The random variable is specified by the function. Examples of solving problems on the topic “Random variables”

In probability theory one has to deal with random variables, all of whose values ​​cannot be enumerated. For example, it is impossible to take and “iterate” all the values ​​of the random variable $X$ - the service time of the clock, since time can be measured in hours, minutes, seconds, milliseconds, etc. You can only specify a certain interval within which the values ​​of the random variable lie.

Continuous random variable is a random variable whose values ​​completely fill a certain interval.

Distribution function of a continuous random variable

Since it is not possible to enumerate all the values ​​of a continuous random variable, it can be specified using 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, $F\left(x\right )=P\left(X< x\right)$.

Properties of the distribution function:

1 . $0\le F\left(x\right)\le 1$.

2 . The probability that the random variable $X$ will take 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 1
0,\ x\le 0\\
x,\ 0< x\le 1\\
1,\ x>1
\end(matrix)\right.$. The probability of a random variable $X$ falling into the interval $\left(0.3;0.7\right)$ can be found as the difference between the values ​​of the distribution function $F\left(x\right)$ at the ends of this interval, that is:

$$P\left(0.3< X < 0,7\right)=F\left(0,7\right)-F\left(0,3\right)=0,7-0,3=0,4.$$

Probability distribution density

The function $f\left(x\right)=(F)"(x)$ is called the probability distribution density, that is, it is the first-order derivative taken from the distribution function $F\left(x\right)$ itself.

Properties of the function $f\left(x\right)$.

1 . $f\left(x\right)\ge 0$.

2 . $\int^x_(-\infty )(f\left(t\right)dt)=F\left(x\right)$.

3 . The probability that the random variable $X$ will take values ​​from the interval $\left(\alpha ;\ \beta \right)$ is $P\left(\alpha< X < \beta \right)=\int^{\beta }_{\alpha }{f\left(x\right)dx}$. Геометрически это означает, что вероятность попадания случайной величины $X$ в интервал $\left(\alpha ;\ \beta \right)$ равна площади криволинейной трапеции, которая будет ограничена графиком функции $f\left(x\right)$, прямыми $x=\alpha ,\ x=\beta $ и осью $Ox$.

4 . $\int^(+\infty )_(-\infty )(f\left(x\right))=1$.

Example 2 . A continuous random variable $X$ is defined by the following distribution function $F(x)=\left\(\begin(matrix)
0,\ x\le 0\\
x,\ 0< x\le 1\\
1,\ x>1
\end(matrix)\right.$. Then the density function $f\left(x\right)=(F)"(x)=\left\(\begin(matrix)
0,\x\le 0\\
1,\ 0 < x\le 1\\
0.\x>1
\end(matrix)\right.$

Expectation of a continuous random variable

The mathematical expectation of a continuous random variable $X$ is calculated using the formula

$$M\left(X\right)=\int^(+\infty )_(-\infty )(xf\left(x\right)dx).$$

Example 3 . Let's find $M\left(X\right)$ for the random variable $X$ from example $2$.

$$M\left(X\right)=\int^(+\infty )_(-\infty )(xf\left(x\right)\ dx)=\int^1_0(x\ dx)=(( x^2)\over (2))\bigg|_0^1=((1)\over (2)).$$

Variance of a continuous random variable

The variance of a continuous random variable $X$ is calculated by the formula

$$D\left(X\right)=\int^(+\infty )_(-\infty )(x^2f\left(x\right)\ dx)-(\left)^2.$$

Example 4 . Let's find $D\left(X\right)$ for the random variable $X$ from example $2$.

$$D\left(X\right)=\int^(+\infty )_(-\infty )(x^2f\left(x\right)\ dx)-(\left)^2=\int^ 1_0(x^2\ dx)-(\left(((1)\over (2))\right))^2=((x^3)\over (3))\bigg|_0^1-( (1)\over (4))=((1)\over (3))-((1)\over (4))=((1)\over(12)).$$

The probability distribution density of a continuous random variable (differential distribution function) is the first derivative of the integral distribution function: f(x)=F’(X). From this definition and the properties of the distribution function it follows that

The mathematical expectation of a continuous random variable X is the number

The variance of a continuous random variable X is determined by the equality

Example 79. Time distribution density T assembling REA on the production line

Find coefficient A, the distribution function of the REA assembly time and the probability that the assembly time will be within the interval (0.1A).

Solution. Based on the property of the distribution function of a random variable

Integrating by parts twice, we get

The distribution function is equal to

The probability that the assembly time of the REA will not exceed the limits (0; 1/λ):

Example 80. Probability density of deviation of the output resistance of the electronic equipment unit from the nominal value R 0 within the tolerance range 2δ is described by the law

Find the mathematical expectation and variance of the resistance deviation from the nominal value.

Solution.

Since the integrand is odd and the limits of integration are symmetrical about the origin, the integral is equal to 0.

Hence, M{R} = 0.

By making a substitution r = a sin x, we get

Example 81. The distribution density of a continuous random variable X is given:

Find: 1. F(x); 2. M(X); 3. D(X).

Solution. 1. To find F(x) we use the formula

If
, That

A

If
, That

If
, then f(x)=0, and

3.

Integrating by parts twice we get:

, Then

82. Find f(x), M(X), D(X) in problems 74, 75.

83. The distribution density of a continuous random variable X is given:

Find the distribution function F(x).

84. The distribution density of a continuous random variable X is given on the entire Ox axis by the equality
. Find the constant parameter C.

85. The random variable X in the interval (-3, 3) is given by the distribution density
; outside this interval

a) Find the variance of X;

b) which is more likely: the result of the test will be X<1 или X>1?

86. Find the variance of the random variable X given by the distribution function

87. A random variable is given by a distribution function

Find the expectation, variance and standard deviation of X.

§8. Uniform and exponential distributions

The distribution of a continuous random variable X is called uniform if on the interval (a,b), which contains all possible values ​​of X, the density remains constant, and outside this interval it is zero, i.e.

An exponential distribution is the probability distribution of a continuous random variable X, which is described by the density

where λ is a constant positive value. Exponential law distribution function

The mathematical expectation and variance are respectively equal

;
;

Example 88. The ammeter scale division value is 0.10A. Ammeter readings are rounded to the nearest whole division. Find the probability that during the counting an error will be made that exceeds 0.02A.

Solution. The rounding error can be considered as a random variable X, which is distributed uniformly in the interval (0;0.1) between two integer divisions. Hence,

Then
.

Example 89. The duration of failure-free operation of an element has an exponential distribution. Find the probability that during a time period of t=100 hours: a) the element will fail; b) the element will not fail.

Solution. a) By definition
, therefore it determines the probability of element failure in time t, therefore

b) The event “the element will not fail” is the opposite of the one considered, therefore its probability

90. The radio-electronic unit is assembled on a production line, assembly cycle is 2 minutes. The finished block is removed from the conveyor for control and adjustment at an arbitrary point in time within the clock cycle. Find the mathematical expectation and standard deviation of the time the finished block is on the conveyor. The time a block spends on the conveyor obeys the law of uniform distribution of random variables.

91. The probability of failure of a REA within a certain time is expressed by the formula . Determine the average operating time of the electronic equipment before failure.

92. The communications satellite being developed must have a mean time between failures of 5 years. Considering the real time between failures to be a random exponentially distributed value, determine the probability that

a) the satellite will operate for less than 5 years,

b) the satellite will operate for at least 10 years,

c) the satellite will fail within the 6th year.

93. A certain tenant bought four incandescent light bulbs with an average service life of 1000 hours. He installed one of them in a table lamp, and kept the rest in reserve in case the lamp burns out. Define:

a) the expected total service life of the four lamps,

b) the probability that the four lamps will operate for a total of 5000 hours or more,

c) the probability that the total service life of all lamps will not exceed 2000 hours.

94. The scale division value of a measuring device is 0.2. Instrument readings are rounded to the nearest whole division. Find the probability that an error will be made during the counting: a) less than 0.04; b) large 0.05.

95. Buses on a certain route run strictly on schedule. Movement interval 5 min. Find the probability that a passenger arriving at a stop will wait less than 3 minutes for the next bus.

96. Find the mathematical expectation of a random variable X, distributed uniformly in the interval (2, 8).

97. Find the variance and standard deviation of a random variable X, distributed uniformly in the interval (2, 8).

98. Two independently operating elements are tested. The duration of the failure-free operation of the first element has an exponential distribution
, second
. Find the probability that during a time duration t=6 hours: a) both elements will fail; b) both elements will not fail; c) only one element will fail; d) at least one element will fail.

To find the distribution function of a discrete random variable, you must use this calculator. Task 1. The distribution density of a continuous random variable X has the form:
Find:
a) parameter A;
b) distribution function F(x) ;
c) the probability of a random variable X falling into the interval;
d) mathematical expectation MX and variance DX.
Draw a graph of the functions f(x) and F(x).

Task 2. Find the variance of the random variable X given by the integral function.

Task 3. Find the mathematical expectation of the random variable X given the distribution function.

Task 4. The probability density of some random variable is given as follows: f(x) = A/x 4 (x = 1; +∞)
Find coefficient A, distribution function F(x), mathematical expectation and variance, as well as the probability that the random variable will take a value in the interval. Draw graphs f(x) and F(x).

Task. The distribution function of some continuous random variable is given as follows:

Determine parameters a and b, find an expression for the probability density f(x), mathematical expectation and variance, as well as the probability that the random variable will take a value in the interval. Draw graphs of f(x) and F(x).

Let's find the distribution density function as a derivative of the distribution function.

Knowing that

let's find parameter a:


or 3a=1, whence a = 1/3
We find the parameter b from the following properties:
F(4) = a*4 + b = 1
1/3*4 + b = 1 whence b = -1/3
Therefore, the distribution function has the form: F(x) = (x-1)/3

Example No. 1. The probability distribution density f(x) of a continuous random variable X is given. Required:

1. Determine coefficient A.
2. find the distribution function F(x) .
3. Schematically construct graphs of F(x) and f(x).
4. find the mathematical expectation and variance of X.
5. find the probability that X will take a value from the interval (2;3).

The random variable X is specified by the distribution density f(x):

Chapter 1. Discrete random variable

§ 1. Concepts of a random variable.

Distribution law of a discrete random variable.

Definition : Random is a quantity that, as a result of testing, takes only one value out of a possible set of its values, unknown in advance and depending on random reasons.

There are two types of random variables: discrete and continuous.

Definition : The random variable X is called discrete (discontinuous) if the set of its values ​​is finite or infinite but countable.

In other words, the possible values ​​of a discrete random variable can be renumbered.

A random variable can be described using its distribution law.

Definition : Distribution law of a discrete random variable call the correspondence between possible values ​​of a random variable and their probabilities.

The distribution law of a discrete random variable X can be specified in the form of a table, in the first row of which all possible values ​​of the random variable are indicated in ascending order, and in the second row the corresponding probabilities of these values, i.e.

where р1+ р2+…+ рn=1

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

If the set of possible values ​​of a random variable is infinite, then the series p1+ p2+…+ pn+… converges and its sum is equal to 1.

The distribution law of a discrete random variable X can be depicted graphically, for which a broken line is constructed in a rectangular coordinate system, connecting sequentially points with coordinates (xi; pi), i=1,2,…n. The resulting line is called distribution polygon (Fig. 1).

Organic chemistry" href="/text/category/organicheskaya_hiimya/" rel="bookmark">organic chemistry are 0.7 and 0.8, respectively. Draw up a distribution law for the random variable X - the number of exams that the student will pass.

Solution. The considered random variable X as a result of the exam can take one of the following values: x1=0, x2=1, x3=2.

Let's find the probability of these values. Let's denote the events:

https://pandia.ru/text/78/455/images/image004_81.jpg" width="259" height="66 src=">

So, the distribution law of the random variable X is given by the table:

Control: 0.6+0.38+0.56=1.

§ 2. Distribution function

A complete description of a random variable is also given by the distribution function.

Definition: Distribution function of a discrete random variable X is called a function F(x), which determines for each value x the probability that the random variable X will take a value less than x:

F(x)=P(X<х)

Geometrically, the distribution function is interpreted as the probability that the random variable X will take the value that is represented on the number line by a point lying to the left of point x.

1)0≤ F(x) ≤1;

2) F(x) is a non-decreasing function on (-∞;+∞);

3) F(x) - continuous on the left at points x= xi (i=1,2,...n) and continuous at all other points;

4) F(-∞)=P (X<-∞)=0 как вероятность невозможного события Х<-∞,

F(+∞)=P(X<+∞)=1 как вероятность достоверного события Х<-∞.

If the distribution law of a discrete random variable X is given in the form of a table:

then the distribution function F(x) is determined by the formula:

https://pandia.ru/text/78/455/images/image007_76.gif" height="110">

0 for x≤ x1,

р1 at x1< х≤ x2,

F(x)= р1 + р2 at x2< х≤ х3

1 for x>xn.

Its graph is shown in Fig. 2:

§ 3. Numerical characteristics of a discrete random variable.

One of the important numerical characteristics is the mathematical expectation.

Definition: Mathematical expectation M(X) discrete random variable X is the sum of the products of all its values ​​and their corresponding probabilities:

M(X) = ∑ xiрi= x1р1 + x2р2+…+ xnрn

The mathematical expectation serves as a characteristic of the average value of a random variable.

Properties of mathematical expectation:

1)M(C)=C, where C is a constant value;

2)M(C X)=C M(X),

3)M(X±Y)=M(X) ±M(Y);

4)M(X Y)=M(X) M(Y), where X, Y are independent random variables;

5)M(X±C)=M(X)±C, where C is a constant value;

To characterize the degree of dispersion of possible values ​​of a discrete random variable around its mean value, dispersion is used.

Definition: Variance D ( X ) random variable X is the mathematical expectation of the squared deviation of the random variable from its mathematical expectation:

Dispersion properties:

1)D(C)=0, where C is a constant value;

2)D(X)>0, where X is a random variable;

3)D(C X)=C2 D(X), where C is a constant value;

4)D(X+Y)=D(X)+D(Y), where X, Y are independent random variables;

To calculate variance it is often convenient to use the formula:

D(X)=M(X2)-(M(X))2,

where M(X)=∑ xi2рi= x12р1 + x22р2+…+ xn2рn

The variance D(X) has the dimension of a squared random variable, which is not always convenient. Therefore, the value √D(X) is also used as an indicator of the dispersion of possible values ​​of a random variable.

Definition: Standard deviation σ(X) random variable X is called the square root of the variance:

Task No. 2. The discrete random variable X is specified by the distribution law:

Find P2, the distribution function F(x) and plot its graph, as well as M(X), D(X), σ(X).

Solution: Since the sum of the probabilities of possible values ​​of the random variable X is equal to 1, then

Р2=1- (0.1+0.3+0.2+0.3)=0.1

Let's find the distribution function F(x)=P(X

Geometrically, this equality can be interpreted as follows: F(x) is the probability that the random variable will take the value that is represented on the number axis by a point lying to the left of point x.

If x≤-1, then F(x)=0, since there is not a single value of this random variable on (-∞;x);

If -1<х≤0, то F(х)=Р(Х=-1)=0,1, т. к. в промежуток (-∞;х) попадает только одно значение x1=-1;

If 0<х≤1, то F(х)=Р(Х=-1)+ Р(Х=0)=0,1+0,1=0,2, т. к. в промежуток

(-∞;x) there are two values ​​x1=-1 and x2=0;

If 1<х≤2, то F(х)=Р(Х=-1) + Р(Х=0)+ Р(Х=1)= 0,1+0,1+0,3=0,5, т. к. в промежуток (-∞;х) попадают три значения x1=-1, x2=0 и x3=1;

If 2<х≤3, то F(х)=Р(Х=-1) + Р(Х=0)+ Р(Х=1)+ Р(Х=2)= 0,1+0,1+0,3+0,2=0,7, т. к. в промежуток (-∞;х) попадают четыре значения x1=-1, x2=0,x3=1 и х4=2;

If x>3, then F(x)=P(X=-1) + P(X=0)+ P(X=1)+ P(X=2)+P(X=3)= 0.1 +0.1+0.3+0.2+0.3=1, because four values ​​x1=-1, x2=0, x3=1, x4=2 fall into the interval (-∞;x) and x5=3.

https://pandia.ru/text/78/455/images/image006_89.gif" width="14 height=2" height="2"> 0 at x≤-1,

0.1 at -1<х≤0,

0.2 at 0<х≤1,

F(x)= 0.5 at 1<х≤2,

0.7 at 2<х≤3,

1 at x>3

Let us represent the function F(x) graphically (Fig. 3):

https://pandia.ru/text/78/455/images/image014_24.jpg" width="158 height=29" height="29">≈1.2845.

§ 4. Binomial distribution law

discrete random variable, Poisson's law.

Definition: Binomial is called the law of distribution of a discrete random variable X - the number of occurrences of event A in n independent repeated trials, in each of which event A may occur with probability p or not occur with probability q = 1-p. Then P(X=m) - the probability of occurrence of event A exactly m times in n trials is calculated using the Bernoulli formula:

Р(Х=m)=Сmnpmqn-m

The mathematical expectation, dispersion and standard deviation of a random variable X distributed according to a binary law are found, respectively, using the formulas:

https://pandia.ru/text/78/455/images/image016_31.gif" width="26"> The probability of event A - “rolling out a five” in each trial is the same and equal to 1/6, i.e. . P(A)=p=1/6, then P(A)=1-p=q=5/6, where

- “failure to get an A.”

The random variable X can take the following values: 0;1;2;3.

We find the probability of each of the possible values ​​of X using Bernoulli’s formula:

Р(Х=0)=Р3(0)=С03р0q3=1 (1/6)0 (5/6)3=125/216;

Р(Х=1)=Р3(1)=С13р1q2=3 (1/6)1 (5/6)2=75/216;

Р(Х=2)=Р3(2)=С23р2q =3 (1/6)2 (5/6)1=15/216;

Р(Х=3)=Р3(3)=С33р3q0=1 (1/6)3 (5/6)0=1/216.

That. the distribution law of the random variable X has the form:

Control: 125/216+75/216+15/216+1/216=1.

Let's find the numerical characteristics of the random variable X:

M(X)=np=3 (1/6)=1/2,

D(X)=npq=3 (1/6) (5/6)=5/12,

Task No. 4. An automatic machine stamps parts. The probability that a manufactured part will be defective is 0.002. Find the probability that among 1000 selected parts there will be:

a) 5 defective;

b) at least one is defective.

Solution: The number n=1000 is large, the probability of producing a defective part p=0.002 is small, and the events under consideration (the part turns out to be defective) are independent, therefore the Poisson formula holds:

Рn(m)= e- λ λm

Let's find λ=np=1000 0.002=2.

a) Find the probability that there will be 5 defective parts (m=5):

Р1000(5)= e-2 25 = 32 0,13534 = 0,0361

b) Find the probability that there will be at least one defective part.

Event A - “at least one of the selected parts is defective” is the opposite of the event - “all selected parts are not defective.” Therefore, P(A) = 1-P(). Hence the desired probability is equal to: P(A)=1-P1000(0)=1- e-2 20 = 1- e-2=1-0.13534≈0.865.

Tasks for independent work.

1.1

1.2. The dispersed random variable X is specified by the distribution law:

Find p4, the distribution function F(X) and plot its graph, as well as M(X), D(X), σ(X).

1.3. There are 9 markers in the box, 2 of which no longer write. Take 3 markers at random. Random variable X is the number of writing markers among those taken. Draw up a law of distribution of a random variable.

1.4. There are 6 textbooks randomly arranged on a library shelf, 4 of which are bound. The librarian takes 4 textbooks at random. Random variable X is the number of bound textbooks among those taken. Draw up a law of distribution of a random variable.

1.5. There are two tasks on the ticket. The probability of correctly solving the first problem is 0.9, the second is 0.7. Random variable X is the number of correctly solved problems in the ticket. Draw up a distribution law, calculate the mathematical expectation and variance of this random variable, and also find the distribution function F(x) and build its graph.

1.6. Three shooters are shooting at a target. The probability of hitting the target with one shot is 0.5 for the first shooter, 0.8 for the second, and 0.7 for the third. Random variable X is the number of hits on the target if the shooters fire one shot at a time. Find the distribution law, M(X),D(X).

1.7. A basketball player throws the ball into the basket with a probability of hitting each shot of 0.8. For each hit, he receives 10 points, and if he misses, no points are awarded to him. Draw up a distribution law for the random variable X - the number of points received by a basketball player in 3 shots. Find M(X),D(X), as well as the probability that he gets more than 10 points.

1.8. Letters are written on the cards, a total of 5 vowels and 3 consonants. 3 cards are chosen at random, and each time the taken card is returned back. Random variable X is the number of vowels among those taken. Draw up a distribution law and find M(X),D(X),σ(X).

1.9. On average, under 60% of contracts, the insurance company pays insurance amounts in connection with the occurrence of an insured event. Draw up a distribution law for the random variable X - the number of contracts for which the insurance amount was paid among four contracts selected at random. Find the numerical characteristics of this quantity.

1.10. The radio station sends call signs (no more than four) at certain intervals until two-way communication is established. The probability of receiving a response to a call sign is 0.3. Random variable X is the number of call signs sent. Draw up a distribution law and find F(x).

1.11. There are 3 keys, of which only one fits the lock. Draw up a law for the distribution of the random variable X-number of attempts to open the lock, if the tried key does not participate in subsequent attempts. Find M(X),D(X).

1.12. Consecutive independent tests of three devices are carried out for reliability. Each subsequent device is tested only if the previous one turned out to be reliable. The probability of passing the test for each device is 0.9. Draw up a distribution law for the random variable X-number of tested devices.

1.13 .Discrete random variable X has three possible values: x1=1, x2, x3, and x1<х2<х3. Вероятность того, что Х примет значения х1 и х2, соответственно равны 0,3 и 0,2. Известно, что М(Х)=2,2, D(X)=0,76. Составить закон распределения случайной величины.

1.14. The electronic device block contains 100 identical elements. The probability of failure of each element during time T is 0.002. The elements work independently. Find the probability that no more than two elements will fail during time T.

1.15. The textbook was published in a circulation of 50,000 copies. The probability that the textbook is bound incorrectly is 0.0002. Find the probability that the circulation contains:

a) four defective books,

b) less than two defective books.

1 .16. The number of calls arriving at the PBX every minute is distributed according to Poisson's law with the parameter λ=1.5. Find the probability that in a minute the following will arrive:

a) two calls;

b) at least one call.

1.17.

Find M(Z),D(Z) if Z=3X+Y.

1.18. The laws of distribution of two independent random variables are given:

Find M(Z),D(Z) if Z=X+2Y.

Answers:

https://pandia.ru/text/78/455/images/image007_76.gif" height="110"> 1.1. p3=0.4; 0 at x≤-2,

0.3 at -2<х≤0,

F(x)= 0.5 at 0<х≤2,

0.9 at 2<х≤5,

1 at x>5

1.2. p4=0.1; 0 at x≤-1,

0.3 at -1<х≤0,

0.4 at 0<х≤1,

F(x)= 0.6 at 1<х≤2,

0.7 at 2<х≤3,

1 at x>3

M(X)=1; D(X)=2.6; σ(X) ≈1.612.

https://pandia.ru/text/78/455/images/image025_24.gif" width="2 height=98" height="98"> 0 at x≤0,

0.03 at 0<х≤1,

F(x)= 0.37 at 1<х≤2,

1 for x>2

M(X)=2; D(X)=0.62

M(X)=2.4; D(X)=0.48, P(X>10)=0.896

1. 8 .

M(X)=15/8; D(X)=45/64; σ(X) ≈

M(X)=2.4; D(X)=0.96

https://pandia.ru/text/78/455/images/image008_71.gif" width="14"> 1.11.

M(X)=2; D(X)=2/3

1.14. 1.22 e-0.2≈0.999

1.15. a)0.0189; b) 0.00049

1.16. a)0.0702; b)0.77687

1.17. 3,8; 14,2

1.18. 11,2; 4.

Chapter 2. Continuous random variable

Definition: Continuous They call a quantity all possible values ​​of which completely fill a finite or infinite span of the number line.

Obviously, the number of possible values ​​of a continuous random variable is infinite.

A continuous random variable can be specified using a distribution function.

Definition: F distribution function a continuous random variable X is called a function F(x), which determines for each value xhttps://pandia.ru/text/78/455/images/image028_11.jpg" width="14" height="13">R

The distribution function is sometimes called the cumulative distribution function.

Properties of the distribution function:

1)1≤ F(x) ≤1

2) For a continuous random variable, the distribution function is continuous at any point and differentiable everywhere, except, perhaps, at individual points.

3) The probability of a random variable X falling into one of the intervals (a;b), [a;b], [a;b], is equal to the difference between the values ​​of the function F(x) at points a and b, i.e. R(a)<Х

4) The probability that a continuous random variable X will take one separate value is 0.

5) F(-∞)=0, F(+∞)=1

Specifying a continuous random variable using a distribution function is not the only way. Let us introduce the concept of probability distribution density (distribution density).

Definition : Probability distribution density f ( x ) of a continuous random variable X is the derivative of its distribution function, i.e.:

The probability density function is sometimes called the differential distribution function or differential distribution law.

The graph of the probability density distribution f(x) is called probability distribution curve .

Properties of probability density distribution:

1) f(x) ≥0, at xhttps://pandia.ru/text/78/455/images/image029_10.jpg" width="285" height="141">.gif" width="14" height ="62 src="> 0 at x≤2,

f(x)= c(x-2) at 2<х≤6,

0 for x>6.

Find: a) the value of c; b) distribution function F(x) and plot it; c) P(3≤x<5)

Solution:

+ ∞

a) We find the value of c from the normalization condition: ∫ f(x)dx=1.

Therefore, -∞

https://pandia.ru/text/78/455/images/image032_23.gif" height="38 src="> -∞ 2 2 x

if 2<х≤6, то F(x)= ∫ 0dx+∫ 1/8(х-2)dx=1/8(х2/2-2х) = 1/8(х2/2-2х - (4/2-4))=

1/8(x2/2-2x+2)=1/16(x-2)2;

Gif" width="14" height="62"> 0 at x≤2,

F(x)= (x-2)2/16 at 2<х≤6,

1 for x>6.

The graph of the function F(x) is shown in Fig. 3

https://pandia.ru/text/78/455/images/image034_23.gif" width="14" height="62 src="> 0 at x≤0,

F(x)= (3 arctan x)/π at 0<х≤√3,

1 for x>√3.

Find the differential distribution function f(x)

Solution: Since f(x)= F’(x), then

https://pandia.ru/text/78/455/images/image011_36.jpg" width="118" height="24">

All properties of mathematical expectation and dispersion, discussed earlier for dispersed random variables, are also valid for continuous ones.

Task No. 3. The random variable X is specified by the differential function f(x):

https://pandia.ru/text/78/455/images/image036_19.gif" height="38"> -∞ 2

X3/9 + x2/6 = 8/9-0+9/6-4/6=31/18,

https://pandia.ru/text/78/455/images/image032_23.gif" height="38"> +∞

D(X)= ∫ x2 f(x)dx-(M(x))2=∫ x2 x/3 dx+∫1/3x2 dx=(31/18)2=x4/12 + x3/9 -

- (31/18)2=16/12-0+27/9-8/9-(31/18)2=31/9- (31/18)2==31/9(1-31/36)=155/324,

https://pandia.ru/text/78/455/images/image032_23.gif" height="38">

P(1<х<5)= ∫ f(x)dx=∫ х/3 dx+∫ 1/3 dx+∫ 0 dx= х2/6 +1/3х =

4/6-1/6+1-2/3=5/6.

Problems for independent solution.

2.1. A continuous random variable X is specified by the distribution function:

0 at x≤0,

F(x)= https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤ π/6,

F(x)= - cos 3x at π/6<х≤ π/3,

1 for x> π/3.

Find the differential distribution function f(x), and also

Р(2π /9<Х< π /2).

2.3.

0 at x≤2,

f(x)= c x at 2<х≤4,

0 for x>4.

2.4. A continuous random variable X is specified by the distribution density:

0 at x≤0,

f(x)= c √x at 0<х≤1,

0 for x>1.

Find: a) number c; b) M(X), D(X).

2.5.

https://pandia.ru/text/78/455/images/image041_3.jpg" width="36" height="39"> at x,

0 at x.

Find: a) F(x) and plot it; b) M(X),D(X), σ(X); c) the probability that in four independent trials the value of X will take exactly 2 times the value belonging to the interval (1;4).

2.6. The probability distribution density of a continuous random variable X is given:

f(x)= 2(x-2) at x,

0 at x.

Find: a) F(x) and plot it; b) M(X),D(X), σ (X); c) the probability that in three independent trials the value of X will take exactly 2 times the value belonging to the segment .

2.7. The function f(x) is given as:

https://pandia.ru/text/78/455/images/image045_4.jpg" width="43" height="38 src=">.jpg" width="16" height="15">[-√ 3/2; √3/2].

2.8. The function f(x) is given as:

https://pandia.ru/text/78/455/images/image046_5.jpg" width="45" height="36 src="> .jpg" width="16" height="15">[- π /4 ; π /4].

Find: a) the value of the constant c at which the function will be the probability density of some random variable X; b) distribution function F(x).

2.9. The random variable X, concentrated on the interval (3;7), is specified by the distribution function F(x)= . Find the probability that

random variable X will take the value: a) less than 5, b) not less than 7.

2.10. Random variable X, concentrated on the interval (-1;4),

is given by the distribution function F(x)= . Find the probability that

the random variable X will take the value: a) less than 2, b) not less than 4.

2.11.

https://pandia.ru/text/78/455/images/image049_6.jpg" width="43" height="44 src="> .jpg" width="16" height="15">.

Find: a) number c; b) M(X); c) probability P(X> M(X)).

2.12. The random variable is specified by the differential distribution function:

https://pandia.ru/text/78/455/images/image050_3.jpg" width="60" height="38 src=">.jpg" width="16 height=15" height="15"> .

Find: a) M(X); b) probability P(X≤M(X))

2.13. The Rem distribution is given by the probability density:

https://pandia.ru/text/78/455/images/image052_5.jpg" width="46" height="37"> for x ≥0.

Prove that f(x) is indeed a probability density function.

2.14. The probability distribution density of a continuous random variable X is given:

https://pandia.ru/text/78/455/images/image054_3.jpg" width="174" height="136 src=">(Fig. 4) (Fig.5)

2.16. The random variable X is distributed according to the “right triangle” law in the interval (0;4) (Fig. 5). Find an analytical expression for the probability density f(x) on the entire number line.

Answers

0 at x≤0,

f(x)= https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤ π/6,

F(x)= 3sin 3x at π/6<х≤ π/3,

0 for x> π/3. A continuous random variable X has a uniform distribution law on a certain interval (a;b), to which all possible values ​​of X belong, if the probability distribution density f(x) is constant on this interval and equal to 0 outside it, i.e.

0 for x≤a,

f(x)= for a<х

0 for x≥b.

The graph of the function f(x) is shown in Fig. 1

https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤a,

F(x)= https://pandia.ru/text/78/455/images/image077_3.jpg" width="30" height="37">, D(X)=, σ(X)=.

Task No. 1. The random variable X is uniformly distributed on the segment. Find:

a) probability distribution density f(x) and plot it;

b) the distribution function F(x) and plot it;

c) M(X),D(X), σ(X).

Solution: Using the formulas discussed above, with a=3, b=7, we find:

https://pandia.ru/text/78/455/images/image081_2.jpg" width="22" height="39"> at 3≤х≤7,

0 for x>7

Let's build its graph (Fig. 3):

https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86 src="> 0 at x≤3,

F(x)= https://pandia.ru/text/78/455/images/image084_3.jpg" width="203" height="119 src=">Fig. 4

D(X) = ==https://pandia.ru/text/78/455/images/image089_1.jpg" width="37" height="43">==https://pandia.ru/text/ 78/455/images/image092_10.gif" width="14" height="49 src="> 0 at x<0,

f(x)= λе-λх for x≥0.

The distribution function of a random variable X, distributed according to the exponential law, is given by the formula:

https://pandia.ru/text/78/455/images/image094_4.jpg" width="191" height="126 src=">fig..jpg" width="22" height="30"> , D(X)=, σ (Х)=

Thus, the mathematical expectation and the standard deviation of the exponential distribution are equal to each other.

The probability of X falling into the interval (a;b) is calculated by the formula:

P(a<Х

Task No. 2. The average failure-free operation time of the device is 100 hours. Assuming that the failure-free operation time of the device has an exponential distribution law, find:

a) probability distribution density;

b) distribution function;

c) the probability that the device’s failure-free operation time will exceed 120 hours.

Solution: According to the condition, the mathematical distribution M(X)=https://pandia.ru/text/78/455/images/image098_10.gif" height="43 src="> 0 at x<0,

a) f(x)= 0.01e -0.01x for x≥0.

b) F(x)= 0 at x<0,

1-e -0.01x at x≥0.

c) We find the desired probability using the distribution function:

P(X>120)=1-F(120)=1-(1- e -1.2)= e -1.2≈0.3.

§ 3.Normal distribution law

Definition: A continuous random variable X has normal distribution law (Gauss's law), if its distribution density has the form:

,

where m=M(X), σ2=D(X), σ>0.

The normal distribution curve is called normal or Gaussian curve (Fig.7)

The normal curve is symmetrical with respect to the straight line x=m, has a maximum at x=a, equal to .

The distribution function of a random variable X, distributed according to the normal law, is expressed through the Laplace function Ф (x) according to the formula:

,

where is the Laplace function.

Comment: The function Ф(x) is odd (Ф(-х)=-Ф(х)), in addition, for x>5 we can assume Ф(х) ≈1/2.

The graph of the distribution function F(x) is shown in Fig. 8

https://pandia.ru/text/78/455/images/image106_4.jpg" width="218" height="33">

The probability that the absolute value of the deviation is less than a positive number δ is calculated by the formula:

In particular, for m=0 the following equality holds:

"Three Sigma Rule"

If a random variable X has a normal distribution law with parameters m and σ, then it is almost certain that its value lies in the interval (a-3σ; a+3σ), because

https://pandia.ru/text/78/455/images/image110_2.jpg" width="157" height="57 src=">a)

b) Let's use the formula:

https://pandia.ru/text/78/455/images/image112_2.jpg" width="369" height="38 src=">

From the table of function values ​​Ф(х) we find Ф(1.5)=0.4332, Ф(1)=0.3413.

So, the desired probability:

P(28

Tasks for independent work

3.1. The random variable X is uniformly distributed in the interval (-3;5). Find:

b) distribution function F(x);

c) numerical characteristics;

d) probability P(4<х<6).

3.2. The random variable X is uniformly distributed on the segment. Find:

a) distribution density f(x);

b) distribution function F(x);

c) numerical characteristics;

d) probability P(3≤х≤6).

3.3. There is an automatic traffic light on the highway, in which the green light is on for 2 minutes, yellow for 3 seconds, red for 30 seconds, etc. A car drives along the highway at a random moment in time. Find the probability that a car will pass a traffic light without stopping.

3.4. Subway trains run regularly at intervals of 2 minutes. A passenger enters the platform at a random time. What is the probability that a passenger will have to wait more than 50 seconds for a train? Find the mathematical expectation of the random variable X - the waiting time for the train.

3.5. Find the variance and standard deviation of the exponential distribution given by the distribution function:

F(x)= 0 at x<0,

1st-8x for x≥0.

3.6. A continuous random variable X is specified by the probability distribution density:

f(x)= 0 at x<0,

0.7 e-0.7x at x≥0.

a) Name the distribution law of the random variable under consideration.

b) Find the distribution function F(X) and the numerical characteristics of the random variable X.

3.7. The random variable X is distributed according to the exponential law specified by the probability distribution density:

f(x)= 0 at x<0,

0.4 e-0.4 x at x≥0.

Find the probability that as a result of the test X will take a value from the interval (2.5;5).

3.8. A continuous random variable X is distributed according to the exponential law specified by the distribution function:

F(x)= 0 at x<0,

1st-0.6x at x≥0

Find the probability that, as a result of the test, X will take a value from the segment.

3.9. The expected value and standard deviation of a normally distributed random variable are 8 and 2, respectively. Find:

a) distribution density f(x);

b) the probability that as a result of the test X will take a value from the interval (10;14).

3.10. The random variable X is normally distributed with a mathematical expectation of 3.5 and a variance of 0.04. Find:

a) distribution density f(x);

b) the probability that as a result of the test X will take a value from the segment .

3.11. The random variable X is normally distributed with M(X)=0 and D(X)=1. Which of the events: |X|≤0.6 or |X|≥0.6 is more likely?

3.12. The random variable X is distributed normally with M(X)=0 and D(X)=1. From which interval (-0.5;-0.1) or (1;2) is it more likely to take a value during one test?

3.13. The current price per share can be modeled using the normal distribution law with M(X)=10 den. units and σ (X)=0.3 den. units Find:

a) the probability that the current share price will be from 9.8 den. units up to 10.4 days units;

b) using the “three sigma rule”, find the boundaries within which the current stock price will be located.

3.14. The substance is weighed without systematic errors. Random weighing errors are subject to the normal law with the mean square ratio σ=5g. Find the probability that in four independent experiments an error in three weighings will not occur in absolute value 3r.

3.15. The random variable X is normally distributed with M(X)=12.6. The probability of a random variable falling into the interval (11.4;13.8) is 0.6826. Find the standard deviation σ.

3.16. The random variable X is distributed normally with M(X)=12 and D(X)=36. Find the interval into which the random variable X will fall as a result of the test with a probability of 0.9973.

3.17. A part manufactured by an automatic machine is considered defective if the deviation X of its controlled parameter from the nominal value exceeds modulo 2 units of measurement. It is assumed that the random variable X is normally distributed with M(X)=0 and σ(X)=0.7. What percentage of defective parts does the machine produce?

3.18. The X parameter of the part is distributed normally with a mathematical expectation of 2 equal to the nominal value and a standard deviation of 0.014. Find the probability that the deviation of X from the nominal value will not exceed 1% of the nominal value.

Answers

https://pandia.ru/text/78/455/images/image116_9.gif" width="14" height="110 src=">

b) 0 for x≤-3,

F(x)= left">

3.10. a)f(x)= ,

b) Р(3.1≤Х≤3.7) ≈0.8185.

3.11. |x|≥0.6.

3.12. (-0,5;-0,1).

3.13. a) P(9.8≤Х≤10.4) ≈0.6562.

3.14. 0,111.

3.15. σ=1.2.

3.16. (-6;30).

3.17. 0,4%.

9. Continuous random variable, its numerical characteristics

A continuous random variable can be specified using two functions. Integral probability distribution function of random variable X is called a function defined by the equality
.

The integral function provides a general way to specify both discrete and continuous random variables. In the case of a continuous random variable. All events: have the same probability, equal to the increment of the integral function on this interval, i.e.. For example, for the discrete random variable specified in example 26, we have:

Thus, the graph of the integral function of the function under consideration is a union of two rays and three segments parallel to the Ox axis.

Example 27. Continuous random variable X is specified by the integral probability distribution function

.

Construct a graph of the integral function and find the probability that, as a result of the test, the random variable X will take a value in the interval (0.5;1.5).

Solution. On the interval
the graph is the straight line y = 0. In the interval from 0 to 2 there is a parabola given by the equation
. On the interval
The graph is the straight line y = 1.

The probability that the random variable X as a result of the test will take a value in the interval (0.5;1.5) is found using the formula.

Thus, .

Properties of the integral probability distribution function:

It is convenient to specify the distribution law of a continuous random variable using another function, namely, probability density function
.

The probability that the value assumed by the random variable X falls within the interval
, is determined by the equality
.

The graph of the function is called distribution curve. Geometrically, the probability of a random variable X falling into the interval is equal to the area of ​​the corresponding curvilinear trapezoid bounded by the distribution curve, the Ox axis and straight lines
.

Properties of the probability density function:

9.1. Numerical characteristics of continuous random variables

Expectation(average value) of a continuous random variable X is determined by the equality
.

M(X) is denoted by A. The mathematical expectation of a continuous random variable has properties similar to those of a discrete variable:

Variance discrete random variable X is the mathematical expectation of the squared deviation of the random variable from its mathematical expectation, i.e. . For a continuous random variable, the variance is given by the formula
.

The dispersion has the following properties:

The last property is very convenient to use to find the variance of a continuous random variable.

The concept of standard deviation is introduced similarly. The standard deviation of the continuous random variable X is called the square root of the variance, i.e.
.

Example 28. A continuous random variable X is specified by a probability density function
in the interval (10;12), outside this interval the value of the function is 0. Find 1) the value of the parameter A, 2) mathematical expectation M(X), variance
, standard deviation, 3) integral function
and build graphs of integral and differential functions.

1). To find the parameter A use the formula
. We'll get it. Thus,
.

2). To find the mathematical expectation, we use the formula: , from which it follows that
.

We will find the variance using the formula:
, i.e. .

Let's find the standard deviation using the formula: , from which we get that
.

3). The integral function is expressed through the probability density function as follows:
. Hence,
at
, = 0 at
u = 1 at
.

The graphs of these functions are presented in Fig. 4. and fig. 5.

Fig.4 Fig.5.

9.2. Uniform probability distribution of a continuous random variable

Probability distribution of a continuous random variable X evenly on the interval if its probability density is constant on this interval and equal to zero outside this interval, i.e. . It is easy to show that in this case
.

If the interval
is contained in the interval, then
.

Example 29. An instantaneous signal event must occur between one o'clock and five o'clock. The signal waiting time is a random variable X. Find the probability that the signal will be detected between two and three o'clock in the afternoon.

Solution. The random variable X has a uniform distribution, and using the formula we find that the probability that the signal will be between 2 and 3 o'clock in the afternoon is equal to
.

In educational and other literature it is often denoted in the literature through
.

9.3. Normal probability distribution of a continuous random variable

The probability distribution of a continuous random variable is called normal if its probability distribution law is determined by the probability density
. For such quantities A– mathematical expectation,
- standard deviation.

Theorem. Probability of a normally distributed continuous random variable falling into a given interval
determined by the formula
, Where
- Laplace function.

A consequence of this theorem is the three sigma rule, i.e. It is almost certain that a normally distributed, continuous random variable X takes its values ​​in the interval
. This rule can be derived from the formula
, which is a special case of the formulated theorem.

Example 30. The operating life of the TV is a random variable X, subject to the normal distribution law, with a warranty period of 15 years and a standard deviation of 3 years. Find the probability that the TV will last from 10 to 20 years.

Solution. According to the conditions of the problem, the mathematical expectation A= 15, standard deviation.

Let's find . Thus, the probability of the TV operating from 10 to 20 years is more than 0.9.

9.4. Chebyshev's inequality

Takes place Chebyshev's lemma. If a random variable X takes only non-negative values ​​and has a mathematical expectation, then for any positive V
.

Considering that , as the sum of the probabilities of opposite events, we obtain that
.

Chebyshev's theorem. If the random variable X has finite variance
and mathematical expectation M(X), then for any positive inequality is true
.

Whence it follows that
.

Example 31. A batch of parts has been produced. The average length of the parts is 100 cm, and the standard deviation is 0.4 cm. Estimate below the probability that the length of a part taken at random will be at least 99 cm. and no more than 101cm.

Solution. Variance. The mathematical expectation is 100. Therefore, to estimate from below the probability of the event in question
let us apply Chebyshev's inequality, in which
, Then
.

10. Elements of mathematical statistics

Statistical aggregate name a set of homogeneous objects or phenomena. Number n elements of this set is called the volume of the collection. Observed values trait X is called options. If the options are arranged in increasing sequence, then we get discrete variation series. In the case of grouping, the option by intervals turns out to be interval variation series. Under frequency t characteristic values ​​understand the number of members of the population with a given variant.

The ratio of frequency to volume of a statistical population is called relative frequency sign:
.

The relationship between the variants of a variation series and their frequencies is called statistical distribution of the sample. A graphical representation of the statistical distribution can be polygon frequency

Example 32. By surveying 25 first-year students, the following data about their age was obtained:
. Compile a statistical distribution of students by age, find the range of variation, construct a frequency polygon and compile a series of distributions of relative frequencies.

Solution. Using the data obtained from the survey, we will create a statistical distribution of the sample

The range of the variation sample is 23 – 17 = 6. To construct a frequency polygon, construct points with coordinates
and connect them in series.

The relative frequency distribution series has the form:

10.1.Numerical characteristics of the variation series

Let the sample be given by a series of frequency distributions of feature X:

The sum of all frequencies is equal p.

Arithmetic mean of the sample name the quantity
.

Variance or the measure of dispersion of the values ​​of a characteristic X in relation to its arithmetic mean is called the value
. The standard deviation is the square root of the variance, i.e. .

The ratio of the standard deviation to the arithmetic mean of the sample, expressed as a percentage, is called coefficient of variation:
.

Empirical relative frequency distribution function call a function that determines for each value the relative frequency of the event
, i.e.
, Where - number of options, smaller X, A n– sample size.

Example 33. Under the conditions of example 32, find the numerical characteristics
.

Solution. Let's find the arithmetic mean of the sample using the formula, then .

The variance of trait X is found by the formula: , i.e. . The standard deviation of the sample is
. The coefficient of variation is
.

10.2. Probability estimation by relative frequency. Confidence interval

Let it be carried out n independent trials, in each of which the probability of occurrence of event A is constant and equal to r. In this case, the probability that the relative frequency will differ from the probability of the occurrence of event A in each trial in absolute value is no more than by , is approximately equal to twice the value of the Laplace integral function:
.

Interval estimation call such an estimate, which is determined by two numbers that are the ends of the interval covering the estimated parameter of the statistical population.

Confidence intervalis an interval that, with a given confidence probability covers the estimated parameter of the statistical population. Considering the formula in which we replace the unknown quantity r to its approximate value obtained from the sample data, we obtain:
. This formula is used to estimate probability by relative frequency. Numbers
And
called lower and, respectively, upper trust boundaries, - the maximum error for a given confidence probability
.

Example 34. The factory workshop produces light bulbs. When checking 625 lamps, 40 were found to be defective. Find, with a confidence probability of 0.95, the boundaries within which the percentage of defective light bulbs produced by the factory workshop lies.

Solution. According to the conditions of the task. We use the formula
. Using Table 2 of the appendix, we find the value of the argument, in which the value of the Laplace integral function is equal to 0.475. We get that
. Thus, . Therefore, we can say with a probability of 0.95 that the share of defects produced by the workshop is high, namely, it varies from 6.2% to 6.6%.

10.3. Parameter estimation in statistics

Let the quantitative characteristic X of the entire population under study (general population) have a normal distribution.

If the standard deviation is known, then the confidence interval covering the mathematical expectation A

, Where n– sample size, - sample arithmetic mean, t is the argument of the Laplace integral function, at which
. In this case, the number
called estimation accuracy.

If the standard deviation is unknown, then from the sample data it is possible to construct a random variable that has a Student distribution with n– 1 degrees of freedom, which is determined by only one parameter n and does not depend on unknowns A And . Student's t-distribution even for small samples
gives quite satisfactory ratings. Then the confidence interval covering the mathematical expectation A of this feature with a given confidence probability is found from the condition

, where S is the corrected mean square, - Student’s coefficient, found from the data
from table 3 of the appendix.

The confidence interval covering the standard deviation of this characteristic with a confidence probability is found using the formulas: and , where
found from the table of values q according to data.

10.4. Statistical methods for studying dependencies between random variables

The correlation dependence of Y on X is the functional dependence of the conditional average from X. Equation
represents the regression equation of Y on X, and
- regression equation of X on Y.

The correlation dependence can be linear or curvilinear. In the case of a linear correlation dependence, the equation of the straight regression line has the form:
, where the slope A straight line of regression Y on X is called the sample regression coefficient Y on X and is denoted
.

For small samples, the data is not grouped, the parameters
are found using the least squares method from the system of normal equations:

, Where n– number of observations of values ​​of pairs of interrelated quantities.

Sample linear correlation coefficient shows the close relationship between Y and X. The correlation coefficient is found using the formula
, and
, namely:

The sample equation of the straight regression line Y on X has the form:

.

With a large number of observations of characteristics X and Y, a correlation table with two inputs is compiled, with the same value X observed times, same meaning at observed times, same pair
observed once.

Example 35. A table of observations of signs X and Y is given.

Find the sample equation of the straight regression line Y on X.

Solution. The relationship between the studied characteristics can be expressed by the equation of a straight line of regression of Y on X: . To calculate the coefficients of the equation, we will create a calculation table: