The normal distribution law is most often encountered in practice. The main feature that distinguishes it from other laws is that it is a limiting law, to which other laws of distribution approach under very common typical conditions (see Chapter 6).
Definition. A continuous random variable X hasnormal distribution law (Gauss's law)with parameters a and a 2, if its probability density has the form
The term “normal” is not entirely appropriate. Many signs obey the normal law, for example, the height of a person, the range of a projectile, etc. But if any characteristic obeys a distribution law different from normal, then this does not at all mean that the phenomenon associated with this characteristic is “abnormal.”
The normal distribution curve is called normal, or Gaussian, crooked. In Fig. 4.6, A, 6 the normal curve fd, (x) with parameters yio 2 are given, i.e. I[a] a 2), and the graph of the distribution function of the random variable X, which has a normal law. Let us pay attention to the fact that the normal curve is symmetrical with respect to the straight line x = a, has a maximum at the point X= A,
equal 
, i.e. 
And two inflection points x = a±
with ordinate 
It can be noted that in the normal law density expression, the parameters are indicated by the letters A and st 2, which we use to denote the mathematical expectation M(X) and variance OH). This coincidence is not accidental. Let us consider a theorem that establishes the probabilistic theoretical meaning of the parameters of the normal law.
Theorem. The mathematical expectation of a random variable X distributed according to a normal law is equal to the parameter a of this law, those.
A its dispersion - to the parameter a 2, i.e.
Expectation of a random variable X:
Let's change the variable by putting
Then 
the limits of integration do not change
and therefore
(the first integral is equal to zero as the integral of an odd function over an interval symmetric with respect to the origin, and the second integral 
- Euler - Poisson integral).
Variance of a random variable X:
Let's make the same change of variable x = a + o^2 t, as in the calculation of the previous integral. Then
Applying the method of integration by parts, we obtain
Let's find out how the normal curve will change when the parameters change A and with 2 (or a). If a = const, and the parameter changes a (a x a 3), i.e. the center of symmetry of the distribution, then the normal curve will shift along the abscissa axis without changing its shape (Fig. 4.7).
If a = const and the parameter a 2 (or a) changes, then the ordinate changes
curve maximum 
As a increases, the ordinate of the maximum
the curve decreases, but since the area under any distribution curve must remain equal to one, the curve becomes flatter, stretching along the x-axis; when decreasing su, on the contrary, the normal curve extends upward while simultaneously compressing from the sides. In Fig. Figure 4.8 shows normal curves with parameters a 1 (o 2 and a 3, where o, A(aka mathematical expectation) characterizes the position of the center, and parameter a 2 (aka dispersion) characterizes the shape of the normal curve.
Normal distribution law of a random variable X with parameters A= 0, st 2 = 1, i.e. X ~ N( 0; 1), called standard or normalized and the corresponding normal curve is standard or normalized.
The difficulty of directly finding the distribution function of a random variable distributed according to the normal law according to formula (3.23) and the probability of its falling on a certain interval according to formula (3.22) is associated with the fact that the integral of the function (4.26) is “uncollectible” in elementary functions . Therefore they are expressed through the function
- function (probability integral) Laplace, for which the tables have been compiled. Let us recall that we have already encountered the Laplace function when considering the Moivre-Laplace integral theorem (see section 2.3). Its properties were also discussed there. Geometrically, the Laplace function Ф(.с) represents the area under the standard normal curve on the segment [-X; X] (Fig. 4.9) 1 .
Rice. 4.10
Rice. 4.9
Theorem. The distribution function of a random variable X, distributed according to the normal law, is expressed through the Laplace functionФ(х) according to the formula
According to formula (3.23), the distribution function is:
Let us make a change of variable, setting at X-> -oo? -» -00, therefore
1 Along with the probability integral of the form (4.29), representing the function Ф(х), its expressions are also used in the literature in the form of other tabulated functions:
representing the areas iodine of the standard normal curve, respectively, at the intervals (0; x], (-oo; x], [-x>/2; Chl/2 .
First integral
(due to the parity of the integrand and the fact that the Euler - Poisson integral is equal to [To).
The second integral, taking into account formula (4.29), is
Geometrically, the distribution function represents the area under the normal curve on the interval (-co, x) (Fig. 4.10). As we see, it consists of two parts: the first, on the interval (-oo, A), equal to 1/2, i.e. half of the entire area under the normal curve, and the second, on the interval (i, x),
equal to 
Let us consider the properties of a random variable distributed according to a normal law.
1. The probability of hitting a random variable X distributed according to a normal law is V interval[x 1(x 2 ], equal to
Considering that, according to property (3.20), the probability P(x,
where and Г 2 are determined by formula (4.33) (Fig. 4.11). ?
2. The probability that the deviation of a random variable X, distributed according to a normal law, from the mathematical expectation a will not exceed the value A > 0 ( in absolute value) is equal to
and also the oddity property of the Laplace function, we obtain
Where? =D/o (Fig. 4.12). ?
In Fig. 4.11 and 4.12 provide a geometric interpretation of the properties of the normal law.
Comment. Discussed in Chap. 2 the approximate integral formula of Moivre - Laplace (2.10) follows from the property (4.32) of a normally distributed random variable at x ( = a, x 2 = b ) a = pr
And 
So
as a binomial law of distribution of a random variable X = t with parameters P And R, for which this formula was obtained, with n -> OS tends to the normal law (see Chapter 6).
Similar are the consequences (2.13), (2.14) and (2.16) of the Moivre-Laplace integral formula for the number X = t occurrence of an event in P independent testing and its frequency t/n follow from properties (4.32) and (4.34) of the normal law.
Let us calculate the probabilities using formula (4.34) P(X-a e) at various values of D (we use Table II of the appendices). We get
This is where the “three sigma rule” comes from.
If a random variable X has a normal distribution law with parameters a and a 2, i.e. M(a; a 2), then it is almost certain that its values lie in the interval(a - For, A+ For).
Violation of the “three sigma rule”, i.e. deviation of a normally distributed random variable X more than 3 (but absolute value), is an almost impossible event, since its probability is very low:
Note that the deviation D in, at which 
, called
probable deviation. For the normal law D in « 0.675a, i.e. per interval (A - 0.675a, A+ 0.675a) accounts for half of the total area under the normal curve.
Let's find the skewness coefficient and kurtosis of a random variable X, distributed according to a normal law.
Obviously, due to the symmetry of the normal curve relative to the vertical line x = a, passing through the distribution center a = M(X), coefficient of asymmetry of the normal distribution A = 0.
Kurtosis of a normally distributed random variable X we find using formula (3.37), i.e. 
where we took into account that the central moment of the 4th order, found by formula (3.30) taking into account definition (4.26), i.e.
(we omit the calculation of the integral).
Thus, the kurtosis of a normal distribution is zero and the steepness of other distributions is determined in relation to the normal one (we already mentioned this in paragraph 3.7).
O Example 4.9. Assuming that the height of men of a certain age group is a normally distributed random variable X with parameters A= 173 and a 2 =36:
- 1) Find: a) the expression of the probability density and distribution function of the random variable X; b) the shares of suits of 4th height (176-182 cm) and 3rd height (170-176 cm), which must be provided in the total production volume for a given age group; c) quantile x 07 and the 10% point of the random variable X.
- 2) Formulate the “three sigma rule” for a random variable X. Solution. 1, a) Using formulas (4.26) and (4.30) we write
1, b) The share of suits of the 4th height (176-182 cm) in the total production volume will be determined by formula (4.32) as the probability
(Fig. 4.14), since according to formulas (4.33)
The share of suits of the 3rd height (170-176 cm) could be determined similarly to formula (4.32), but it is easier to do this using formula (4.34), given that this interval is symmetrical with respect to the mathematical expectation A = M(X) = 173, i.e. inequality 170 X X -173|
(see Fig. 4.14;.
1, c) Quantile x 07(see paragraph 3.7) random variable X we find from equation (3.29) taking into account formula (4.30):
where 
According to the table We find 11 applications I- 0.524 and
This means that 70% of men in this age group are up to 176 cm tall.
- The 10% point is the ego quantile x 09 = 181 cm (located similarly), i.e. 10% of men are at least 181 cm tall.
- 2) It is almost certain that the height of men in this age group lies within the boundaries of A- Z = 173 - 3 6 = 155 to a + Zet = 173 + 3 - 6 = = 191 (cm), i.e. 155
Due to the features of the normal distribution law noted at the beginning of the section (and in Chapter 6), it occupies a central place in the theory and practice of probabilistic statistical methods. The great theoretical significance of the normal law is that with its help a number of important distributions are obtained, which are discussed below.
- Arrows in Fig. 4.11-4.13 the conventional areas and corresponding figures under the normal curve are marked.
- The values of the Laplace function Ф(х) are determined from the table. II applications.
Definition 3. X has normal distribution law (Gauss's law), if its distribution density has the form:
Where m = M(X), σ 2 = D(X), σ > 0 .
The normal distribution curve is called normal or Gaussian curve(Fig. 6.7).
A normal curve is symmetrical about a straight line x = m, has a maximum at the point x = m, equal .
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:
F( x) – Laplace function.
Comment. Function Ф( X) is odd (Ф(- X) = -F( X)), in addition, when X> 5 can be considered Ф( X) ≈ 1/2.
Table of values of the function Ф( X) is given in the appendix (Table P 2.2).
Distribution function graph F(x) is shown in Fig. 6.8.
The probability that the random variable X will take values belonging to the interval ( a;b) are calculated by the formula:
R(a< X < b ) = .
The probability that the absolute value of the deviation of a random variable from its mathematical expectation is less than a positive number δ is calculated by the formula:
P(| X - m| .
In particular, when m=0 the equality is true:
P(| X | .
"Three Sigma Rule"
If the random variable X has a normal distribution law with parameters m and σ, then it is almost certain that its values lie in the interval ( m 3σ; m+ 3σ), since P(| X - m| = 0,9973.
Problem 6.3. Random value X is distributed normally with a mathematical expectation of 32 and a variance of 16. Find: a) probability distribution density f(x); X will take the value from the interval (28;38).
Solution: By condition m= 32, σ 2 = 16, therefore, σ = 4, then
A)
b) Let's use the formula:
R(a< X )= .
Substituting a= 28, b= 38, m= 32, σ= 4, we get
R(28< X < 38)= F(1.5) F(1)
According to the table of function values Φ( X) we find Ф(1,5) = 0.4332, Ф(1) = 0.3413.
So, the desired probability:
P(28 Tasks 6.1.
Random value X uniformly distributed in the interval (-3;5). Find: a) distribution density f(x);
b) distribution functions F(x);
c) numerical characteristics; d) probability R(4<X<6). 6.2.
Random value X uniformly distributed on the segment. Find: a) distribution density f(x);
b) distribution function F(x);
c) numerical characteristics; d) probability R(3≤X≤6). 6.3.
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 a highway at a random moment in time. Find the probability that a car will pass a traffic light without stopping. 6.4.
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 a random variable X- train waiting time. 6.5.
Find the variance and standard deviation of the exponential distribution given by the distribution function: 6.6.
Continuous random variable X given by the probability distribution density: a) Name the distribution law of the random variable under consideration. b) Find the distribution function F(x) and numerical characteristics of a random variable X. 6.7.
Random value X distributed according to the exponential law specified by the probability distribution density: X will take a value from the interval (2.5;5). 6.8.
Continuous random variable X distributed according to the exponential law specified by the distribution function: Find the probability that as a result of the test X will take the value from the segment . 6.9.
The expected value and standard deviation of a normally distributed random variable are 8 and 2, respectively. Find: a) density distributions f(x); b) the probability that as a result of the test X will take a value from the interval (10;14). 6.10.
Random value X distributed normally 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 the value from the segment . 6.11.
Random value X normally distributed with M(X) =
0 and D(X)=
1. Which of the events: | X|≤0.6 or | X|≥0.6 is more likely? 6.12.
Random value X normally distributed 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? 6.13.
The current price per share can be modeled using a normal distribution with M(X)=
10 days 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. 6.14.
The substance is weighed without systematic errors. Random weighing errors are subject to the normal law with a standard deviation σ= 5g. Find the probability that in four independent experiments the error in three weighings will not exceed 3 g in absolute value. 6.15.
Random value X 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. 6.16.
Random value X normally distributed with M(X) = 12 and D(X) = 36. Find the interval in which, with a probability of 0.9973, the random variable will fall as a result of the test X. 6.17.
A part produced by an automatic machine is considered defective if the deviation X its controlled parameter exceeds the nominal value by 2 units of measurement. It is assumed that the random variable X normally distributed with M(X) = 0 and σ( X) = 0.7. What percentage of defective parts does the machine produce? 3.18.
Parameter X parts are 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 X of the nominal value will not exceed 1% of the nominal value. Answers V) M(X)=1, D(X)=16/3, σ( X)= 4/ , d)1/8. V) M(X)=4,5, D(X) =2 , σ ( X)= , d)3/5. 6.3.
40/51. 6.4.
7/12, M(X)=1. 6.5.
D(X) = 1/64, σ ( X)=1/8 6.6.
M(X)=1 , D(X) =2 , σ ( X)= 1 . 6.7.
P(2.5<X<5)=e -1 e -2 ≈0,2325 6.8.
P(2≤ X≤5)=0,252. b) R(10 < X < 14) ≈ 0,1574. b) R(3,1 ≤ X ≤ 3,7) ≈ 0,8185. 6.11.
|x|≥0,6. 6.12.
(-0,5; -0,1). 6.13.
a) P(9.8 ≤ X ≤ 10.4) ≈ 0.6562 6.14.
0,111. b) (9.1; 10.9). 6.15.
σ = 1.2. 6.16.
(-6; 30). 6.17.
0,4 %.
Chapter 6. Continuous random variables. § 1. Density and distribution function of a continuous random variable. The set of values of a continuous random variable is uncountable and usually represents some finite or infinite interval. A random variable x(w) defined in a probability space (W, S, P) is called continuous(absolutely continuous) W, if there is a non-negative function such that for any x the distribution function Fx(x) can be represented as an integral The function is called a function probability distribution densities. The definition implies the properties of the distribution density function: 1..gif" width="97" height="51"> 3. At points of continuity, the distribution density is equal to the derivative of the distribution function: . 4. The distribution density determines the law of distribution of a random variable, since it determines the probability of a random variable falling into the interval: 5. The probability that a continuous random variable will take a specific value is zero: . Therefore, the following equalities are valid: The graph of the distribution density function is called distribution curve, and the area bounded by the distribution curve and the x-axis is equal to unity. Then, geometrically, the value of the distribution function Fx(x) at point x0 is the area bounded by the distribution curve and the x-axis and lying to the left of point x0. Task 1. The density function of a continuous random variable has the form: Determine the constant C, construct the distribution function Fx(x) and calculate the probability. Solution. The constant C is found from the condition We have: To construct the distribution function Fx(x), note that the interval divides the range of values of the argument x (numeric axis) into three parts: https://pandia.ru/text/78/107/images/image017_17.gif" width="264 " height="49"> since the density x on the semi-axis is zero. In the second case Finally, in the last case, when x>2, Since the density vanishes on the semi-axis. So, the distribution function is obtained Probability Let's calculate using the formula. Thus,
§ 2. Numerical characteristics of a continuous random variable Expected value for continuously distributed random variables is determined by the formula https://pandia.ru/text/78/107/images/image028_11.gif" width="205" height="56 src=">, if the integral on the right converges absolutely. Dispersion x can be calculated using the formula All properties of mathematical expectation and dispersion given in Chapter 5 for discrete random variables are also valid for continuous random variables. Problem 2.
For the random variable x from Problem 1, calculate the mathematical expectation and variance .
Solution. And that means https://pandia.ru/text/78/107/images/image035_9.gif" width="184" height="69 src="> For a uniform distribution density graph, see Fig. . Fig.6.2. Distribution function and distribution density. uniform law The distribution function Fx(x) of a uniformly distributed random variable is equal to Fx(x)= Expectation and variance; . Exponential (exponential) distribution. A continuous random variable x taking non-negative values has an exponential distribution with parameter l>0 if the probability density distribution of the random variable is equal to рx(x)= Rice. 6.3. Distribution function and distribution density of the exponential law. The distribution function of the exponential distribution has the form Fx(x)=https://pandia.ru/text/78/107/images/image041_8.gif" width="17" height="41">.gif" width="13" height="15"> and if its distribution density is equal to Through denotes the set of all random variables distributed according to a normal law with parameters parameters and . The distribution function of a normally distributed random variable is equal to Rice. 6.4. Distribution function and normal distribution density The parameters of the normal distribution are the mathematical expectation https://pandia.ru/text/78/107/images/image048_6.gif" width="64 height=24" height="24"> In the special case when https://pandia.ru/text/78/107/images/image050_6.gif" width="44" height="21 src="> normal distribution is called standard, and the class of such distributions is denoted by https://pandia.ru/text/78/107/images/image052_6.gif" width="119" height="49">, and the distribution function Such an integral cannot be calculated analytically (it is not taken in “quadratures”), and therefore tables have been compiled for the function. The function is related to the Laplace function introduced in Chapter 4 by the following relation Therefore, the probability of a normally distributed random variable falling into an interval can be calculated using the formula A non-negative random variable x is called lognormally distributed if its logarithm h=lnx obeys the normal law. The expected value and variance of a lognormally distributed random variable are Mx= and Dx=. Task 3. Let a random variable be given https://pandia.ru/text/78/107/images/image065_5.gif" width="81" height="23">. Solution. Here https://pandia.ru/text/78/107/images/image068_5.gif" width="573" height="45"> Laplace distribution is given by the function fx(x)=https://pandia.ru/text/78/107/images/image070_5.gif" width="23" height="41"> and the kurtosis is gx=3. Fig.6.5. Laplace distribution density function. Random variable x is distributed over Weibull's law, if it has a distribution density function equal to https://pandia.ru/text/78/107/images/image072_5.gif" width="189" height="53"> The Weibull distribution governs the failure-free operation times of many technical devices. In problems of this profile, an important characteristic is the failure rate (mortality rate) l(t) of the studied elements of age t, determined by the relation l(t)=. If a=1, then the Weibull distribution turns into an exponential distribution, and if a=2 - into the so-called distribution Rayleigh. Mathematical expectation of the Weibull distribution: -https://pandia.ru/text/78/107/images/image075_4.gif" width="219" height="45 src=">, where Г(а) is the Euler function. . In various problems of applied statistics, so-called “truncated” distributions are often encountered. For example, tax authorities are interested in the distribution of income of those individuals whose annual income exceeds a certain threshold c0 established by tax laws. These distributions turn out to approximately coincide with the Pareto distribution. Pareto distribution given by functions Fx(x)=P(x Here https://pandia.ru/text/78/107/images/image081_4.gif" width="60" height="21 src=">. Task 4. The random variable is uniformly distributed on the segment. Find the density of a random variable. Solution. From the problem conditions it follows that Next, the function is a monotone and differentiable function on an interval and has an inverse function § 5. Pair of continuous random variables Let two continuous random variables x and h be given. Then the pair (x, h) defines a “random” point on the plane. The pair (x, h) is called random vector or two-dimensional random variable. Joint distribution function random variables x and h and the function is called F(x, y)=Phttps://pandia.ru/text/78/107/images/image093_3.gif" width="173" height="25">. joint density probability distribution of random variables x and h is called a function such that The meaning of this definition of joint distribution density is as follows. The probability that a “random point” (x, h) will fall into a region on a plane is calculated as the volume of a three-dimensional figure – a “curvilinear” cylinder bounded by the surface https://pandia.ru/text/78/107/images/image098_3. gif" width="211" height="39 src="> The simplest example of a joint distribution of two random variables is the two-dimensional uniform distribution on the setA.
Let a bounded set M be given with area. It is defined as the distribution of the pair (x, h), defined by the following joint density: Task 5. Let a two-dimensional random vector (x, h) be uniformly distributed inside the triangle. Calculate the probability of the inequality x>h. Solution. The area of the indicated triangle is equal to (see Fig. No.?). By virtue of the definition of a two-dimensional uniform distribution, the joint density of random variables x, h is equal to An event corresponds to a set On the half-plane B, the joint density is zero outside the set https://pandia.ru/text/78/107/images/image102_2.gif" width="15" height="17">. Thus, the half-plane B is divided into two sets and https://pandia.ru/text/78/107/images/image110_1.gif" width="17" height="23"> and , and the second integral is equal to zero, since the joint density there is equal to zero. That's why If the joint distribution density for a pair (x, h) is given, then the densities of both components x and h are called private densities and are calculated using the formulas: https://pandia.ru/text/78/107/images/image116_1.gif" width="224" height="23 src="> For continuously distributed random variables with densities рx(х), рh(у), independence means that Task 6. In the conditions of the previous problem, determine whether the components of the random vector x and h are independent? Solution. Let us calculate the partial densities and . We have: https://pandia.ru/text/78/107/images/image119_1.gif" width="283" height="61 src="> Obviously, in our case https://pandia.ru/text/78/107/images/image121_1.gif" width="64" height="25"> is the joint density of the quantities x and h, and j(x, y) is a function of two arguments, then https://pandia.ru/text/78/107/images/image123_1.gif" width="184" height="152 src="> Task 7. In the conditions of the previous problem, calculate . Solution. According to the above formula we have: Representing the triangle as https://pandia.ru/text/78/107/images/image127_1.gif" width="479" height="59"> § 5. Density of the sum of two continuous random variables Let x and h be independent random variables with densities https://pandia.ru/text/78/107/images/image128_1.gif" width="43" height="25">. The density of the random variable x + h is calculated by formula convolution https://pandia.ru/text/78/107/images/image130_0.gif" width="39" height="19 src=">. Calculate the density of the sum. Solution. Since x and h are distributed according to the exponential law with the parameter , their densities are equal Hence, https://pandia.ru/text/78/107/images/image134_0.gif" width="339 height=51" height="51"> If x<0, то в этой формуле аргумент https://pandia.ru/text/78/107/images/image136_0.gif" width="65" height="25">is negative, and therefore . Therefore, if https://pandia.ru/text/78/107/images/image140_0.gif" width="359 height=101" height="101"> Thus we got the answer: https://pandia.ru/text/78/107/images/image142_0.gif" width="40" height="41 "> normally distributed with parameters 0 and 1. Random variables x1 and x2 are independent and have normal distributions with parameters a1, and a2, respectively. Prove that x1 + x2 has a normal distribution. The random variables x1, x2, ... xn are distributed and independent and have the same density function Find the distribution function and density of distribution of values: a) h1 = min (x1, x2, ...xn) ; b) h(2) = max (x1,x2, ... xn) Random variables x1, x2, ... xn are independent and uniformly distributed on the interval [a, b]. Find distribution functions and density functions of distributions of quantities x(1) = min (x1,x2, ... xn) and x(2)= max(x1, x2, ...xn). Prove that Mhttps://pandia.ru/text/78/107/images/image147_0.gif" width="176" height="47">. The random variable is distributed according to Cauchy's law Find: a) coefficient a; b) distribution function; c) the probability of falling into the interval (-1, 1). Show that the mathematical expectation of x does not exist. The random variable is subject to Laplace's law with the parameter l (l>0): Find the coefficient a; construct distribution density graphs and distribution functions; find Mx and Dx; find the probabilities of events (|x|< и {çxç<}. Случайная величина x подчинена закону Симпсона на отрезке [-а, а], т. е. график её плотности распределения имеет вид: Write a formula for the distribution density, find Mx and Dx. Computational tasks. A random point A has a uniform distribution in a circle of radius R. Find the mathematical expectation and variance of the distance r of the point to the center of the circle. Show that the value r2 is uniformly distributed on the segment. Find the constant c, the distribution density h = and the probability P (0.25 The failure-free operation time of a computer is distributed according to an exponential law with the parameter l = 0.05 (failures per hour), i.e., it has a density function p(x) = Solving a certain problem requires trouble-free operation of the machine for 15 minutes. If a failure occurs while solving a problem, the error is detected only after the solution is completed, and the problem is solved again. Find: a) the probability that during the solution of the problem not a single failure will occur; b) the average time in which the problem will be solved. A rod 24 cm long is broken into two parts; We will assume that the break point is distributed evenly along the entire length of the rod. What is the average length of most of the rod? A piece of length 12 cm is randomly cut into two parts. The cut point is evenly distributed along the entire length of the segment. What is the average length of the small part of the segment? The random variable is uniformly distributed on the segment. Find the distribution density of the random variable a) h1 = 2x + 1; b) h2 =-ln(1-x); c) h3 = . Show that if x has a continuous distribution function F(x) = P(x Find the density function and distribution function of the sum of two independent quantities x and h with uniform distribution laws on the segments and, respectively. The random variables x and h are independent and uniformly distributed on the segments and, respectively. Calculate the density of the sum x+h. The random variables x and h are independent and uniformly distributed on the segments and, respectively. Calculate the density of the sum x+h. The random variables x and h are independent and uniformly distributed on the segments and, respectively. Calculate the density of the sum x+h. Random variables are independent and have an exponential distribution with density As is known, random variable
is called a variable quantity that can take on certain values depending on the case. Random variables are denoted by capital letters of the Latin alphabet (X, Y, Z), and their values are denoted by corresponding lowercase letters (x, y, z). Random variables are divided into discontinuous (discrete) and continuous. Discrete random variable
is a random variable that takes only a finite or infinite (countable) set of values with certain non-zero probabilities. Distribution law of a discrete random variable
is a function that connects the values of a random variable with their corresponding probabilities. The distribution law can be specified in one of the following ways. 1
. The distribution law can be given by the table:
where λ>0, k = 0, 1, 2, … . V) by using distribution function F(x)
, which determines for each value x the probability that the random variable X will take a value less than x, i.e. F(x) = P(X< x). Properties of the function F(x) 3
. The distribution law can be specified graphically
– distribution polygon (polygon) (see problem 3). Note that to solve some problems it is not necessary to know the distribution law. In some cases, it is enough to know one or several numbers that reflect the most important features of the distribution law. This can be a number that has the meaning of the “average value” of a random variable, or a number showing the average size of the deviation of a random variable from its mean value. Numbers of this kind are called numerical characteristics of a random variable. Basic numerical characteristics of a discrete random variable
: 1000 lottery tickets were issued: 5 of them will win 500 rubles, 10 will win 100 rubles, 20 will win 50 rubles, 50 will win 10 rubles. Determine the law of probability distribution of the random variable X - winnings per ticket. Solution.
According to the conditions of the problem, the following values of the random variable X are possible: 0, 10, 50, 100 and 500. The number of tickets without winning is 1000 – (5+10+20+50) = 915, then P(X=0) = 915/1000 = 0.915. Similarly, we find all other probabilities: P(X=0) = 50/1000=0.05, P(X=50) = 20/1000=0.02, P(X=100) = 10/1000=0.01 , P(X=500) = 5/1000=0.005. Let us present the resulting law in the form of a table: Let's find the mathematical expectation of the value X: M(X) = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = (1+ 2+3+4+5+6)/6 = 21/6 = 3.5 The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Draw up a distribution law for the number of failed elements in one experiment, construct a distribution polygon. Find the distribution function F(x) and plot it. Find the mathematical expectation, variance and standard deviation of a discrete random variable. Solution.
1.
The discrete random variable X = (the number of failed elements in one experiment) has the following possible values: x 1 = 0 (none of the device elements failed), x 2 = 1 (one element failed), x 3 = 2 (two elements failed ) and x 4 =3 (three elements failed). Failures of elements are independent of each other, the probabilities of failure of each element are equal, therefore it is applicable Bernoulli's formula
. Considering that, according to the condition, n=3, p=0.1, q=1-p=0.9, we determine the probabilities of the values: Thus, the desired binomial distribution law of X has the form: We plot the possible values of x i along the abscissa axis, and the corresponding probabilities p i along the ordinate axis. Let's construct points M 1 (0; 0.729), M 2 (1; 0.243), M 3 (2; 0.027), M 4 (3; 0.001). By connecting these points with straight line segments, we obtain the desired distribution polygon. 3.
Let's find the distribution function F(x) = Р(Х Graph of function F(x) 4.
For binomial distribution X: We can highlight the most common laws of distribution of discrete random variables: For given distributions of discrete random variables, the calculation of the probabilities of their values, as well as numerical characteristics (mathematical expectation, variance, etc.) is carried out using certain “formulas”. Therefore, it is very important to know these types of distributions and their basic properties. A discrete random variable $X$ is subject to the binomial probability distribution law if it takes values $0,\ 1,\ 2,\ \dots ,\ n$ with probabilities $P\left(X=k\right)=C^k_n\cdot p^k\cdot (\left(1-p\right))^(n-k)$. In fact, the random variable $X$ is the number of occurrences of event $A$ in $n$ independent trials. Law of probability distribution of random variable $X$: $\begin(array)(|c|c|) For such a random variable, the mathematical expectation is $M\left(X\right)=np$, the variance is $D\left(X\right)=np\left(1-p\right)$. Example
. The family has two children. Assuming the probabilities of having a boy and a girl equal to $0.5$, find the law of distribution of the random variable $\xi$ - the number of boys in the family. Let the random variable $\xi $ be the number of boys in the family. Values that $\xi can take:\ 0,\ 1,\ 2$. The probabilities of these values can be found using the formula $P\left(\xi =k\right)=C^k_n\cdot p^k\cdot (\left(1-p\right))^(n-k)$, where $n =2$ is the number of independent trials, $p=0.5$ is the probability of an event occurring in a series of $n$ trials. We get: $P\left(\xi =0\right)=C^0_2\cdot (0,5)^0\cdot (\left(1-0,5\right))^(2-0)=(0, 5)^2=0.25;$ $P\left(\xi =1\right)=C^1_2\cdot 0.5\cdot (\left(1-0.5\right))^(2-1)=2\cdot 0.5\ cdot 0.5=0.5;$ $P\left(\xi =2\right)=C^2_2\cdot (0.5)^2\cdot (\left(1-0.5\right))^(2-2)=(0, 5)^2=0.25.$ Then the distribution law of the random variable $\xi $ is the correspondence between the values $0,\ 1,\ 2$ and their probabilities, that is: $\begin(array)(|c|c|) The sum of the probabilities in the distribution law should be equal to $1$, that is, $\sum _(i=1)^(n)P(\xi _((\rm i)))=0.25+0.5+0, 25=$1. Expectation $M\left(\xi \right)=np=2\cdot 0.5=1$, variance $D\left(\xi \right)=np\left(1-p\right)=2\ cdot 0.5\cdot 0.5=0.5$, standard deviation $\sigma \left(\xi \right)=\sqrt(D\left(\xi \right))=\sqrt(0.5 )\approx $0.707. If a discrete random variable $X$ can only take non-negative integer values $0,\ 1,\ 2,\ \dots ,\ n$ with probabilities $P\left(X=k\right)=(((\lambda )^k )\over (k}\cdot e^{-\lambda }$, то говорят, что она подчинена закону распределения Пуассона с параметром $\lambda $. Для такой случайной величины математическое ожидание и дисперсия равны между собой и равны параметру $\lambda $, то есть $M\left(X\right)=D\left(X\right)=\lambda $.!} Comment. The peculiarity of this distribution is that, based on experimental data, we find estimates $M\left(X\right),\ D\left(X\right)$, if the obtained estimates are close to each other, then we have reason to assert that the random variable is subject to the Poisson distribution law. Example
. Examples of random variables subject to the Poisson distribution law can be: the number of cars that will be served by a gas station tomorrow; number of defective items in manufactured products. Example
. The factory sent $500$ of products to the base. The probability of damage to the product in transit is $0.002$. Find the law of distribution of the random variable $X$ equal to the number of damaged products; what is $M\left(X\right),\ D\left(X\right)$. Let the discrete random variable $X$ be the number of damaged products. Such a random variable is subject to the Poisson distribution law with the parameter $\lambda =np=500\cdot 0.002=1$. The probabilities of the values are equal to $P\left(X=k\right)=(((\lambda )^k)\over (k}\cdot e^{-\lambda }$. Очевидно, что все вероятности всех значений $X=0,\ 1,\ \dots ,\ 500$ перечислить невозможно, поэтому мы ограничимся лишь первыми несколькими значениями.!} $P\left(X=0\right)=((1^0)\over (0}\cdot e^{-1}=0,368;$!} $P\left(X=1\right)=((1^1)\over (1}\cdot e^{-1}=0,368;$!} $P\left(X=2\right)=((1^2)\over (2}\cdot e^{-1}=0,184;$!} $P\left(X=3\right)=((1^3)\over (3}\cdot e^{-1}=0,061;$!} $P\left(X=4\right)=((1^4)\over (4}\cdot e^{-1}=0,015;$!} $P\left(X=5\right)=((1^5)\over (5}\cdot e^{-1}=0,003;$!} $P\left(X=6\right)=((1^6)\over (6}\cdot e^{-1}=0,001;$!} $P\left(X=k\right)=(((\lambda )^k)\over (k}\cdot e^{-\lambda }$!} Distribution law of random variable $X$: $\begin(array)(|c|c|) For such a random variable, the mathematical expectation and variance are equal to each other and equal to the parameter $\lambda $, that is, $M\left(X\right)=D\left(X\right)=\lambda =1$. If a discrete random variable $X$ can only take natural values $1,\ 2,\ \dots ,\ n$ with probabilities $P\left(X=k\right)=p(\left(1-p\right)) ^(k-1),\ k=1,\ 2,\ 3,\ \dots $, then they say that such a random variable $X$ is subject to the geometric law of probability distribution. In fact, the geometric distribution is a Bernoulli test until the first success. Example
. Examples of random variables that have a geometric distribution can be: the number of shots before the first hit on the target; number of device tests until the first failure; the number of coin tosses until the first head comes up, etc. The mathematical expectation and variance of a random variable subject to geometric distribution are respectively equal to $M\left(X\right)=1/p$, $D\left(X\right)=\left(1-p\right)/p^ $2. Example
. On the way of fish movement to the spawning site there is a $4$ lock. The probability of fish passing through each lock is $p=3/5$. Construct a series of distribution of the random variable $X$ - the number of locks passed by the fish before the first detention at the lock. Find $M\left(X\right),\ D\left(X\right),\ \sigma \left(X\right)$. Let the random variable $X$ be the number of locks passed by the fish before the first arrest at the lock. Such a random variable is subject to the geometric law of probability distribution. Values that the random variable $X can take:$ 1, 2, 3, 4. The probabilities of these values are calculated using the formula: $P\left(X=k\right)=pq^(k-1)$, where: $ p=2/5$ - probability of fish being detained through the lock, $q=1-p=3/5$ - probability of fish passing through the lock, $k=1,\ 2,\ 3,\ 4$. $P\left(X=1\right)=((2)\over (5))\cdot (\left(((3)\over (5))\right))^0=((2)\ over (5))=0.4;$ $P\left(X=2\right)=((2)\over (5))\cdot ((3)\over (5))=((6)\over (25))=0.24; $ $P\left(X=3\right)=((2)\over (5))\cdot (\left(((3)\over (5))\right))^2=((2)\ over (5))\cdot ((9)\over (25))=((18)\over (125))=0.144;$ $P\left(X=4\right)=((2)\over (5))\cdot (\left(((3)\over (5))\right))^3+(\left(( (3)\over (5))\right))^4=((27)\over (125))=0.216.$ $\begin(array)(|c|c|) Expected value: $M\left(X\right)=\sum^n_(i=1)(x_ip_i)=1\cdot 0.4+2\cdot 0.24+3\cdot 0.144+4\cdot 0.216=2.176.$ Dispersion: $D\left(X\right)=\sum^n_(i=1)(p_i(\left(x_i-M\left(X\right)\right))^2=)0.4\cdot (\ left(1-2,176\right))^2+0,24\cdot (\left(2-2,176\right))^2+0,144\cdot (\left(3-2,176\right))^2+$ $+\0.216\cdot (\left(4-2,176\right))^2\approx 1.377.$ Standard deviation: $\sigma \left(X\right)=\sqrt(D\left(X\right))=\sqrt(1,377)\approx 1,173.$ If $N$ objects, among which $m$ objects have a given property. $n$ objects are randomly retrieved without returning, among which there were $k$ objects that have a given property. The hypergeometric distribution makes it possible to estimate the probability that exactly $k$ objects in the sample have a given property. Let the random variable $X$ be the number of objects in the sample that have a given property. Then the probabilities of the values of the random variable $X$: $P\left(X=k\right)=((C^k_mC^(n-k)_(N-m))\over (C^n_N))$ Comment. The statistical function HYPERGEOMET of the Excel $f_x$ function wizard allows you to determine the probability that a certain number of tests will be successful. $f_x\to$ statistical$\to$ HYPERGEOMET$\to$ OK. A dialog box will appear that you need to fill out. In the column Number_of_successes_in_sample indicate the value $k$. sample_size equals $n$. In the column Number_of_successes_in_together indicate the value $m$. population_size equals $N$. The mathematical expectation and variance of a discrete random variable $X$, subject to the geometric distribution law, are respectively equal to $M\left(X\right)=nm/N$, $D\left(X\right)=((nm\left(1 -((m)\over (N))\right)\left(1-((n)\over (N))\right))\over (N-1))$. Example
. The bank's credit department employs 5 specialists with higher financial education and 3 specialists with higher legal education. The bank's management decided to send 3 specialists to improve their qualifications, selecting them in random order. a) Make a distribution series for the number of specialists with higher financial education who can be sent to improve their skills; b) Find the numerical characteristics of this distribution. Let the random variable $X$ be the number of specialists with higher financial education among the three selected ones. Values that $X can take: 0,\ 1,\ 2,\ 3$. This random variable $X$ is distributed according to a hypergeometric distribution with the following parameters: $N=8$ - population size, $m=5$ - number of successes in the population, $n=3$ - sample size, $k=0,\ 1, \2,\3$ - number of successes in the sample. Then the probabilities $P\left(X=k\right)$ can be calculated using the formula: $P(X=k)=(C_(m)^(k) \cdot C_(N-m)^(n-k) \over C_( N)^(n) ) $. We have: $P\left(X=0\right)=((C^0_5\cdot C^3_3)\over (C^3_8))=((1)\over (56))\approx 0.018;$ $P\left(X=1\right)=((C^1_5\cdot C^2_3)\over (C^3_8))=((15)\over (56))\approx 0.268;$ $P\left(X=2\right)=((C^2_5\cdot C^1_3)\over (C^3_8))=((15)\over (28))\approx 0.536;$ $P\left(X=3\right)=((C^3_5\cdot C^0_3)\over (C^3_8))=((5)\over (28))\approx 0.179.$ Then the distribution series of the random variable $X$: $\begin(array)(|c|c|) Let us calculate the numerical characteristics of the random variable $X$ using the general formulas of the hypergeometric distribution. $M\left(X\right)=((nm)\over (N))=((3\cdot 5)\over (8))=((15)\over (8))=1,875.$ $D\left(X\right)=((nm\left(1-((m)\over (N))\right)\left(1-((n)\over (N))\right)) \over (N-1))=((3\cdot 5\cdot \left(1-((5)\over (8))\right)\cdot \left(1-((3)\over (8 ))\right))\over (8-1))=((225)\over (448))\approx 0.502.$ $\sigma \left(X\right)=\sqrt(D\left(X\right))=\sqrt(0.502)\approx 0.7085.$
whence C=3/8.
, and also, as in the discrete case, according to the formula https://pandia.ru/text/78/107/images/image031_11.gif" width="123" height="49 src=">.
.
.
,
. In the case of arbitrary parameter values https://pandia.ru/text/78/107/images/image043_5.gif" width="21" height="21 src="> the distribution function of a random variable is related to the Laplace function using the relation:
.
.
, whose derivative is equal to Therefore,
.
on a plane, i.e. a half-plane. Then the probability
.
.
The distribution density of a random variable has the form: 
Calculate the constant C, the distribution function F(x), and the probability 
The distribution density of a random variable has the form: 
Calculate the constant C, the distribution function F(x), and the probability 
The distribution density of a random variable has the form: 
Calculate the constant C, the distribution function F(x), , variance and probability. A random variable has a distribution function 
Calculate the density of a random variable, mathematical expectation, variance and probability Check that the function = 
may be a distribution function of a random variable. Find the numerical characteristics of this quantity: Mx and Dx. The random variable is uniformly distributed on the segment. Write down the distribution density. Find the distribution function. Find the probability of a random variable falling on the segment and on the segment. The distribution density x is equal to
.
.
. Find the distribution density of their sum. Find the distribution of the sum of independent random variables x and h, where x has a uniform distribution on the interval, and h has an exponential distribution with parameter l. Find P 
, if x has: a) normal distribution with parameters a and s2; b) exponential distribution with parameter l; c) uniform distribution on the segment [-1;1]. The joint distribution of x, h is squared uniform
K = (x, y): |x| +|y|£ 2). Find probability 
. Are x and h independent? A pair of random variables x and h are uniformly distributed inside the triangle K=. Calculate the densities x and h. Are these random variables independent? Find the probability. Random variables x and h are independent and uniformly distributed on the segments and [-1,1]. Find the probability. A two-dimensional random variable (x, h) is uniformly distributed in a square with vertices (2,0), (0,2), (-2, 0), (0,-2). Find the value of the joint distribution function at point (1, -1). A random vector (x, h) is uniformly distributed inside a circle of radius 3 centered at the origin. Write an expression for the joint distribution density. Determine whether these random variables are dependent. Calculate probability. A pair of random variables x and h are uniformly distributed inside a trapezoid with vertices at points (-6,0), (-3,4), (3,4), (6,0). Find the joint distribution density for this pair of random variables and the density of the components. Are x and h dependent? A random pair (x, h) is uniformly distributed inside a semicircle. Find the densities x and h, investigate the question of their dependence. The joint density of two random variables x and h is equal to 
.
Find the densities x, h. Investigate the question of the dependence of x and h. A random pair (x, h) is uniformly distributed on the set. Find the densities x and h, investigate the question of their dependence. Find M(xh). Random variables x and h are independent and distributed according to the exponential law with the parameter Find
For binomial distribution M(X)=np, for Poisson distribution M(X)=λ
For binomial distribution D(X)=npq, for Poisson distribution D(X)=λExamples of solving problems on the topic “The law of distribution of a discrete random variable”
Task 1.
Task 3.
P 3 (0) = C 3 0 p 0 q 3-0 = q 3 = 0.9 3 = 0.729;
P 3 (1) = C 3 1 p 1 q 3-1 = 3*0.1*0.9 2 = 0.243;
P 3 (2) = C 3 2 p 2 q 3-2 = 3*0.1 2 *0.9 = 0.027;
P 3 (3) = C 3 3 p 3 q 3-3 = p 3 =0.1 3 = 0.001;
Check: ∑p i = 0.729+0.243+0.027+0.001=1.
for 0< x ≤1 имеем F(x) = Р(Х<1) = Р(Х = 0) = 0,729;
for 1< x ≤ 2 F(x) = Р(Х<2) = Р(Х=0) + Р(Х=1) =0,729+ 0,243 = 0,972;
for 2< x ≤ 3 F(x) = Р(Х<3) = Р(Х = 0) + Р(Х = 1) + Р(Х = 2) = 0,972+0,027 = 0,999;
for x > 3 there will be F(x) = 1, because the event is reliable. 
- mathematical expectation M(X) = np = 3*0.1 = 0.3;
- variance D(X) = npq = 3*0.1*0.9 = 0.27;
- standard deviation σ(X) = √D(X) = √0.27 ≈ 0.52.
1. Binomial distribution law.
\hline
X_i & 0 & 1 & \dots & n \\
\hline
p_i & P_n\left(0\right) & P_n\left(1\right) & \dots & P_n\left(n\right) \\
\hline
\end(array)$
\hline
\xi & 0 & 1 & 2 \\
\hline
P(\xi) & 0.25 & 0.5 & 0.25 \\
\hline
\end(array)$2. Poisson distribution law.
\hline
X_i & 0 & 1 & 2 & 3 & 4 & 5 & 6 & ... & k \\
\hline
P_i & 0.368; & 0.368 & 0.184 & 0.061 & 0.015 & 0.003 & 0.001 & ... & (((\lambda )^k)\over (k}\cdot e^{-\lambda } \\!}
\hline
\end(array)$3. Geometric distribution law.
\hline
X_i & 1 & 2 & 3 & 4 \\
\hline
P\left(X_i\right) & 0.4 & 0.24 & 0.144 & 0.216 \\
\hline
\end(array)$4. Hypergeometric distribution law.
\hline
X_i & 0 & 1 & 2 & 3 \\
\hline
p_i & 0.018 & 0.268 & 0.536 & 0.179 \\
\hline
\end(array)$