A continuous random variable x is given by a distribution density. Mathematical expectation of a continuous random variable
Chapter 1. Discrete random variable
§ 1. The concept of a random variable.
The law of distribution of a discrete random variable.
Definition : Random is a quantity that, as a result of the test, takes only one value out of a possible set of its values, unknown in advance and depending on random causes.
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.
You can describe a random variable using its distribution law.
Definition : The distribution law of a discrete random variable called the correspondence between the possible values of a random variable and their probabilities.
The distribution law of a discrete random variable X can be given in the form of a table, in the first line of which all possible values of the random variable are indicated in ascending order, and in the second line the corresponding probabilities of these values, i.e.
where р1+ р2+…+ рn=1
Such a table is called a series of distribution of a discrete random variable.
If the set of possible values of a random variable is infinite, then the series р1+ р2+…+ рn+… converges and its sum is equal to 1.
The distribution law of a discrete random variable X can be depicted graphically, for which a polygonal line is built in a rectangular coordinate system, connecting successively 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"> of organic chemistry are 0.7 and 0.8, respectively. Draw up the law of distribution of the random variable X - the number of exams that the student will pass.
Solution. As a result of the exam, the considered random variable X can take one of the following values: x1=0, x2=1, x3=2.
Let's find the probability of these values. 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: The distribution function of a discrete random variable X the function F(x) is called, which determines for each value x the probability that the random variable X takes 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 depicted on the number line by a point to the left of the point x.
1)0≤F(x)≤1;
2) F(x) is a non-decreasing function on (-∞;+∞);
3) F(x) - continuous from the left at the 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,
p1 at x1< х≤ x2,
F(x)= p1 + p2 at x2< х≤ х3
1 for x> xn.
Its graph is shown in Fig. 2:
§ 3. Numerical characteristics of a discrete random variable.
Mathematical expectation is one of the important numerical characteristics.
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
Expected value 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) \u003d 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, variance is used.
Definition: dispersion 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 the variance, it is often convenient to use the formula:
D(X)=M(X2)-(M(X))2,
where М(Х)=∑ xi2рi= x12р1 + x22р2+…+ xn2рn
The variance D(X) has the dimension of the square of a 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 number 2. The discrete random variable X is given 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 the possible values of the random variable X is equal to 1, then
Р2=1- (0.1+0.3+0.2+0.3)=0.1
Find the distribution function F(x)=P(X Geometrically, this equality can be interpreted as follows: F(x) is the probability that a random variable will take the value that is depicted on the real axis by a point to the left of 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 (-∞;х) two values x1=-1 and x2=0 fall; 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, since 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 for 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 for x>3 Let's 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
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 Р(Х=m)-probability of occurrence of event A exactly m times in n trials is calculated by the Bernoulli formula: P(X=m)=Сmnpmqn-m Mathematical expectation, variance and mean standard deviation a random variable X distributed according to a binary law is found, respectively, by the formulas: https://pandia.ru/text/78/455/images/image016_31.gif" width="26"> The probability of the event A - "getting five" in each test is the same and equal to 1/6, i.e. P(A)=p=1/6, then P(A)=1-p=q=5/6, where - "drops are not five." Random variable X can take values: 0;1;2;3. We find the probability of each of the possible values of X using the Bernoulli formula: P(X=0)=P3(0)=C03p0q3=1 (1/6)0 (5/6)3=125/216; P(X=1)=P3(1)=C13p1q2=3 (1/6)1 (5/6)2=75/216; P(X=2)=P3(2)=C23p2q=3(1/6)2(5/6)1=15/216; P(X=3)=P3(3)=C33p3q0=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 numerical characteristics random variable X: M(X)=np=3 (1/6)=1/2, D(X)=npq=3 (1/6) (5/6)=5/12, Task number 4. 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 manufacturing a defective part p=0.002 is small, and the events under consideration (the part turns out to be defective) are independent, so the Poisson formula takes place: 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): P1000(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) \u003d 1-P (). Hence the desired probability is equal to: Р(А)=1-Р1000(0)=1- e-2
20
\u003d 1-e-2 \u003d 1-0.13534≈0.865. Tasks for independent work.
1.1
1.2.
The dispersed random variable X is given 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 felt-tip pens in the box, 2 of which no longer write. At random, take 3 felt-tip pens. Random variable X - the number of writing felt-tip pens among those taken. Compose the law of distribution of a random variable. 1.4.
There are 6 textbooks randomly placed on the library shelf, 4 of them are bound. The librarian takes 4 textbooks at random. Random variable X is the number of bound textbooks among those taken. Compose the law of distribution of a random variable. 1.5.
The ticket has two tasks. Probability right decision the first task is equal to 0.9, the second-0.7. The random variable X is the number of correctly solved problems in the ticket. Compose 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 shoot at a target. The probability of hitting the target with one shot for the first shooter is 0.5, for the second - 0.8, for the third - 0.7. The random variable X is the number of hits on the target if the shooters make one shot each. Find the distribution law, M(X),D(X). 1.7.
A basketball player throws the ball into the basket with a probability of hitting on each throw 0.8. For each hit, he receives 10 points, and in case of a miss, he is not awarded points. Compose the law of distribution of a random variable X-number of points received by a basketball player for 3 throws. Find M(X),D(X) and also the probability that he will get more than 10 points. 1.8.
Letters are written on the cards, only 5 vowels and 3 consonants. 3 cards are chosen at random, and each time the card taken is returned back. Random variable X is the number of vowels among those taken. Compose a distribution law and find M(X),D(X),σ(X). 1.9.
On average, 60% of contracts Insurance Company pays sums insured in connection with the occurrence of an insured event. Draw up a distribution law for a random variable X - the number of contracts for which the sum insured was paid out among four randomly selected contracts. Find the numerical characteristics of this quantity. 1.10.
The radio station at certain intervals sends call signs (no more than four) until two-way communication is established. The probability of receiving a response to a call sign is 0.3. Random variable X-number of sent callsigns. Compose the distribution law and find F(x). 1.11.
There are 3 keys, of which only one fits the lock. Draw up a distribution law for 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.
Sequential independent tests of three devices for reliability are carried out. Each subsequent device is tested only if the previous one turned out to be reliable. The probability of passing the test for each instrument is 0.9. Compile the law of distribution of 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 block of the electronic device contains 100 identical elements. The probability of failure of each element during the time T is equal to 0.002. The elements work independently. Find the probability that no more than two elements will fail in time T. 1.15.
The textbook was published in 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 the Poisson law with the parameter λ=1.5. Find the probability that in a minute there will be: 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 for x≤-2, 0.3 at -2<х≤0, F(x)= 0.5 at 0<х≤2, 0.9 at 2<х≤5, 1 for x>5 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 for 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 for 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(Х) ≈ 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.
1.22e-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
name the value, all possible values of which completely fill the finite or infinite interval of the numerical axis. 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 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. Distribution function properties:
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 that a random variable X falls 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. P(a<Х 4) The probability that a continuous random variable X will take one single value is 0. 5) F(-∞)=0, F(+∞)=1 Specifying a continuous random variable using a distribution function is not the only one. Let us introduce the concept of probability distribution density (distribution density). Definition
:
Probability density
f
(
x
)
continuous random variable X is the derivative of its distribution function, i.e.: The probability distribution density is sometimes called the differential distribution function or the differential distribution law. The graph of the density of the probability distribution f(x) is called probability distribution curve
.
Probability density properties:
1) f(x) ≥0, when xhttps://pandia.ru/text/78/455/images/image029_10.jpg" width="285" height="141">.gif" width="14" height ="62 src="> 0 for x≤2, f(x)= c(x-2) at 2<х≤6, 0 for x>6. Find: a) the value of c; b) the distribution function F(x) and build its graph; c) Р(3≤х<5) Solution:
+
∞ a) 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 for x≤2, F (x) \u003d (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 for x≤0, F (x) \u003d (3 arctg x) / π at 0<х≤√3, 1 for x>√3. Find the differential distribution function f(x) Solution:
Since f (x) \u003d F '(x), then https://pandia.ru/text/78/455/images/image011_36.jpg" width="118" height="24"> All the properties of mathematical expectation and dispersion considered earlier for dispersed random variables are also valid for continuous ones. Task number 3. The random variable X is given 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. Tasks for independent solution.
2.1.
A continuous random variable X is given by a distribution function: 0 for x≤0, F(x)= https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤ π/6, F(х)= - 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 for x≤2, f(x)= with x at 2<х≤4, 0 for x>4. 2.4.
A continuous random variable X is given by the distribution density: 0 for x≤0, f(х)= с √х at 0<х≤1, 0 for x>1. Find: a) the number c; b) M(X), D(X). 2.5.
https://pandia.ru/text/78/455/images/image041_3.jpg" width="36" height="39"> for x, 0 at x . Find: a) F(x) and plot its graph; b) M(X),D(X), σ(X); c) the probability that in four independent trials the value 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) \u003d 2 (x-2) for x, 0 at x . Find: a) F(x) and plot its graph; b) M(X),D(X), σ(X); c) the probability that in three independent tests the value X will take exactly 2 times the value belonging to the interval . 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 /four ; 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.
Random variable Х, concentrated on the interval (3;7), is given by the distribution function F(х)= . 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), given by the distribution function F(x)= . Find the probability that 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) the number c; b) M(X); c) probability P(X > M(X)). 2.12.
The random variable is given 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 Р(Х≤М(Х)) 2.13.
The Time 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 distribution. 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) 2.16.
The random variable X is distributed according to the law " right triangle» in the interval (0;4) (Fig.5). Find an analytical expression for the probability density f(x) on the entire real axis. Answers
0 for 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 is 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. one F(х)= https://pandia.ru/text/78/455/images/image077_3.jpg" width="30" height="37">, D(X)=, σ(Х)=. Task number 1. The random variable X is uniformly distributed on the segment . Find: a) the probability distribution density f(x) and build its graph; b) the distribution function F(x) and build its graph; 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 for x≤3, F(х)= 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 for x<0, f(х)= λе-λх at х≥0. The distribution function of a random variable X, distributed according to an 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(X)= 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: Р(a<Х Task number 2. The average uptime of the device is 100 hours. Assuming that the uptime of the device has an exponential distribution law, find: a) probability distribution density; b) distribution function; c) the probability that the time of failure-free operation of the device will exceed 120 hours. Solution:
By condition, the mathematical distribution M(X)=https://pandia.ru/text/78/455/images/image098_10.gif" height="43 src="> 0 for x<0, a) f(x)= 0.01e -0.01x for x≥0. b) F(x)= 0 for 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 (Gaussian 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 distribution function of a random variable X, distributed according to the normal law, is expressed through the Laplace function Ф (х) according to the formula: where is the Laplace function. Comment:
The function Ф(х) is odd (Ф(-х)=-Ф(х)), besides, if x>5, we can consider Ф(х) ≈1/2. The graph of the distribution function F(x) is shown in fig. eight 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 positive numberδ is calculated by the formula: In particular, for m=0 the equality is true: "Three Sigma Rule"
If the random variable X has a normal distribution law with the parameters m and σ, then it is practically 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="> According to the table of values of the function Ф(х) we find Ф(1.5)=0.4332, Ф(1)=0.3413. So the desired probability is: 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 Р(3≤х≤6). 3.3.
An automatic traffic light is installed on the highway, in which the green light is on for 2 minutes for vehicles, yellow for 3 seconds and red for 30 seconds, etc. The car passes along the highway at a random time. Find the probability that the car passes the traffic light without stopping. 3.4.
Subway trains run regularly at intervals of 2 minutes. The passenger enters the platform at a random time. What is the probability that the passenger will have to wait more than 50 seconds for the train? Find the mathematical expectation of a random variable X - the train's waiting time. 3.5.
Find the variance and standard deviation of the exponential distribution given by the distribution function: F(x)= 0 at x<0, 1-e-8x for x≥0. 3.6.
A continuous random variable X is given by the probability distribution density: f(x)=0 at x<0, 0.7 e-0.7x at x≥0. a) Name the law of distribution of the considered random variable. 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, given 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 given 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 interval . 3.9.
The mathematical expectation 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.
Random variable X is normally distributed with mean 3.5 and variance 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 interval . 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 has a higher probability? 3.12.
The random variable X is normally distributed with M(X)=0 and D(X)=1. From which interval (-0.5;-0.1) or (1;2) in one test will it take on a value with a greater probability? 3.13.
The current price per share can be modeled using a normal distribution with M(X)=10den. 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 den. units; b) using the "rule of three sigma" to find the boundaries in which the current price of the stock will be located. 3.14.
The substance is weighed without systematic errors. Random weighing errors are subject to the normal law with the root-mean-square ratio σ=5r. Find the probability that in four independent experiments the 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 normally distributed with M(X)=12 and D(X)=36. Find the interval in which, with a probability of 0.9973, the random variable X will fall as a result of the test. 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 2 units of measurement in modulo . 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 give out? 3.18.
The detail parameter X is normally distributed 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 face value modulo does not exceed 1% of the face 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) Р(9.8≤Х≤10.4) ≈0.6562. 3.14.
0,111. 3.15.
σ=1.2. 3.16.
(-6;30). 3.17.
0,4%. distribution function random variable X called a function F(X) expressing for each X the probability that the random variable X takes on a value less than X:. Function F(X) is sometimes called integral distribution function, or integral distribution law. Random value X called continuous, if its distribution function is continuous at any point and differentiable everywhere, except perhaps at individual points. Examples continuous random variables: the diameter of the part that the turner grinds to a given size, the height of a person, the range of the projectile, etc. Theorem. The probability of any single value of a continuous random variable is zero Consequence. If a X is a continuous random variable, then the probability that the random variable falls into the interval If a continuous random variable X can only take values between a before b(where a and b are some constants), then its distribution function is equal to zero for all values All properties of distribution functions of discrete random variables are also satisfied for distribution functions of continuous random variables. Specifying a continuous random variable using a distribution function is not the only one. Probability Density
(distribution density or density)
R(X) continuous random variable X is called the derivative of its distribution function Probability Density R(X), as well as the distribution function F(X), is one of the forms of the distribution law, but unlike the distribution function, it exists only for continuous random variables. The probability density is sometimes called differential function, or differential distribution law. A probability density plot is called a distribution curve. Properties probability density of a continuous random variable: Rice. 8.1 Rice. 8.2 4.
Geometrically, the properties of the probability density mean that its graph - the distribution curve - lies not below the abscissa axis, and the total area of the figure bounded by the distribution curve and the abscissa axis is equal to one. Example 8.1. The minute hand of an electric clock moves in leaps every minute. You glanced at your watch. They are showing a minutes. Then for you the true time at the moment will be a random variable. Find its distribution function. Solution. Obviously, the true time distribution function is 0 for all O Rice. 8.3 Example 8.2. Is the distribution function of some random variable the function Solution. All values of this function belong to the interval Equalities also hold: Therefore, the function Example 8.3. Is the distribution function of some random variable the function Solution. This function is not a distribution function of a random variable, since in the interval Rice. 8.5 Example 8.4. Random value X given by the distribution function Find coefficient a and the probability density of the random variable X. Determine the probability of inequality Solution. The distribution density is equal to the first derivative of the distribution function Coefficient a define using equality The same result could be obtained using the continuity of the function Consequently, Therefore, the probability density has the form Probability Example 8.5. Random value X has a probability density (Cauchy's law) Find coefficient a and the probability that the random variable X will take some value from the interval Solution. Let's find the coefficient a from equality Consequently, So, The probability that the random variable X will take some value from the interval Find the distribution function of this random variable P Rice. 8.6 Solution. Using the graph, we write down the analytical expression for the probability distribution density of a given random variable Let's find the distribution function. If a If a If a If a Therefore, the distribution function has the form In probability theory, one has to deal with random variables, all of whose values cannot be sorted out. For example, it is impossible to take and "sort through" 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 a random variable are located. Continuous random variable is a random variable whose values completely fill a certain interval. Since it is not possible to sort through all the values of a continuous random variable, it can be specified using the distribution function. distribution function random variable $X$ is a function $F\left(x\right)$, which determines the probability that the random variable $X$ takes a value less than some fixed value $x$, i.e. $F\left(x\right)$ )=P\left(X< x\right)$. Distribution function properties: 1
. $0\le F\left(x\right)\le 1$. 2
. The probability that the random variable $X$ takes values from the interval $\left(\alpha ;\ \beta \right)$ is equal to the difference between the values of the distribution function at the ends of this interval: $P\left(\alpha< X < \beta \right)=F\left(\beta \right)-F\left(\alpha \right)$. 3
. $F\left(x\right)$ - non-decreasing. 4
. $(\mathop(lim)_(x\to -\infty ) F\left(x\right)=0\ ),\ (\mathop(lim)_(x\to +\infty ) F\left(x \right)=1\ )$. Example 1
$$P\left(0,3< X < 0,7\right)=F\left(0,7\right)-F\left(0,3\right)=0,7-0,3=0,4.$$ 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 a random variable $X$ takes 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 given by the following distribution function $F(x)=\left\(\begin(matrix) The mathematical expectation of a continuous random variable $X$ is calculated by the formula $$M\left(X\right)=\int^(+\infty )_(-\infty )(xf\left(x\right)dx).$$ Example 3
. Find $M\left(X\right)$ for the random variable $X$ from example $2$. $$M\left(X\right)=\int^(+\infty )_(-\infty )(xf\left(x\right)\ dx)=\int^1_0(x\ dx)=(( x^2)\over (2))\bigg|_0^1=((1)\over (2)).$$ 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 $2$. $$D\left(X\right)=\int^(+\infty )_(-\infty )(x^2f\left(x\right)\ dx)-(\left)^2=\int^ 1_0(x^2\ dx)-(\left(((1)\over (2))\right))^2=((x^3)\over (3))\bigg|_0^1-( (1)\over (4))=((1)\over (3))-((1)\over (4))=((1)\over(12)).$$ Examples of solving problems on the topic "Random variables".
A task 1
. There are 100 tickets issued in the lottery. One win of 50 USD was played. and ten wins of $10 each. Find the law of distribution of the value X - the cost of a possible gain. Solution. Possible values of X: x 1
= 0; x 2
= 10 and x 3
= 50. Since there are 89 “empty” tickets, then p 1
= 0.89, the probability of winning is 10 c.u. (10 tickets) – p 2
= 0.10 and for a win of 50 c.u. –p 3
= 0.01. In this way: 0,89
0,10
0,01
Easy to control: . A task 2.
The probability that the buyer has familiarized himself with the advertisement of the product in advance is 0.6 (p = 0.6). Selective quality control of advertising is carried out by polling buyers before the first one who has studied the advertisement in advance. Make a series of distribution of the number of interviewed buyers. Solution. According to the condition of the problem p = 0.6. From: q=1 -p = 0.4. Substituting these values, we get: and construct a distribution series: pi 0,24
A task 3.
A computer consists of three independently operating elements: a system unit, a monitor, and a keyboard. With a single sharp increase in voltage, the probability of failure of each element is 0.1. Based on the Bernoulli distribution, draw up the distribution law for the number of failed elements during a power surge in the network. Solution. Consider Bernoulli distribution(or binomial): the probability that in n tests, event A will appear exactly k once: q n p n AT let's get back to the task. Possible values of X (number of failures): x 0 =0 - none of the elements failed; x 1 =1 - failure of one element; x 2 =2 - failure of two elements; x 3 =3 - failure of all elements. Since, by condition, p = 0.1, then q = 1 – p = 0.9. Using the Bernoulli formula, we get , ,
, .
Control: . Therefore, the desired distribution law: 0,729
0,243
0,027
0,001
Task 4. Produced 5000 rounds. The probability that one cartridge is defective Solution. Applicable Poisson distribution: this distribution is used to determine the probability that, given a very large number of trials (mass trials), in each of which the probability of event A is very small, event A will occur k times: Here n \u003d 5000, p \u003d 0.0002, k \u003d 3. We find , then the desired probability: Task 5. When firing before the first hit with the probability of hitting p
= 0.6 for a shot, you need to find the probability that the hit will occur on the third shot. Solution. Let us apply the geometric distribution: let independent trials be performed, in each of which the event A has a probability of occurrence p (and non-occurrence q = 1 - p). Trials end as soon as event A occurs. Under such conditions, the probability that event A will occur on the kth test is determined by the formula: . Here p = 0.6; q \u003d 1 - 0.6 \u003d 0.4; k \u003d 3. Therefore, . Task 6. Let the law of distribution of a random variable X be given: Find the mathematical expectation. Solution. . Note that the probabilistic meaning of the mathematical expectation is the average value of a random variable. Task 7. Find the variance of a random variable X with the following distribution law: Solution. Here
.
The law of distribution of the square of X 2
:
X 2
Required variance: . Dispersion characterizes the degree of deviation (scattering) of a random variable from its mathematical expectation. Task 8. Let the random variable be given by the distribution: 10m Find its numerical characteristics. Solution: m, m 2
,
M 2
, m. About a random variable X, one can say either - its mathematical expectation is 6.4 m with a variance of 13.04 m 2
, or - its mathematical expectation is 6.4 m with a deviation of m. The second formulation is obviously clearer. A task 9.
Random value X given by the distribution function: Find the probability that, as a result of the test, the value X will take on a value contained in the interval Solution. The probability that X will take a value from a given interval is equal to the increment of the integral function in this interval, i.e. . In our case and , therefore A task 10.
Discrete random variable X given by the distribution law: Find distribution function F(x ) and build its graph. Solution. Since the distribution function at ; at ; at ; at ; Relevant chart: Task 11. Continuous random variable X given by the differential distribution function: Find the probability of hitting X to interval Solution. Note that this is a special case of the exponential distribution law. Let's use the formula: A task 12.
Find the numerical characteristics of a discrete random variable X given by the distribution law: –5
X 2 : x2 .
,
where The values of this function are found using a table. In our case: . According to the table we find:, therefore: To find the distribution function of a discrete random variable you need to use this calculator. Exercise 1. The distribution density of a continuous random variable X has the form: Task 2. Find the variance of the random variable X given by the integral function. Task 3. Find the mathematical expectation of a random variable X given function distribution. Task 4. The probability density of some random variable is given as follows: f(x) = A/x 4 (x = 1; +∞) A task. The distribution function of some continuous random variable is given as follows: Determine the parameters a and b , find the expression for the probability density f(x) , the mathematical expectation and variance, as well as the probability that the random variable takes a value in the interval . Plot f(x) and F(x) graphs. Let's find the distribution density function as a derivative of the distribution function. Example #1. The probability distribution density f(x) of a continuous random variable X is given. Required: The random variable X is given by the distribution density f(x):
1.2.
p4=0.1; 0 for x≤-1,
(fig.5)
https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤a,
,
The normal curve is symmetrical with respect to the straight line x=m, has a maximum at x=a equal to .
,
.
does not depend on whether this interval is open or closed, i.e.
and unit for values 
.For a continuous random variable
.
.
and unit for 
. Time flows evenly. Therefore, the probability that the true time is less a+ 0.5 min equals 0.5, since it is equally likely that the a less or more than half a minute. The probability that the true time is less a+ 0.25 min, equals 0.25 (the probability of this time is three times less than the probability that the true time is greater than a+ 0.25 min, and their sum is equal to one, as the sum of the probabilities of opposite events). Arguing similarly, we find that the probability that the true time is less than a+ 0.6 min equals 0.6. In general, the probability that the true time is less than a
+ + α
min 
, is equal to α
. Therefore, the real time distribution function has the following expression:
on is continuous everywhere, and its derivative is continuous at all points except for two: x = a and x = a+ 1. The graph of this function looks like (Fig. 8.3):
, i.e. 
. Function F(X) is non-decreasing: in the interval 
it is constant, equal to zero, in the interval 
increases between 
is also constant, equal to one (see Fig. 8.4). The function is continuous at every point X 0 area of its definition - a gap 
, so it is continuous on the left, i.e. equality
,
.
,
.
satisfies all the properties characteristic of the distribution function. So this function 
is the distribution function of some random variable X.
it is decreasing and not continuous. The graph of the function is shown in Fig. 8.5.
.
,
.
at the point 
,
.
.
random hits X within a given interval is calculated by the formula
.
. Find the distribution function of this random variable.
,
.
.
, is equal to
example 8.6. Probability density plot of a random variable X shown in fig. 8.6 (Simpson's law). Write the expression for the probability density and the distribution function of this random variable.
, then 
.
, then .
, then
, then
Distribution function of a continuous random variable
0,\ x\le 0\\
x,\0< x\le 1\\
1,\x>1
\end(matrix)\right.$. The probability that a random variable $X$ falls 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, i.e.:Probability density
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.$Mathematical expectation of a continuous random variable
Dispersion of a continuous random variable
, or:
. What is the probability that there will be exactly 3 defective cartridges in the entire batch?
, where .
.
.
.
.
for 
, then
.
.
is the Laplace function.
Find:
a) parameter A ;
b) distribution function F(x) ;
c) the probability of hitting a random variable X in the interval ;
d) mathematical expectation MX and variance DX .
Plot the functions f(x) and F(x) .
Find coefficient A , distribution function F(x) , mathematical expectation and variance, as well as the probability that a random variable takes a value in the interval . Plot f(x) and F(x) graphs.
Knowing that
find the 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 is: F(x) = (x-1)/3
Dispersion.
1 / 9 4 3 - (1 / 9 1 3) - (5 / 2) 2 = 3 / 4
Find the probability that a random variable takes a value in the interval
P(2< x< 3) = F(3) – F(2) = (1/3*3 - 1/3) - (1/3*2 - 1/3) = 1/3
f(x) = A*sqrt(x), 1 ≤ x ≤ 4.
Solution:
Find the parameter A from the condition: 
or
14/3*A-1=0
Where,
A = 3 / 14 
The distribution function can be found by the formula. 
