Examples of solving problems on the topic "Random variables".

A task 1 . There are 100 tickets issued in the lottery. One win of 50 USD was played. and ten wins of $10 each. Find the law of distribution of the value X - the cost of a possible gain.

Solution. Possible values ​​of X: x 1 = 0; x 2 = 10 and x 3 = 50. Since there are 89 “empty” tickets, then p 1 = 0.89, the probability of winning is 10 c.u. (10 tickets) – p 2 = 0.10 and for a win of 50 c.u. –p 3 = 0.01. In this way:

	0,89	0,10	0,01

Easy to control: .

A task 2. The probability that the buyer has familiarized himself with the advertisement of the product in advance is 0.6 (p = 0.6). Selective quality control of advertising is carried out by polling buyers before the first one who has studied the advertisement in advance. Make a series of distribution of the number of interviewed buyers.

Solution. According to the condition of the problem p = 0.6. From: q=1 -p = 0.4. Substituting these values, we get: and construct a distribution series:

pi		0,24

A task 3. A computer consists of three independently operating elements: a system unit, a monitor, and a keyboard. With a single sharp increase in voltage, the probability of failure of each element is 0.1. Based on the Bernoulli distribution, draw up the distribution law for the number of failed elements during a power surge in the network.

Solution. Consider Bernoulli distribution(or binomial): the probability that in n tests, event A will appear exactly k once: , or:

	q n					p n

AT let's get back to the task.

Possible values ​​of X (number of failures):

x 0 =0 - none of the elements failed;

x 1 =1 - failure of one element;

x 2 =2 - failure of two elements;

x 3 =3 - failure of all elements.

Since, by condition, p = 0.1, then q = 1 – p = 0.9. Using the Bernoulli formula, we get

, ,

, .

Control: .

Therefore, the desired distribution law:

	0,729	0,243	0,027	0,001

Task 4. Produced 5000 rounds. The probability that one cartridge is defective . What is the probability that there will be exactly 3 defective cartridges in the entire batch?

Solution. Applicable Poisson distribution: this distribution is used to determine the probability that, given a very large

number of trials (mass trials), in each of which the probability of event A is very small, event A will occur k times: , where .

Here n \u003d 5000, p \u003d 0.0002, k \u003d 3. We find , then the desired probability: .

Task 5. When firing before the first hit with the probability of hitting p = 0.6 for a shot, you need to find the probability that the hit will occur on the third shot.

Solution. Let us apply the geometric distribution: let independent trials be made, in each of which the event A has a probability of occurrence p (and non-occurrence q = 1 - p). Trials end as soon as event A occurs.

Under such conditions, the probability that event A will occur on the kth test is determined by the formula: . Here p = 0.6; q \u003d 1 - 0.6 \u003d 0.4; k \u003d 3. Therefore, .

Task 6. Let the law of distribution of a random variable X be given:

Find the mathematical expectation.

Solution. .

Note that the probabilistic meaning of the mathematical expectation is the average value of a random variable.

Task 7. Find the variance of a random variable X with the following distribution law:

Solution. Here .

The law of distribution of the square of X 2 :

X 2

Required variance: .

Dispersion characterizes the degree of deviation (scattering) of a random variable from its mathematical expectation.

Task 8. Let the random variable be given by the distribution:

			10m

Find its numerical characteristics.

Solution: m, m 2 ,

M 2 , m.

About a random variable X, one can say either - its mathematical expectation is 6.4 m with a variance of 13.04 m 2 , or - its mathematical expectation is 6.4 m with a deviation of m. The second formulation is obviously clearer.

A task 9. Random value X given by the distribution function:
.

Find the probability that, as a result of the test, the value X will take on a value contained in the interval .

Solution. The probability that X will take a value from a given interval is equal to the increment of the integral function in this interval, i.e. . In our case and , therefore

.

A task 10. Discrete random variable X given by the distribution law:

Find distribution function F(x ) and build its graph.

Solution. Since the distribution function

for , then

at ;

at ;

at ;

at ;

Relevant chart:

Task 11. Continuous random variable X given by the differential distribution function: .

Find the probability of hitting X to interval

Solution. Note that this is a special case of the exponential distribution law.

Let's use the formula: .

A task 12. Find the numerical characteristics of a discrete random variable X given by the distribution law:

	–5
	X 2 : x2 . , where is the Laplace function. The values ​​of this function are found using a table. In our case: . According to the table we find:, therefore:

Definition.Dispersion (scattering) Discrete random variable is called the mathematical expectation of the squared deviation of the random variable from its mathematical expectation:

Example. For the example above, we find

The mathematical expectation of a random variable is:

Possible values ​​of the squared deviation:

; ;

The dispersion is:

However, in practice, this method of calculating the variance is inconvenient, because leads to cumbersome calculations for a large number of values ​​of a random variable. Therefore, another method is used.

Variance Calculation

Theorem. The variance is equal to the difference between the mathematical expectation of the square of the random variable X and the square of its mathematical expectation:

Proof. Taking into account the fact that the mathematical expectation and the square of the mathematical expectation are constant values, we can write:

Let's apply this formula to the example above:

X
x2
p	0,0778	0,2592	0,3456	0,2304	0,0768	0,0102

Dispersion Properties

1) The dispersion of a constant value is zero:

2) The constant factor can be taken out of the dispersion sign by squaring it:

.

3) The variance of the sum of two independent random variables is equal to the sum of the variances of these variables:

4) The variance of the difference of two independent random variables is equal to the sum of the variances of these variables:

The validity of this equality follows from property 2.

Theorem. The variance of the number of occurrences of event A in n independent trials, in each of which the probability of occurrence of the event is constant, is equal to the product of the number of trials by the probability of occurrence and the probability of the event not occurring in each trial:

Example. The plant produces 96% of first grade products and 4% of second grade products. 1000 items are chosen at random. Let X- the number of products of the first grade in this sample. Find the distribution law, mathematical expectation and variance of a random variable.

Thus, the distribution law can be considered binomial.

Example. Find the variance of a discrete random variable X– number of occurrences of the event BUT in two independent trials, if the probabilities of the occurrence of this event in each trial are equal and it is known that

Because random value X distributed according to the binomial law, then

Example. Independent tests are performed with the same probability of occurrence of the event BUT in every test. Find the probability of an event occurring BUT if the variance of the number of occurrences of the event in three independent trials is 0.63.

According to the dispersion formula of the binomial law, we obtain:

;

Example. A device consisting of four independently operating devices is being tested. The probabilities of failure of each of the devices are equal, respectively ; ; . Find the mathematical expectation and variance of the number of failed devices.

Taking the number of failed devices as a random variable, we see that this random variable can take on the values ​​0, 1, 2, 3, or 4.

To draw up a distribution law for this random variable, it is necessary to determine the corresponding probabilities. Let's accept .

1) Not a single device failed:

2) One of the devices failed.

Random variable A quantity is called that, as a result of tests carried out under the same conditions, takes on different, generally speaking, values, depending on random factors that are not taken into account. Examples of random variables: the number of points dropped on a dice, the number of defective products in a batch, the deviation of the point of impact of the projectile from the target, the device's uptime, etc. Distinguish between discrete and continuous random variables. Discrete A random variable is called, the possible values ​​of which form a countable set, finite or infinite (i.e., such a set, the elements of which can be numbered).

Continuous A random variable is called, the possible values ​​of which continuously fill some finite or infinite interval of the numerical axis. The number of values ​​of a continuous random variable is always infinite.

Random variables will be denoted by capital letters of the end of the Latin alphabet: X, Y, . ; values ​​of a random variable - in lowercase letters: X, y. . In this way, X Denotes the entire set of possible values ​​of a random variable, and X - Some specific meaning.

distribution law A discrete random variable is a correspondence given in any form between the possible values ​​of a random variable and their probabilities.

Let the possible values ​​of the random variable X Are . As a result of the test, the random variable will take one of these values, i.e. One event from a complete group of pairwise incompatible events will occur.

Let also the probabilities of these events be known:

Distribution law of a random variable X It can be written in the form of a table called Near distribution Discrete random variable:

random variables. Discrete random variable.
Expected value

The second section on probability theory dedicated random variables , which invisibly accompanied us literally in every article on the topic. And the time has come to clearly articulate what it is:

Random called value, which as a result of the test will take one and only one a numerical value that depends on random factors and is not predictable in advance.

Random variables are usually designate through * , and their values ​​in the corresponding small letters with subscripts, for example, .

* Sometimes used as well as Greek letters

We came across an example on first lesson in probability theory, where we actually considered the following random variable:

- the number of points that will fall out after throwing a dice.

This test will result in one and only the line, which one is not predictable (tricks are not considered); in this case, the random variable can take one of the following values:

- the number of boys among 10 newborns.

It is quite clear that this number is not known in advance, and in the next ten children born there may be:

Or boys - one and only one of the listed options.

And, in order to keep in shape, a little physical education:

- long jump distance (in some units).

Even the master of sports is not able to predict it 🙂

However, what are your hypotheses?

As soon as set of real numbers infinite, then the random variable can take infinitely many values ​​from some interval. And this is its fundamental difference from the previous examples.

In this way, it is advisable to divide random variables into 2 large groups:

1) Discrete (intermittent) random variable - takes separately taken, isolated values. The number of these values certainly or infinite but countable.

... incomprehensible terms were drawn? Urgently repeat basics of algebra!

2) Continuous random variable - takes all numeric values ​​from some finite or infinite range.

Note : abbreviations DSV and NSV are popular in educational literature

First, let's analyze a discrete random variable, then - continuous.

Distribution law of a discrete random variable

- this is conformity between the possible values ​​of this quantity and their probabilities. Most often, the law is written in a table:

The term is quite common row distribution, but in some situations it sounds ambiguous, and therefore I will adhere to the "law".

And now very important point: since the random variable necessarily will accept one of the values, then the corresponding events form full group and the sum of the probabilities of their occurrence is equal to one:

or, if written folded:

So, for example, the law of the distribution of probabilities of points on a die has the following form:

You may be under the impression that a discrete random variable can only take on "good" integer values. Let's dispel the illusion - they can be anything:

Some game has the following payoff distribution law:

…probably you have been dreaming about such tasks for a long time 🙂 I will tell you a secret - me too. Especially after finishing work on field theory.

Solution: since a random variable can take only one of three values, the corresponding events form full group, which means that the sum of their probabilities is equal to one:

We expose the "partisan":

– thus, the probability of winning conventional units is 0.4.

Control: what you need to make sure.

Answer:

It is not uncommon when the distribution law needs to be compiled independently. For this use classical definition of probability, multiplication / addition theorems for event probabilities and other chips tervera:

There are 50 lottery tickets in the box, 12 of which are winning, and 2 of them win 1000 rubles each, and the rest - 100 rubles each. Draw up a law of distribution of a random variable - the size of the winnings, if one ticket is randomly drawn from the box.

Solution: as you noticed, it is customary to place the values ​​of a random variable in ascending order. Therefore, we start with the smallest winnings, and namely rubles.

In total, there are 50 - 12 = 38 such tickets, and according to classical definition:
is the probability that a randomly drawn ticket will not win.

The rest of the cases are simple. The probability of winning rubles is:

And for :

Checking: - and this is a particularly pleasant moment of such tasks!

Answer: the required payoff distribution law:

The following task for an independent decision:

The probability that the shooter will hit the target is . Make a distribution law for a random variable - the number of hits after 2 shots.

... I knew that you missed him 🙂 We remember multiplication and addition theorems. Solution and answer at the end of the lesson.

The distribution law completely describes a random variable, but in practice it is useful (and sometimes more useful) to know only some of it. numerical characteristics .

Mathematical expectation of a discrete random variable

In simple terms, this average expected value with repeated testing. Let a random variable take values ​​with probabilities, respectively. Then the mathematical expectation of this random variable is equal to sum of works all its values ​​by the corresponding probabilities:

or in folded form:

Let's calculate, for example, the mathematical expectation of a random variable - the number of points dropped on a dice:

What is the probabilistic meaning of the result obtained? If you roll the die enough times, then mean the points dropped will be close to 3.5 - and the more tests you do, the closer. Actually, I already talked about this effect in detail in the lesson about statistical probability.

Now let's recall our hypothetical game:

The question arises: is it even profitable to play this game? ... who has any impressions? So you can’t say “offhand”! But this question can be easily answered by calculating the mathematical expectation, in essence - weighted average probabilities of winning:

Thus, the mathematical expectation of this game losing.

Don't trust impressions - trust numbers!

Yes, here you can win 10 or even 20-30 times in a row, but in the long run we will inevitably be ruined. And I would not advise you to play such games 🙂 Well, maybe only for fun.

From all of the above, it follows that the mathematical expectation is NOT a RANDOM value.

Creative task for independent research:

Mr X plays European roulette according to the following system: he constantly bets 100 rubles on red. Compose the law of distribution of a random variable - its payoff. Calculate the mathematical expectation of winnings and round it up to kopecks. How average does the player lose for every hundred bet?

Reference : European roulette contains 18 red, 18 black and 1 green sector ("zero"). In the event of a “red” falling out, the player is paid a double bet, otherwise it goes to the casino’s income

There are many other roulette systems for which you can create your own probability tables. But this is the case when we do not need any distribution laws and tables, because it is established for certain that the mathematical expectation of the player will be exactly the same. Only changes from system to system dispersion, which we will learn about in part 2 of the lesson.

But before that, it will be useful to stretch your fingers on the keys of the calculator:

The random variable is given by its own probability distribution law:

Find if it is known that . Run a check.

Then we turn to the study dispersion of a discrete random variable, and if possible, RIGHT NOW!!- so as not to lose the thread of the topic.

Solutions and answers:

Example 3 Solution: by condition - the probability of hitting the target. Then:
is the probability of a miss.

Let's make - the law of distribution of hits at two shots:

- not a single hit. By the theorem of multiplication of probabilities of independent events:

- one hit. By theorems of addition of probabilities of incompatible and multiplication of independent events:

- two hits. According to the theorem of multiplication of probabilities of independent events:

Check: 0.09 + 0.42 + 0.49 = 1

Answer :

Note : it was possible to use designations - this is not important.

Example 4 Solution: the player wins 100 rubles in 18 cases out of 37, and therefore the law of distribution of his winnings has the following form:

Let's calculate the mathematical expectation:

Thus, for every hundred wagered, the player loses an average of 2.7 rubles.

Example 5 Solution: by definition of mathematical expectation:

Let's swap parts and make simplifications:

thus:

Let's check:

, which was to be verified.

Answer :

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Discrete random variables

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 $p_1,\dots ,\ p_n$.

$\begin
\hline
X_i & x_1 & x_2 & \dots & x_n \\
\hline
p_i & p_1 & p_2 & \dots & p_n \\
\hline
\end$

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 $X$:

$\begin
\hline
1 & 2 & 3 & 4 & 5 & 6 \\
\hline
1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 \\
\hline
\end$

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

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_ $. 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 $2$.

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:

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 $D\left(X\right)=\sum^n_ —^2$.

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 $2$.

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

Graph of the distribution function $F\left(x\right)$:

Basic laws of distribution

1. Binomial distribution law.

The binomial distribution law describes the probability of occurrence of event A m times in n independent trials, provided that the probability p of occurrence of event A in each trial is constant.

For example, the sales department of a hardware store receives, on average, one order for the purchase of televisions in 10 calls. Write a probability distribution law for the purchase of m TV sets. Construct a polygon of the probability distribution.

In the table, m is the number of orders received by the company for the purchase of a TV set. C n m is the number of combinations of m TVs by n, p is the probability of the occurrence of event A, i.e. ordering a TV, q is the probability that event A will not occur, i.e. not ordering a TV, P m,n is the probability of ordering m TVs out of n. Figure 1 shows the polygon of the probability distribution.

2.Geometric distribution.

The geometric distribution of a random variable has the following form:

P m is the probability of occurrence of event A in trial number m.
p is the probability of occurrence of event A in one trial.
q = 1 - p

Example. A home appliance repair company received a batch of 10 replacement units for washing machines. There are cases when a batch contains 1 defective block. A check is carried out until a defective block is found. It is necessary to draw up a distribution law for the number of checked blocks. The probability that a block may be defective is 0.1. Construct a polygon of the probability distribution.

It can be seen from the table that with an increase in the number m, the probability that a defective block will be detected decreases. The last line (m=10) combines two probabilities: 1 - that the tenth block turned out to be faulty - 0.038742049, 2 - that all the checked blocks turned out to be serviceable - 0.34867844. Since the probability of a block failing is relatively low (p=0.1), the probability of the last event P m (10 tested blocks) is relatively high. Fig.2.

3. Hypergeometric distribution.

The hypergeometric distribution of a random variable has the following form:

For example, to draw up a law of distribution of 7 guessed numbers out of 49. In this example, the total numbers N=49, n=7 numbers were removed, M is the total numbers that have a given property, i.e. correctly guessed numbers, m is the number of correctly guessed numbers among the withdrawn ones.

The table shows that the probability of guessing one number m=1 is higher than when m=0. However, then the probability begins to decline rapidly. Thus, the probability of guessing 4 numbers is already less than 0.005, and 5 is negligible.

4. Poisson distribution law.

A random variable X has a Poisson distribution if its distribution law has the form:

Np = const
n is the number of trials tending to infinity
p is the probability of the event occurring, tending to zero
m is the number of occurrences of event A

For example, on average, a TV company receives about 100 calls per day. The probability of ordering a brand A TV is 0.08; B - 0.06 and C - 0.04. Draw up the law of distribution of orders for the purchase of TV sets of brands A, B and C. Construct a polygon of probability distribution.

From the condition we have: m=100, ? 1=8, ? 2=6, ? 3 =4 (?10)

(table is not complete)

If n is large enough to go to infinity and p goes to zero so that the product np goes to a constant number, then this law is an approximation to the binomial distribution law. It can be seen from the graph that the greater the probability p, the closer the curve is to the m axis, i.e. more gentle. (Fig.4)

It should be noted that the binomial, geometric, hypergeometric and Poisson distribution laws express the probability distribution of a discrete random variable.

5. Uniform distribution law.

If the probability density? (x) is a constant value on a certain interval, then the distribution law is called uniform. Figure 5 shows the graphs of the probability distribution function and the probability density of the uniform distribution law.

6. Normal distribution law (Gauss law).

Among the laws of distribution of continuous random variables, the most common is the normal distribution law. A random variable is distributed according to the normal distribution law if its probability density has the form:

where
a is the mathematical expectation of a random variable
? — standard deviation

The graph of the probability density of a random variable with a normal distribution law is symmetrical with respect to the straight line x=a, i.e. x equal to the mathematical expectation. Thus, if x=a, then the curve has a maximum equal to:

When the value of the mathematical expectation changes, the curve will shift along the Ox axis. The graph (Fig. 6) shows that at x=3 the curve has a maximum, because the mathematical expectation is 3. If the mathematical expectation takes a different value, for example, a=6, then the curve will have a maximum at x=6. Speaking of the standard deviation, as you can see from the graph, the larger the standard deviation, the smaller the maximum value of the probability density of a random variable.

A function that expresses the distribution of a random variable on the interval (-?, x), and having a normal distribution law, is expressed through the Laplace function according to the following formula:

Those. the probability of a random variable X consists of two parts: the probability where x takes values ​​from minus infinity to a, equal to 0.5, and the second part is from a to x. (Fig.7)

Learning Together

Useful materials for students, diploma and term papers to order

Lesson: the law of distribution of a discrete random variable

The distribution law of a discrete random variable is the correspondence between possible values ​​and their probabilities. It can be specified tabularly, graphically and analytically.

What is a random variable is discussed in this lesson.

With the tabular way of setting, the first row of the table contains possible values, and the second their probabilities, that is

This quantity is called the distribution series. discrete random variable.

X=x1, X=x2, X=xn form a complete group, since in one trial the random variable will take one and only one possible value. Therefore, the sum of their probabilities is equal to one, that is, p1 + p2 + pn = 1 or

If the set of values ​​of X is infinite, then Example 1. There are 100 tickets issued in a cash lottery. One win of 1000 rubles and 10 of 100 rubles are played. Find the law of distribution of a random variable X - the cost of a possible win for the owner of one lottery ticket.

The desired distribution law has the form:

Control; 0.01+0.1+0.89=1.
With a graphical method of setting the distribution law, points are built on the coordinate plane (Xi: Pi), and then they are connected by straight line segments. The resulting broken line is called distribution polygon. For example 1, the distribution polygon is shown in Figure 1.

In the analytical method of setting the distribution law, a formula is indicated that relates the probabilities of a random variable with its possible values.

Examples of discrete distributions

Binomial distribution

Let n trials be made, in each of which the event A occurs with a constant probability p, therefore, does not occur with a constant probability q = 1- p. Consider a random variable X- the number of occurrences of event A in these n trials. The possible values ​​of X are x1 = 0 , x2 = 1,…, xn+1 = n . The probability of these possible

The law of distribution of a discrete random variable is called Windows XP Word 2003 Excel 2003 The laws of distribution of discrete random variables The law of distribution of a discrete random variable is any relation that establishes a relationship between the possible values ​​of a random variable and […]

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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 ​​- 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.

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.