Draw up the law of distribution of a random variable upon return. Distribution law of a random variable

Random variable A quantity is called which, as a result of tests carried out under the same conditions, takes on different, generally speaking, values ​​depending on random factors not taken into account. Examples of random variables: the number of points drawn per dice, the number of defective products in a batch, the deviation of the point of impact of the projectile from the target, the time of failure-free operation of the device, etc. There are discrete and continuous random variables. Discrete Called random variable, possible values which form a countable set, finite or infinite (that is, a set whose elements can be numbered).

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

We will denote random variables in capital letters end Latin alphabet: X, Y, . ; random variable values ​​– lowercase letters: X, y,. . Thus, X Denotes the entire set of possible values ​​of a random variable, and X - Some of its specific meaning.

Law of distribution A discrete random variable is a correspondence specified 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 the probabilities of these events also be known:

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

Random variables. Discrete random variable.
Expectation

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

Random called size, which as a result of the test will take one and only one a numerical value that depends on random factors and is unpredictable in advance.

Random variables are usually denote through * , and their meanings are written in corresponding small letters with subscripts, for example, .

* Sometimes Greek letters are also used

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

– the number of points that will appear after throwing the dice.

As a result of this test, it will fall out one and only the line, which one exactly, cannot be predicted (we do not consider tricks); in this case, the random variable can take one of the following values:

– the number of boys among 10 newborns.

It is absolutely clear that this number is not known in advance, and the next ten children born may include:

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

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

– long jump distance (in some units).

Even a master of sports cannot predict it :)

However, your hypotheses?

As soon as many real numbers infinitely, then the random variable can take infinitely many values ​​from a certain interval. And this is what it consists of fundamental difference from previous examples.

Thus, It is advisable to divide random variables into 2 large groups:

1) Discrete (intermittent) random variable – takes individual, isolated values. Number of these values Certainly or infinite but countable.

...are there any unclear terms? We urgently repeat algebra basics!

2) Continuous random variable – accepts All numeric values from some finite or infinite interval.

Note : V educational literature popular abbreviations DSV and NSV

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

Distribution law of a discrete random variable

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

The term is used quite often row distribution, but in some situations it sounds ambiguous, and so I will stick 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 condensed:

So, for example, the law of probability distribution of points rolled 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 next law winning distribution:

…you’ve probably dreamed of such tasks for a long time 🙂 I’ll tell you a secret – me too. Especially after finishing work on field theory.

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

Exposing the “partisan”:

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

Control: that’s what we needed to make sure of.

Answer:

It is not uncommon when you need to draw up a distribution law yourself. For this they use classical definition of probability, multiplication/addition theorems for event probabilities and other chips tervera:

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

Solution: as you noticed, the values ​​of a random variable are usually placed in in ascending order. Therefore, we start with the smallest winnings, namely rubles.

There are 50 such tickets in total - 12 = 38, and according to classical definition :
– the probability that a randomly drawn ticket will be a loser.

In other cases everything is simple. The probability of winning rubles is:

And for:

Check: – and this is special nice moment such tasks!

Answer: the desired law of distribution of winnings:

The following task is for you to solve on your own:

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

...I knew that you missed him :) Let's remember multiplication and addition theorems. The solution and answer are at the end of the lesson.

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

Expectation of a discrete random variable

Speaking in simple language, This average expected value when testing is repeated many times. Let the random variable take values ​​with probabilities accordingly. Then mathematical expectation of this random variable is equal to sum of products all its values ​​to the corresponding probabilities:

or collapsed:

Let us calculate, for example, the mathematical expectation of a random variable - the number of points rolled on a die:

What is the probabilistic meaning of the result obtained? If you roll the dice enough times, then average value The points dropped will be close to 3.5 - and the more tests you carry out, the closer. Actually, I already spoke in detail about this effect in the lesson about statistical probability.

Now let's remember our hypothetical game:

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

Thus, the mathematical expectation of this game losing.

Don't trust your impressions - trust the numbers!

Yes, here you can win 10 and even 20-30 times in a row, but in the long run we will face inevitable ruin. And I wouldn’t advise you to play such games :) Well, maybe only for fun.

From all of the above it follows that the mathematical expectation is no longer a RANDOM value.

Creative task for independent research:

Mr. X plays European roulette next system: constantly bets 100 rubles on “red”. Draw up a law of distribution of a random variable - its winnings. Calculate the mathematical expectation of winnings and round it to the nearest kopeck. How many on average Does the player lose for every hundred he bet?

Reference : European roulette contains 18 red, 18 black and 1 green sector (“zero”). If “red” is rolled out, the player is paid double the 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 has been established for certain that the player’s mathematical expectation will be exactly the same. The only thing that changes from system to system is dispersion, which we will learn about in the 2nd part of the lesson.

But first, it will be useful to stretch your fingers on the calculator keys:

A random variable is specified by its probability distribution law:

Find if it is known that . Perform check.

Then let's move on to studying variance 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:
– probability of miss.

Let’s compose the law of hit distribution for two shots:

- not a single hit. By probability multiplication theorem independent events :

- one hit. By theorems for 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 : you could use notations - it doesn’t matter.

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 bet, the player loses on average 2.7 rubles.

Example 5. Solution: by definition of mathematical expectation:

Let's swap parts and make simplifications:

Thus:

Let's check:

, which was what needed to be checked.

Answer :

(Go to main page)

High-quality works without plagiarism – Zaochnik.com

www.mathprofi.ru

Discrete random variables

Random variable A variable is called a variable that, as a result of each test, takes on one previously unknown value, depending on random reasons. Random variables are denoted in capitals in Latin letters: $X,\ Y,\ Z,\ \dots $ According to their type, random variables can be discrete And continuous.

Discrete random variable- this is a random variable whose values ​​can be no more than countable, that is, either finite or countable. By countability we mean that the values ​​of a random variable can be numbered.

Example 1 . Here are 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 emblems dropped when tossing a coin, here the possible values ​​are $0,\ 1,\ \dots ,\ n$.

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

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

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

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

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

Example 2 . Let the random variable $X$ be the number of points rolled when tossing a die. Such a random variable $X$ can take following values$1,\ 2,\ 3,\ 4,\ 5,\ 6$. The probabilities of all these values ​​are equal to $1/6$. Then the law of probability distribution of 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 in the law of distribution of a discrete random variable $X$ the events $1,\ 2,\ \dots ,\ 6$ form a complete group of events, then the sum of the probabilities must be equal to one, that is, $\sum

2. Mathematical expectation of a discrete random variable.

Expectation of a random variable sets its “central” meaning. 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, that is: $M\left(X\right)=\sum ^n_ $. In English-language literature, another notation $E\left(X\right)$ is used.

Properties of mathematical expectation$M\left(X\right)$:

1. $M\left(X\right)$ is contained between the smallest and highest values 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 sign of the mathematical expectation: $M\left(CX\right)=CM\left(X\right)$.
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 $M\left(X\right)$ lies 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 disperse differently around their average values. For example, in two student groups GPA for the exam in probability theory it turned out to be equal to 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 a numerical characteristic of a random variable that would show the spread of the values ​​of the random variable around its mathematical expectation. This characteristic is dispersion.

Variance of a discrete random variable$X$ is equal to:

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 using the formula $D\left(X\right)=\sum^n_ —^2$.

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

1. The variance is always greater than or equal to zero, i.e. $D\left(X\right)\ge 0$.
2. The variance of the constant is 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 between 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's 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 the random variable $X$ 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 called a function $F\left(x\right)$, which determines the probability that the random variable $X$ will take a value less than some fixed value $x$, that is, $F\left(x\right )=P\left(X 6$, 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 the occurrence of event A m times in n independent tests, provided that the probability p of the occurrence of event A in each trial is constant.

For example, the sales department of a home appliance store receives on average one order for the purchase of televisions out of 10 calls. Draw up a law of probability distribution for the purchase of m televisions. Construct a probability distribution polygon.

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

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 event A occurring in one trial.
q = 1 - p

Example. A household appliance repair company received a batch of 10 spare units for washing machines. There are cases when 1 block in a batch turns out to be defective. An inspection is carried out until a defective unit is detected. It is necessary to draw up a distribution law for the number of verified blocks. The probability that a block may be defective is 0.1. Construct a probability distribution polygon.

The table shows that as the number m increases, 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 checked blocks turned out to be good - 0.34867844. Since the probability that the unit will be faulty is relatively low (p = 0.1), then the probability last event P m (10 blocks tested) is relatively high. Fig.2.

3. Hypergeometric distribution.

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

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

The table shows that the probability of guessing one number m=1 is higher than with m=0. However, then the probability begins to decrease 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 tests tending to infinity
p is the probability of an event occurring, tending to zero
m is the number of occurrences of event A

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

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

(the table is not given in full)

If n is large enough to go to infinity and the value of p goes to zero, so the product np goes to constant number, That this law is an approximation to the binomial distribution law. From the graph it is clear that more likely p, the closer the curve is to the m axis, i.e. more flat. (Fig.4)

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

5.Uniform distribution law.

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

6.Normal distribution law (Gauss's law).

Among the laws of distribution of continuous random variables, the most common is normal law distributions. 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
? - average standard deviation

The probability density graph of a random variable that has a normal distribution law is symmetrical with respect to the straight line x=a, i.e. x is 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 about the standard deviation, as can be seen from the graph, the larger the standard deviation, the smaller maximum value probability density of a random variable.

A function that expresses the distribution of a random variable on the interval (-?, x), and has a normal distribution law, is expressed through the Laplace function using 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 - from a to x. (Fig.7)

Let's study together

Useful materials for students, diploma and term papers to order

Lesson: Distribution Law of a Discrete Random Variable

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

What a random variable is is discussed in this lesson.

With the tabular method of specifying, 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 wins of 100 rubles are drawn. Find the distribution law of the random variable X - the cost of a possible win for the owner of one lottery ticket.

The required distribution law has the form:

Control; 0.01+0.1+0.89=1.
At graphically setting the distribution law for coordinate plane construct points (Xi:Pi), and then connect them with straight segments. Received broken line called distribution polygon. For example 1, the distribution polygon is shown in Figure 1.

At analytical way assignments of the distribution law indicate a formula connecting the probabilities of a random variable with its possible values.

Examples of discrete distributions

Binomial distribution

Let n trials be performed, in each of which event A occurs with constant probability p, therefore, does not occur with constant probability q = 1- p. Consider the random variable X- the number of occurrences of event A in these n trials. 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 Laws of distribution of discrete random variables The law of distribution of a discrete random variable is any relationship that establishes a connection between the possible values ​​of a random variable and […]

As is known, random variable called variable quantity, which can take one or another value depending on the case. Random variables are denoted by capital letters of the Latin alphabet (X, Y, Z), and their values ​​are denoted by 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.

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, … .

V) by using distribution function F(x) , which determines for each value x the probability that the random variable X will take a value less than x, i.e. F(x) = P(X< x).

Properties of the function F(x)

3 . The distribution law can be specified graphically – distribution polygon (polygon) (see problem 3).

Note that to solve some problems it is not necessary to know the distribution law. In some cases, it is enough to know one or more numbers that reflect the most important features law of distribution. This can be a number that has the meaning of the “average value” of a random variable, or a number showing the average size of the deviation of a random variable from its mean value. Numbers of this kind are called numerical characteristics of a random variable.

Basic numerical characteristics of a discrete random variable :

• Mathematical expectation (average 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 were issued: 5 of them will win 500 rubles, 10 will win 100 rubles, 20 will win 50 rubles, 50 will win 10 rubles. Determine the law of probability distribution of the random variable X - winnings per ticket.

Solution. According to the conditions 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. Let us present the resulting law in the form of a table:

Let's find the mathematical expectation of the value X: M(X) = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = (1+ 2+3+4+5+6)/6 = 21/6 = 3.5

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, construct 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. The discrete random variable X = (the number of failed elements in one experiment) has the following possible values: x 1 =0 (none of the device elements failed), x 2 =1 (one element failed), x 3 =2 (two elements failed ) and x 4 =3 (three elements failed).

Failures of elements are independent of each other, the probabilities of failure of each element are equal, therefore it is applicable Bernoulli formula . Considering that, according to the condition, n=3, p=0.1, q=1-p=0.9, we determine the probabilities of the values:
P 3 (0) = C 3 0 p 0 q 3-0 = q 3 = 0.9 3 = 0.729;
P 3 (1) = C 3 1 p 1 q 3-1 = 3*0.1*0.9 2 = 0.243;
P 3 (2) = C 3 2 p 2 q 3-2 = 3*0.1 2 *0.9 = 0.027;
P 3 (3) = C 3 3 p 3 q 3-3 = p 3 =0.1 3 = 0.001;
Check: ∑p i = 0.729+0.243+0.027+0.001=1.

Thus, the desired binomial law distribution X has the form:

We plot the possible values ​​of x i along the abscissa axis, and the corresponding probabilities p i along the ordinate axis. Let's construct points M 1 (0; 0.729), M 2 (1; 0.243), M 3 (2; 0.027), M 4 (3; 0.001). By connecting these points with straight line segments, we obtain the desired distribution polygon.

3. Let's find the distribution function F(x) = Р(Х

Graph of function F(x)

4. For binomial distribution X:
- mathematical expectation M(X) = np = 3*0.1 = 0.3;
- variance D(X) = npq = 3*0.1*0.9 = 0.27;
- standard deviation σ(X) = √D(X) = √0.27 ≈ 0.52.

Examples of solving problems on the topic “Random variables”.

Task 1 . There are 100 tickets issued for the lottery. One winning of 50 USD was drawn. and ten wins of 10 USD each. Find the law of distribution of the value X - the cost of possible winnings.

Solution. Possible values ​​for X: x 1 = 0; x 2 = 10 and x 3 = 50. Since there are 89 “empty” tickets, then p 1 = 0.89, probability of winning $10. (10 tickets) – p 2 = 0.10 and to win 50 USD -p 3 = 0.01. Thus:

Easy to control: .

Task 2. The probability that the buyer has read the product advertisement in advance is 0.6 (p=0.6). Selective control of the quality of advertising is carried out by surveying buyers before the first one who has studied the advertising in advance. Draw up a distribution series for the number of buyers surveyed.

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

Task 3. A computer consists of three independently working elements: the system unit, the monitor and the 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 a distribution law for the number of failed elements during a power surge in the network.

Solution. Let's consider Bernoulli distribution(or binomial): the probability that n tests, event A will appear exactly k once: , or:

IN Let's return to the task.

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

x 0 =0 – none of the elements failed;

x 1 =1 – failure of one element;

x 2 =2 – failure of two elements;

x 3 =3 – failure of all elements.

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

, ,

, .

Control: .

Therefore, the required distribution law:

Problem 4. 5,000 rounds produced. 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, for very large

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

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

Problem 5. When firing until the first hit with hit probability p = 0.6 when firing, you need to find the probability that a hit will occur on the third shot.

Solution. Let us apply a geometric distribution: let independent trials be carried out, in each of which event A has a probability of occurrence p (and non-occurrence q = 1 – p). The test ends as soon as event A occurs.

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

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

Find the mathematical expectation.

Solution. .

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

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

Solution. Here .

Distribution law for the squared value of X 2 :

Required variance: .

Dispersion characterizes the measure of deviation (dispersion) of a random variable from its mathematical expectation.

Problem 8. Let a random variable be given by the distribution:

Find its numerical characteristics.

Solution: m, m 2 ,

M 2 , m.

About the random variable X we 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 more clear.

Task 9. Random variable X given by the distribution function:
.

Find the probability that as a result of the test the value X will take the 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

.

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

Find the distribution function F(x ) and plot it.

Solution. Since the distribution function,

For , That

at ;

at ;

at ;

at ;

Relevant chart:

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

Find the hit probability X per interval

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

Let's use the formula: .

Task 12. Find the numerical characteristics of a discrete random variable X specified by the distribution law:

LAW OF DISTRIBUTION AND CHARACTERISTICS

RANDOM VARIABLES

Random variables, their classification and methods of description.

A random quantity is a quantity that, as a result of experiment, can take on one or another value, but which one is not known in advance. For a random variable, therefore, you can only specify values, one of which it will definitely take as a result of experiment. In what follows we will call these values ​​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 corresponding small letters.

There are three types of random variables:

Discrete; Continuous; Mixed.

Discrete is a random variable whose number of possible values ​​forms a countable set. In turn, a set whose elements can be numbered is called countable. The word "discrete" comes from the Latin discretus, meaning "discontinuous, consisting of separate parts".

Example 1. A discrete random variable is the number of defective parts X in a batch of nproducts. 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 a random variable whose possible values ​​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 monthly electricity consumption of an enterprise.

Example 4. A continuous random variable is the error in measuring height using an altimeter. Let it be known from the operating principle 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 completely specified if its possible values ​​are indicated on the numerical axis and the distribution law is established.

Law of distribution of a random variable is a relation that establishes a connection 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, distribution function, probability density, and characteristic function are used as distribution laws.

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

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

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

Such a table is called a distribution series of a discrete random variable. 1

Events X 1, X 2,..., X n, consisting in the fact that as a result of the test, the random variable X will take the values ​​x 1, x 2,...x n, respectively, are inconsistent and the only possible ones (since 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").

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

Example The lottery includes: a car worth 5,000 den. units, 4 TVs costing 250 den. units, 5 video recorders worth 200 den. units A total of 1000 tickets are sold for 7 days. units Draw up a distribution law for the net winnings received by a lottery participant who bought one ticket.

Solution. Possible values ​​of the random variable X - the net winnings per ticket - are equal to 0-7 = -7 money. units (if the ticket did not win), 200-7 = 193, 250-7 = 243, 5000-7 = 4993 den. units (if the ticket has the winnings of a VCR, TV or car, respectively). Considering that out of 1000 tickets the number of non-winners is 990, and the indicated winnings are 5, 4 and 1, respectively, and using the classical definition of probability, we obtain.