. kurtosis theoretical distribution is called the value determined by the equality:

where ? central moment of the fourth order.

For a normal distribution. When deviating from a normal distribution, the asymmetry is positive if the "long" and flatter part of the distribution curve is located to the right of the point on the x-axis corresponding to the mode; if this part of the curve is located to the left of the mode, then the asymmetry is negative (Fig. 1, a, b).

Kurtosis characterizes the “steepness” of the rise of the distribution curve compared to the normal curve: if the kurtosis is positive, then the curve has a higher and sharper peak; in the case of negative kurtosis, the compared curve has a lower and flatter peak.

It should be borne in mind that when using these comparison characteristics, the assumptions about the same values ​​of the mathematical expectation and variance for the normal and theoretical distributions are the reference ones.

Example. Let a discrete random variable X given by the distribution law:

Find: skewness and kurtosis of the theoretical distribution.

Let us first find the mathematical expectation of a random variable:

Then we calculate the initial and central moments of the 2nd, 3rd and 4th orders and :

Now, using the formulas, we find the required values:

In this case, the "long" part of the distribution curve is located to the right of the mode, and the curve itself is somewhat more peaked than the normal curve with the same values ​​of mathematical expectation and variance.

Theorem. For an arbitrary random variable X and any number

?>0 the following inequalities are valid:

Probability of the opposite inequality.

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

Let X– water consumption at the livestock farm (l).

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

That is, not less than 0.96.

For the binomial distribution, the Chebyshev inequality takes the form:

LAWS OF DISTRIBUTION OF RANDOM VARIABLES

LAWS OF DISTRIBUTION OF RANDOM VARIABLES - section Mathematics, PROBABILITY THEORY AND MATHEMATICAL STATISTICS The most common laws are uniform, normal and exponential.

The laws of uniform, normal and exponential distribution of probabilities of continuous random variables are most often encountered.

The probability distribution of a continuous random variable X is called uniform if on the interval (a, b) to which all possible values ​​of X belong, the distribution density remains constant (6.1)

The distribution function has the form:

Normal is the probability distribution of a continuous random variable X, the density of which has the form:

The probability that a random variable X will take a value belonging to the interval (?; ?):

where is the Laplace function, and,

What is the probability that the absolute value of the deviation will be less than a positive number?:

In particular, for a = 0, . (6.7)

The exponential (exponential) is the probability distribution of a continuous random variable X, which is described by the density:

where? is a constant positive value.

The distribution function of the exponential law:

The probability of hitting a continuous random variable X in the interval (a, c), distributed according to the exponential law:

1. The random variable X is uniformly distributed in the interval (-2;N). Find: a) the differential function of the random variable X; b) integral function; c) the probability of a random variable falling into the interval (-1;); d) mathematical expectation, variance and standard deviation of a random variable X.

2. Find the mathematical expectation and variance of a random variable uniformly distributed in the interval: a) (5; 11); b) (-3; 5). Draw graphs of these functions.

3. The random variable X is uniformly distributed on the interval (2; 6), and D(x) = 12. Find the distribution functions of the random variable X. Draw graphs of the functions.

4. The random variable X is distributed according to the law of a right triangle (Fig. 1) in the interval (0; a). Find: a) the differential function of the random variable X; b) integral function; c) probably

probability of hitting a random variable

to interval(); d) mathematical

expectation, variance and mean square

random deviation

5. Random variable X is distributed according to Simpson's law ("isosceles triangle law") (Fig. 2) on the interval (-a; a). Find: a) the differential probability distribution function of the random variable X;

b) an integral function and plot its graph; c) the probability of a random variable falling into the interval (-); d) mathematical expectation, variance and standard deviation of a random variable X.

6. To study the productivity of a certain breed of poultry, the diameter of the eggs is measured. The largest transverse diameter of eggs is a random variable distributed according to the normal law with an average value of 5 cm and a standard deviation of 0.3 cm. Find the probability that: a) the diameter of an egg taken at random will be between 4.7 and 6, 2 cm; b) the deviation of the diameter from the average does not exceed 0.6 cm in absolute value.

7. The weight of fish caught in the pond obeys the normal distribution law with a standard deviation of 150 g and mathematical expectation a = 1000 g. Find the probability that the weight of the fish caught will be: a) from 900 to 1300 g; b) no more than 1500 g; c) not less than 800 g; d) differ from the average weight modulo no more than 200 g; e) draw a graph of the differential function of the random variable X.

8. The yield of winter wheat for the totality of plots is distributed according to the normal law with the following parameters: a = 50 c/ha, = 10 c/ha. Determine: a) what percentage of plots will have a yield of more than 40 c/ha; b) the percentage of plots with a yield of 45 to 60 c/ha.

9. Grain weediness is measured by a sampling method, random measurement errors are subject to a normal distribution law with a standard deviation of 0.2 g and mathematical expectation a = 0. Find the probability that, out of four independent measurements, the error of at least one of them will not exceed in absolute value 0.3 g

10. The amount of grain harvested from each plot of the experimental field is a normally distributed random variable X, which has the mathematical expectation a = 60 kg and the standard deviation is 1.5 kg. Find the interval in which the quantity X will be enclosed with a probability of 0.9906. Write the differential function of this random variable.

11. With a probability of 0.9973, it was found that the absolute deviation of the live weight of a randomly taken head of cattle from the average weight of the animal throughout the herd does not exceed 30 kg. Find the standard deviation of the live weight of livestock, assuming that the distribution of livestock according to live weight obeys the normal law.

12. The yield of vegetables by plots is a normally distributed random variable with a mathematical expectation of 300 centners/ha and a standard deviation of 30 centners/ha. With a probability of 0.9545, determine the boundaries within which the average yield of vegetables in the plots will be.

13. Normally distributed random variable X is given by a differential function:

Determine: a) the probability of a random variable falling into the interval

(3; 9); b) the mode and median of the random variable X.

14. A trading company sells the same type of products from two manufacturers. The service life of products is subject to the normal law. The average service life of products of the first manufacturer is 5.5 thousand hours, and the second - 6 thousand hours. The first manufacturer claims that with a probability of 0.95, the service life of the first manufacturer is in the range from 5 to 6 thousand hours, and the second, with a probability of 0.9, is in the range from 5 to 7 thousand hours. Which manufacturer has the greatest variability in product life.

15. The monthly wages of employees of the enterprise are distributed according to the normal law with the mathematical expectation a = 10 thousand rubles. It is known that 50% of the company's employees receive wages from 8 to 12 thousand rubles. Determine what percentage of the company's employees have a monthly salary of 9 to 18 thousand rubles.

16. Write the density and distribution function of the exponential law if: a) a parameter; b) ; in) . Draw graphs of functions.

17. The random variable X is distributed according to the exponential law, moreover. Find the probability of hitting a random variable X in the interval: a) (0; 1); b) (2; 4). M(X), D(X), (X).

18. Find M(X), D(X), (X) of the exponential law of distribution of a random variable X by a given function:

19. Two independently operating elements are tested. The uptime of the first has a more indicative distribution than the second. Find the probability that during 20 hours: a) both elements will work; b) only one element will fail; c) at least one element fails; d) both elements will fail.

20. The probability that both independent elements will work for 10 days is 0.64. Determine the reliability function for each element if the functions are the same.

21. The average number of errors that the operator makes during an hour of work is 2. Find the probability that in 3 hours of work the operator will make: a) 4 errors; b) at least two mistakes; c) at least one mistake.

22. The average number of calls arriving at the PBX in one minute is three. Find the probability that in 2 minutes there will be: a) 4 calls; b) at least three calls.

23. Random variable X is distributed according to the Cauchy law

Continuous random variables

6. Continuous random variables

6.1. Numerical characteristics of continuous random variables

A continuous variable is a random variable that can take on all values ​​from some finite or infinite interval.

The distribution function is called the function F (x) ? determining the probability that the random variable X as a result of the test will take a value less than x, i.e.

Distribution function properties:

1. The values ​​of the distribution function belong to the segment , i.e.

2. F (x) is a non-decreasing function, i.e. if , then .

The probability that a random variable X will take a value contained in the interval is equal to:

· The probability that a continuous random variable X will take one definite value is equal to zero.

The probability distribution density of a continuous random variable X is called the function - the first derivative of the distribution function.

Probability of hitting a continuous random variable in a given interval:

Finding the distribution function from a known distribution density:

Distribution Density Properties

1. The distribution density is a non-negative function:

2. Normalization condition:

Standard deviation

6.2. Uniform distribution

A probability distribution is called uniform if, on the interval to which all possible values ​​of the random variable belong, the distribution density remains constant.

Probability density of a uniformly distributed random variable

Standard deviation

6.3. Normal distribution

Normal is the probability distribution of a random variable, which is described by the distribution density

a - mathematical expectation

standard deviation

dispersion

Probability of hitting the interval

Where is the Laplace function. This function is tabulated, i.e. the integral does not need to be calculated, it is necessary to use the table.

Probability of deviation of a random variable x from the mathematical expectation

Three sigma rule

If a random variable is normally distributed, then the absolute value of its deviation from the mathematical expectation does not exceed three times the standard deviation.

To be precise, the probability of going beyond the specified interval is 0.27%

Probability of Normal Distribution Online Calculator

6.4. exponential distribution

A random variable X is distributed according to an exponential law if the distribution density has the form

Standard deviation

A distinctive feature of this distribution is that the mathematical expectation is equal to the standard deviation.

Probability Theory. Random events (page 6)

12. Random variables X , if , , , .

13. The probability of manufacturing a defective product is 0.0002. Calculate the probability that an inspector checking the quality of 5000 items will find 4 defective items among them.

X X will take a value belonging to the interval . Plot the functions and .

15. The probability of failure-free operation of the element is distributed according to the exponential law (). Find the probability that the element will work flawlessly for 50 hours.

16. The device consists of 10 independently operating elements. Probability of failure of each element in time T equals 0.05. Using the Chebyshev inequality, estimate the probability that the absolute value of the difference between the number of failed elements and the average number (expectation) of failures over time T will be less than two.

17. Three independent shots were fired at the target (in Fig. 4.1 m, m) without a systematic error () with the expected spread of hit m. Find the probability of at least one hit on the target.

1. How many three-digit numbers can be made from the numbers 0,1,2,3,4,5?

2. The choir consists of 10 members. In how many ways can 6 participants be chosen for 3 days so that each day there will be a different line-up of the choir?

3. In how many ways can a deck of 52 shuffled cards be divided in half so that there are three aces in one half?

4. From a box containing tokens with numbers from 1 to 40, participants in the draw draw tokens. Determine the probability that the number of the first token drawn at random does not contain the number 2.

5. On the test bench, 250 devices are tested under certain conditions. Find the probability that at least one of the devices under test will fail within an hour, if it is known that the probability of failure within an hour of one of these devices is equal to 0.04 and is the same for all devices.

6. There are 10 rifles in the pyramid, 4 of which are equipped with an optical sight. The probability that the shooter will hit the target when fired from a rifle with a telescopic sight is 0.95; for rifles without a telescopic sight, this probability is 0.8. The shooter hit the target with a rifle taken at random. Find the probability that the shooter fired from a rifle with an optical sight.

7. The device consists of 10 nodes. Reliability (probability of failure-free operation over time t for each node is equal to . Nodes fail independently of each other. Find the probability that in time t: a) at least one node will fail; b) exactly two nodes will fail; c) exactly one node will fail; d) at least two nodes will fail.

8. Each of the 16 elements of some device is tested. The probability that an element will pass the test is 0.8. Find the most likely number of elements that will pass the test.

9. Find the probability that the event BUT(gear shift) will occur 70 times on a 243-kilometer track if the probability of shifting per kilometer of that track is 0.25.

10. The probability of hitting the target with one shot is 0.8. Find the probability that with 100 shots the target will be hit at least 75 times and at most 90 times.

X.

12. Random variables X and independent. Find the mathematical expectation and variance of a random variable , if , , , .

13. Manuscript of 1000 pages of typewritten text contains 100 misprints. Find the probability that a randomly selected page contains exactly 2 misprints.

14. Continuous random variable X distributed uniformly with a constant probability density , where Find 1) the parameter and write down the distribution law; 2) Find , ; 3) Find the probability that X will take a value belonging to the interval .

15. The uptime of an element has an exponential distribution (). Find the probability that t= 24 hours the element will not fail.

16. Continuous random variable X distributed according to the normal law . Find , . Find the probability that as a result of the test X will take the value enclosed in the interval .

17. The probability distribution of a discrete two-dimensional random variable is given:

Find the law of distribution of components X and ; their mathematical expectations and ; variances and ; correlation coefficient .

1. How many three-digit numbers can be made from the numbers 1,2, 3, 4, 5 if each of these numbers is used no more than once?

2. Given n points, no 3 of which lie on the same line. How many lines can be drawn by connecting the points in pairs?

How many dominoes can be made using numbers from 0 to 9?

3. What is the probability that a randomly torn sheet from the new calendar corresponds to the first day of the month? (The year is not considered a leap year.)

4. There are 3 telephones in the workshop that work independently of each other.

5. The employment probabilities of each of them are respectively the following: ; ; . Find the probability that at least one phone is free.

6. There are three identical urns. The first urn contains 20 white balls, the second urn contains 10 white and 10 black balls, and the third urn contains 20 black balls. A white ball is drawn from an urn chosen at random. Find the probability that the ball is drawn from the first urn.

7. In some areas in summer, on average, 20% of the days are rainy. What is the probability that in one week: a) there will be at least one rainy day; b) there will be exactly one rainy day; c) the number of rainy days will be no more than four; d) there will be no rainy days.

8. The probability of violation of accuracy in the assembly of the device is 0.32. Determine the most probable number of precision instruments in a batch of 9 pieces.

9. Determine the probability that with 150 shots from a rifle the target will be hit 70 times if the probability of hitting the target with one shot is 0.4.

10. Determine the probability that out of 1000 born children the number of boys will be at least 455 and at most 555, if the probability of the birth of boys is 0.515.

11. The law of distribution of a discrete random variable is given X:

Find: 1) the probability value corresponding to the value ; 2) , , ; 3) distribution function; build her graph. Construct a distribution polygon of a random variable X.

12. Random variables X and independent. Find the mathematical expectation and variance of a random variable , if , , , .

13. The probability of manufacturing a non-standard part is 0.004. Find the probability that among 1000 parts there will be 5 non-standard ones.

14. Continuous random variable X given by the distribution function Find: 1) density function ; 2) , , ; 3) the probability that, as a result of the experiment, the random variable X will take a value belonging to the interval . Build graphs of functions and .km, km. Determine the probability of two hits on the target.

1. Speakers must speak at the meeting BUT, AT, FROM, D. In how many ways can they be placed on the list of speakers so that AT spoke after the speaker BUT?

2. In how many ways can 14 identical balls be placed in 8 boxes?

3. How many five-digit numbers can be made from the numbers 1 to 9?

4. The student came to the exam knowing only 24 of the 32 questions of the program. The examiner asked him 3 questions. Find the probability that the student answered all questions.

5. By the end of the day, there were 60 watermelons left in the store, including 50 ripe ones. The customer chooses 2 watermelons. What is the probability that both watermelons are ripe?

6. In the group of athletes there are 20 runners, 6 jumpers and 4 hammer throwers. The probability that the norm of the master of sports will be fulfilled by a runner is 0.9; jumper - 0.8 and thrower - 0.75. Determine the probability that a randomly called athlete will fulfill the standard of the master of sports.

7. The probability that a rented item will be returned in good condition is 0.8. Determine the probability that out of five things taken: a) three will be returned in good condition; b) all five items will be returned intact; c) at least two items will be returned intact.

8. The probability of a defect in a batch of 500 parts is 0.035. Determine the most likely number of defective parts in this batch.

9. In the production of electric light bulbs, the probability of manufacturing a first grade lamp is assumed to be 0.64. Determine the probability that out of 100 randomly selected electric lamps, 70 will be of the first grade.

10. 400 ore samples are subject to examination. The probability of industrial metal content in each sample is the same and equals 0.8. Find the probability that the number of samples with industrial metal content will be between 290 and 340.

11. The law of distribution of a discrete random variable is given X if X X and ; 4) find out whether these quantities are dependent.

1. In how many ways can 8 guests be seated at a round table so that two famous guests sit side by side?

2. How many different "words" can be formed by rearranging the letters of the word "combinatorics"?

3. How many triangles are there whose side lengths take one of the following values: 4, 5, 6, 7 cm?

4. The letters of the split alphabet are in the envelope: O, P, R, FROM, T. The letters are carefully mixed. Determine the probability that by taking out these letters and putting them side by side, you get the word " SPORT‘.

5. From the first machine, 20% goes to the assembly, from the second 30%, from the third - 50% of the parts. The first machine gives an average of 0.2% of defects, the second - 0.3%, the third - 1%. Find the probability that the part received for assembly is defective.

6. One of the three shooters is called to the line of fire and fires a shot. The target is hit. The probability of hitting the target with one shot for the first shooter is 0.3, for the second - 0.5, for the third - 0.8. Find the probability that the shot was fired by the second shooter.

7. There are 6 motors in the workshop. For each motor, the probability that it is currently on is 0.8. Find the probability that at the moment: a) 4 motors are on; b) at least one motor is on; c) all motors are on.

8. The TV has 12 lamps. Each of them with a probability of 0.4 may fail during the warranty period. Find the most likely number of lamps that failed during the warranty period.

9. The probability of having a boy is 0.515. Find the probability that out of 200 children born boys and girls will be equally divided.

10. The probability that the part did not pass the Quality Control Department check will be . Find the probability that 70 to 100 parts will be unchecked among 400 randomly selected parts.

11. The law of distribution of a discrete random variable is given X:

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Random variable a variable is called which, as a result of each test, takes on one previously unknown value, depending on random causes. Random variables are denoted by capital Latin letters: $X,\ Y,\ Z,\ \dots $ By their type, random variables can be discrete and continuous.

Discrete random variable- this is such a random variable, the values ​​​​of which can be no more than countable, that is, either finite or countable. Countability means that the values ​​of a random variable can be enumerated.

Example 1 . Let us give examples of discrete random variables:

a) the number of hits on the target with $n$ shots, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

b) the number of coats of arms that fell out when tossing a coin, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

c) the number of ships that arrived on board (a countable set of values).

d) the number of calls arriving at the exchange (a countable set of values).

1. Law of probability distribution of a discrete random variable.

A discrete random variable $X$ can take the values ​​$x_1,\dots ,\ x_n$ with probabilities $p\left(x_1\right),\ \dots ,\ p\left(x_n\right)$. The correspondence between these values ​​and their probabilities is called distribution law of a discrete random variable. As a rule, this correspondence is specified using a table, in the first line of which the values ​​of $x_1,\dots ,\ x_n$ are indicated, and in the second line the probabilities corresponding to these values ​​are $p_1,\dots ,\ p_n$.

$\begin(array)(|c|c|)
\hline
X_i & x_1 & x_2 & \dots & x_n \\
\hline
p_i & p_1 & p_2 & \dots & p_n \\
\hline
\end(array)$

Example 2 . Let the random variable $X$ be the number of points rolled when a dice is rolled. Such a random variable $X$ can take the following values ​​$1,\ 2,\ 3,\ 4,\ 5,\ 6$. The probabilities of all these values ​​are equal to $1/6$. Then the probability distribution law for the random variable $X$:

$\begin(array)(|c|c|)
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline

\hline
\end(array)$

Comment. Since the events $1,\ 2,\ \dots ,\ 6$ form a complete group of events in the distribution law of the discrete random variable $X$, the sum of the probabilities must be equal to one, i.e. $\sum(p_i)=1$.

2. Mathematical expectation of a discrete random variable.

Mathematical expectation of a random variable specifies its "central" value. For a discrete random variable, the mathematical expectation is calculated as the sum of the products of the values ​​$x_1,\dots ,\ x_n$ and the probabilities $p_1,\dots ,\ p_n$ corresponding to these values, i.e.: $M\left(X\right)=\sum ^n_(i=1)(p_ix_i)$. In English literature, another notation $E\left(X\right)$ is used.

Expectation Properties$M\left(X\right)$:

1. $M\left(X\right)$ is between the smallest and largest values ​​of the random variable $X$.
2. The mathematical expectation of a constant is equal to the constant itself, i.e. $M\left(C\right)=C$.
3. The constant factor can be taken out of the expectation sign: $M\left(CX\right)=CM\left(X\right)$.
4. The mathematical expectation of the sum of random variables is equal to the sum of their mathematical expectations: $M\left(X+Y\right)=M\left(X\right)+M\left(Y\right)$.
5. The mathematical expectation of the product of independent random variables is equal to the product of their mathematical expectations: $M\left(XY\right)=M\left(X\right)M\left(Y\right)$.

Example 3 . Let's find the mathematical expectation of the random variable $X$ from example $2$.

$$M\left(X\right)=\sum^n_(i=1)(p_ix_i)=1\cdot ((1)\over (6))+2\cdot ((1)\over (6) )+3\cdot ((1)\over (6))+4\cdot ((1)\over (6))+5\cdot ((1)\over (6))+6\cdot ((1 )\over (6))=3.5.$$

We can notice that $M\left(X\right)$ is between the smallest ($1$) and largest ($6$) values ​​of the random variable $X$.

Example 4 . It is known that the mathematical expectation of the random variable $X$ is equal to $M\left(X\right)=2$. Find the mathematical expectation of the random variable $3X+5$.

Using the above properties, we get $M\left(3X+5\right)=M\left(3X\right)+M\left(5\right)=3M\left(X\right)+5=3\cdot 2 +5=11$.

Example 5 . It is known that the mathematical expectation of the random variable $X$ is equal to $M\left(X\right)=4$. Find the mathematical expectation of the random variable $2X-9$.

Using the above properties, we get $M\left(2X-9\right)=M\left(2X\right)-M\left(9\right)=2M\left(X\right)-9=2\cdot 4 -9=-1$.

3. Dispersion of a discrete random variable.

Possible values ​​of random variables with equal mathematical expectations can scatter differently around their average values. For example, in two student groups, the average score for the exam in probability theory turned out to be 4, but in one group everyone turned out to be good students, and in the other group - only C students and excellent students. Therefore, there is a need for such a numerical characteristic of a random variable, which would show the spread of the values ​​of a random variable around its mathematical expectation. This characteristic is dispersion.

Dispersion of a discrete random variable$X$ is:

$$D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2).\ $$

In English literature, the notation $V\left(X\right),\ Var\left(X\right)$ is used. Very often the variance $D\left(X\right)$ is calculated by the formula $D\left(X\right)=\sum^n_(i=1)(p_ix^2_i)-(\left(M\left(X \right)\right))^2$.

Dispersion Properties$D\left(X\right)$:

1. The dispersion is always greater than or equal to zero, i.e. $D\left(X\right)\ge 0$.
2. The dispersion from a constant is equal to zero, i.e. $D\left(C\right)=0$.
3. The constant factor can be taken out of the dispersion sign, provided that it is squared, i.e. $D\left(CX\right)=C^2D\left(X\right)$.
4. The variance of the sum of independent random variables is equal to the sum of their variances, i.e. $D\left(X+Y\right)=D\left(X\right)+D\left(Y\right)$.
5. The variance of the difference of independent random variables is equal to the sum of their variances, i.e. $D\left(X-Y\right)=D\left(X\right)+D\left(Y\right)$.

Example 6 . Let us calculate the variance of the random variable $X$ from example $2$.

$$D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2)=((1)\over (6))\cdot (\left(1-3,5\right))^2+((1)\over (6))\cdot (\left(2-3,5\right))^2+ \dots +((1)\over (6))\cdot (\left(6-3,5\right))^2=((35)\over (12))\approx 2.92.$$

Example 7 . It is known that the variance of the random variable $X$ is equal to $D\left(X\right)=2$. Find the variance of the random variable $4X+1$.

Using the above properties, we find $D\left(4X+1\right)=D\left(4X\right)+D\left(1\right)=4^2D\left(X\right)+0=16D\ left(X\right)=16\cdot 2=32$.

Example 8 . It is known that the variance of $X$ is equal to $D\left(X\right)=3$. Find the variance of the random variable $3-2X$.

Using the above properties, we find $D\left(3-2X\right)=D\left(3\right)+D\left(2X\right)=0+2^2D\left(X\right)=4D\ left(X\right)=4\cdot 3=12$.

4. Distribution function of a discrete random variable.

The method of representing a discrete random variable in the form of a distribution series is not the only one, and most importantly, it is not universal, since a continuous random variable cannot be specified using a distribution series. There is another way to represent a random variable - the distribution function.

distribution function random variable $X$ is a function $F\left(x\right)$, which determines the probability that the random variable $X$ takes a value less than some fixed value $x$, i.e. $F\left(x\right)$ )=P\left(X< x\right)$

Distribution function properties:

1. $0\le F\left(x\right)\le 1$.
2. The probability that the random variable $X$ takes values ​​from the interval $\left(\alpha ;\ \beta \right)$ is equal to the difference between the values ​​of the distribution function at the ends of this interval: $P\left(\alpha< X < \beta \right)=F\left(\beta \right)-F\left(\alpha \right)$
3. $F\left(x\right)$ - non-decreasing.
4. $(\mathop(lim)_(x\to -\infty ) F\left(x\right)=0\ ),\ (\mathop(lim)_(x\to +\infty ) F\left(x \right)=1\ )$.

Example 9 . Let us find the distribution function $F\left(x\right)$ for the distribution law of the discrete random variable $X$ from example $2$.

$\begin(array)(|c|c|)
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
\hline
\end(array)$

If $x\le 1$, then obviously $F\left(x\right)=0$ (including $x=1$ $F\left(1\right)=P\left(X< 1\right)=0$).

If $1< x\le 2$, то $F\left(x\right)=P\left(X=1\right)=1/6$.

If $2< x\le 3$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)=1/6+1/6=1/3$.

If $3< x\le 4$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)=1/6+1/6+1/6=1/2$.

If $4< x\le 5$, то $F\left(X\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)=1/6+1/6+1/6+1/6=2/3$.

If $5< x\le 6$, то $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)=1/6+1/6+1/6+1/6+1/6=5/6$.

If $x > 6$ then $F\left(x\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right) +P\left(X=4\right)+P\left(X=5\right)+P\left(X=6\right)=1/6+1/6+1/6+1/6+ 1/6+1/6=1$.

So $F(x)=\left\(\begin(matrix)
0,\ at\ x\le 1,\\
1/6, at \ 1< x\le 2,\\
1/3,\ at\ 2< x\le 3,\\
1/2, at \ 3< x\le 4,\\
2/3,\ at\ 4< x\le 5,\\
5/6, \ at \ 4< x\le 5,\\
1,\ for \ x > 6.
\end(matrix)\right.$

LAW OF DISTRIBUTION AND CHARACTERISTICS

RANDOM VALUES

Random variables, their classification and methods of description.

A random value is a quantity that, as a result of an experiment, can take on one or another value, but which one is not known in advance. For a random variable, therefore, only values ​​can be specified, one of which it will necessarily take as a result of the experiment. These values ​​will be referred to as 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 ​​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 countable set is a set whose elements can be numbered. The word "discrete" comes from the Latin discretus, which means "discontinuous, consisting of separate parts."

Example 1. A discrete random variable is the number of defective parts X in a batch of nfl. 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 is called a random variable, 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 electricity consumption at the enterprise for a month.

Example 4. A continuous random variable is the error in the height measurement 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.

Law of distribution of random variables.

A random variable is considered to be completely specified if its possible values ​​are indicated on the numerical axis and the distribution law is established.

The law of distribution of a random variable is called a relation that establishes a 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 subject to a given distribution law. A number of probabilities, a distribution function, a probability density, a characteristic function are used as distribution laws.

The distribution law gives a complete probable description of a random variable. According to the distribution law, it is possible to judge before experience which possible values ​​of a random variable will appear more often, and which ones less often.

For a discrete random variable, the distribution law can be given in the form of a table, analytically (in the form of a formula) and graphically.

The simplest form of specifying the law of distribution of a discrete random variable is a table (matrix), which lists in ascending order all possible values ​​of a random variable and their corresponding probabilities, i.e.

Such a table is called a series of distribution of a discrete random variable. one

The events X 1 , X 2 ,..., X n , consisting in the fact that, as a result of the test, the random variable X will take the values ​​x 1 , x 2 ,... x n, respectively, are inconsistent and the only possible ones (because the table lists all possible values ​​of a random variable), i.e. form a complete group. Therefore, the sum of their probabilities is equal to 1. Thus, for any discrete random variable

(This unit is somehow distributed among the values ​​of the random variable, hence the term "distribution").

A distribution series can be displayed graphically if the values ​​of a random variable are plotted along the abscissa axis, and their corresponding probabilities along the ordinate axis. The connection of the obtained points forms a broken line, called a polygon or polygon of the probability distribution (Fig. 1).

Example The lottery is played: a car worth 5000 den. units, 4 TVs worth 250 den. unit, 5 VCRs worth 200 den. units In total, 1000 tickets are sold for 7 den. units Draw up the law of distribution of the net winnings received by the lottery participant who bought one ticket.

Solution. Possible values ​​of the random variable X - net winnings per ticket - are 0-7 = -7 den. units (if the ticket did not win), 200-7 = 193, 250-7 = 243, 5000-7 = 4993 den. units (if the ticket won the VCR, TV or car, respectively). Given 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.