Laws of distribution of random variables. Normal distribution of continuous random variables What parameters characterize the normal distribution

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.

Definition. A continuous random variable X has normal law distribution(Gauss's law )with parameters a and σ 2 if its probability density f(x) looks like:

The normal distribution curve is called normal or Gaussian curve. In Fig. 6.5 a), b) shows a normal curve with parameters A And σ 2 and distribution function graph.

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 to , and two inflection points X = A σ with ordinates.

It can be noted that in the normal law density expression, the distribution parameters are indicated by the letters A And σ 2, which we used to denote the mathematical expectation and dispersion. 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 distribution, i.e.

and its dispersion – to the parameter σ 2, i.e.

Let's find out how the normal curve will change when the parameters change A And σ .

If σ = const, and the parameter changes a (A 1 < A 2 < 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. 6.6).

Rice. 6.6

Rice. 6.7

If A= const and the parameter changes σ , then the ordinate of the curve maximum changes f max(a) = . When increasing σ the ordinate of the maximum decreases, but since the area under any distribution curve must remain equal to unity, the curve becomes flatter, stretching along the x-axis. When decreasing σ On the contrary, the normal curve extends upward while simultaneously compressing from the sides (Fig. 6.7).

So the parameter a characterizes the position, and the parameter σ – the shape of a normal curve.

Normal distribution law of a random variable with parameters a= 0 and σ = 1 is 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 is due to the fact that the integral of the normal distribution function is not expressed through elementary functions. However, it can be calculated through a special function expressing a definite integral of the expression or. This function is called Laplace function, tables have been compiled for it. There are many varieties of this function, for example:

, .

We will use the function

Let us consider the properties of a random variable distributed according to a normal law.

1. The probability of a random variable X, distributed according to a normal law, falling into the interval [α , β ] equal to

Using this formula, we calculate the probabilities for various values δ (using the table of Laplace function values):

at δ = σ = 2Ф(1) = 0.6827;

at δ = 2σ = 2Ф(2) = 0.9545;

at δ = 3σ = 2Ф(3) = 0.9973.

This leads to the so-called “ three sigma rule»:

If a random variable X has a normal distribution law with parameters a and σ, then it is almost certain that its values ​​lie in the interval(a – 3σ ; a + 3σ ).

Example 6.3. Assuming that the height of men of a certain age group is a normally distributed random variable X with parameters A= 173 and σ 2 = 36, find:

1. Expression of probability density and distribution function of a random variable X;

2. The share of suits of the 4th height (176 - 183 cm) and the share of suits of the 3rd height (170 - 176 cm), which must be included in the total production volume for this age group;

3. Formulate the “three sigma rule” for a random variable X.

1. Finding the probability density

and the distribution function of the random variable X

= .

2. We find the proportion of suits of height 4 (176 – 182 cm) as a probability

R(176 ≤ X ≤ 182) = = Ф(1.5) – Ф(0.5).

According to the table of values ​​of the Laplace function ( Appendix 2) we find:

F(1.5) = 0.4332, F(0.5) = 0.1915.

Finally we get

R(176 ≤ X ≤ 182) = 0,4332 – 0,1915 = 0,2417.

The share of suits of the 3rd height (170 – 176 cm) can be found in a similar way. However, it is easier to do this if we take into account that this interval is symmetrical with respect to the mathematical expectation A= 173, i.e. inequality 170 ≤ X≤ 176 is equivalent to inequality │ X– 173│≤ 3. Then

R(170 ≤X ≤176) = R(│X– 173│≤ 3) = 2Ф(3/6) = 2Ф(0.5) = 2·0.1915 = 0.3830.

3. Let us formulate the “three sigma rule” for the random variable X:

It is almost certain that the height of men in this age group ranges from A – 3σ = 173 – 3 6 = 155 to A + 3σ = 173 + 3·6 = 191, i.e. 155 ≤ X ≤ 191. ◄

7. LIMIT THEOREMS OF PROBABILITY THEORY

As already mentioned when studying random variables, it is impossible to predict in advance what value a random variable will take as a result of a single test - it depends on many reasons that cannot be taken into account.

However, when tests are repeated many times, the behavior of the sum of random variables almost loses its random character and becomes natural. The presence of patterns is associated precisely with the mass nature of phenomena that in their totality generate a random variable that is subject to a well-defined law. The essence of the stability of mass phenomena comes down to the following: the specific features of each individual random phenomenon have almost no effect on the average result of the mass of such phenomena; random deviations from the average, inevitable in each individual phenomenon, are mutually canceled out, leveled out, leveled out in the mass.

It is this stability of averages that represents the physical content of the “law of large numbers,” understood in the broad sense of the word: with a very large number of random phenomena, their result practically ceases to be random and can be predicted with a high degree of certainty.

In the narrow sense of the word, the “law of large numbers” in probability theory is understood as a series of mathematical theorems, each of which, for certain conditions, establishes the fact that the average characteristics of a large number of experiments approach certain certain constants.

The law of large numbers plays an important role in the practical applications of probability theory. The property of random variables, under certain conditions, to behave practically like non-random ones allows one to confidently operate with these quantities and predict the results of mass random phenomena with almost complete certainty.

The possibilities of such predictions in the field of mass random phenomena are further expanded by the presence of another group of limit theorems, which concern not the limiting values ​​of random variables, but the limiting laws of distribution. We are talking about a group of theorems known as the “central limit theorem.” The various forms of the central limit theorem differ from each other in the conditions for which this limiting property of the sum of random variables is established.

Various forms of the law of large numbers with various forms of the central limit theorem form a set of so-called limit theorems probability theory. Limit theorems make it possible not only to make scientific forecasts in the field of random phenomena, but also to evaluate the accuracy of these forecasts.

(real, strictly positive)

Normal distribution, also called Gaussian distribution or Gauss - Laplace- probability distribution, which in the one-dimensional case is specified by the probability density function coinciding with the Gaussian function:

where the parameter μ is the expectation (mean value), median and mode of the distribution, and the parameter σ is the standard deviation (σ² is the dispersion) of the distribution.

Thus, the one-dimensional normal distribution is a two-parameter family of distributions. The multivariate case is described in the article “Multivariate normal distribution”.

Standard normal distribution is called a normal distribution with mathematical expectation μ = 0 and standard deviation σ = 1.

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  The importance of the normal distribution in many fields of science (for example, mathematical statistics and statistical physics) follows from the central limit theorem of probability theory. If the result of an observation is the sum of many random weakly interdependent quantities, each of which makes a small contribution relative to the total sum, then as the number of terms increases, the distribution of the centered and normalized result tends to be normal. This law of probability theory results in the widespread distribution of the normal distribution, which was one of the reasons for its name.

  Properties

  Moments

  If random variables X 1 (\displaystyle X_(1)) And X 2 (\displaystyle X_(2)) are independent and have a normal distribution with mathematical expectations μ 1 (\displaystyle \mu _(1)) And μ 2 (\displaystyle \mu _(2)) and variances σ 1 2 (\displaystyle \sigma _(1)^(2)) And σ 2 2 (\displaystyle \sigma _(2)^(2)) accordingly, then X 1 + X 2 (\displaystyle X_(1)+X_(2)) also has a normal distribution with mathematical expectation μ 1 + μ 2 (\displaystyle \mu _(1)+\mu _(2)) and variance σ 1 2 + σ 2 2 . (\displaystyle \sigma _(1)^(2)+\sigma _(2)^(2).) It follows that a normal random variable can be represented as the sum of an arbitrary number of independent normal random variables.

  Maximum entropy

  The normal distribution has the maximum differential entropy among all continuous distributions whose variance does not exceed a given value.

  Modeling normal pseudorandom variables

  The simplest approximate modeling methods are based on the central limit theorem. Namely, if you add several independent identically distributed quantities with finite variance, then the sum will be distributed approximately Fine. For example, if you add 100 independent ones as standard evenly distributed random variables, then the distribution of the sum will be approximately normal.

  For programmatic generation of normally distributed pseudorandom variables, it is preferable to use the Box-Muller transformation. It allows you to generate one normally distributed value based on one uniformly distributed value.

  Normal distribution in nature and applications

  Normal distribution is often found in nature. For example, the following random variables are well modeled by the normal distribution:

  • deviation when shooting.
  • measurement errors (however, the errors of some measuring instruments do not have normal distributions).
  • some characteristics of living organisms in a population.

  This distribution is so widespread because it is an infinitely divisible continuous distribution with finite variance. Therefore, some others approach it in the limit, for example, binomial and Poisson. This distribution models many non-deterministic physical processes.

  Relationship with other distributions

  • The normal distribution is a Pearson type XI distribution.
  • The ratio of a pair of independent standard normally distributed random variables has a Cauchy distribution. That is, if the random variable X (\displaystyle X) represents the relation X = Y / Z (\displaystyle X=Y/Z)(Where Y (\displaystyle Y) And Z (\displaystyle Z)- independent standard normal random variables), then it will have a Cauchy distribution.
  • If z 1 , … , z k (\displaystyle z_(1),\ldots ,z_(k))- jointly independent standard normal random variables, that is z i ∼ N (0 , 1) (\displaystyle z_(i)\sim N\left(0,1\right)), then the random variable x = z 1 2 + … + z k 2 (\displaystyle x=z_(1)^(2)+\ldots +z_(k)^(2)) has a chi-square distribution with k degrees of freedom.
  • If the random variable X (\displaystyle X) is subject to lognormal distribution, then its natural logarithm has a normal distribution. That is, if X ∼ L o g N (μ , σ 2) (\displaystyle X\sim \mathrm (LogN) \left(\mu ,\sigma ^(2)\right)), That Y = ln ⁡ (X) ∼ N (μ , σ 2) (\displaystyle Y=\ln \left(X\right)\sim \mathrm (N) \left(\mu ,\sigma ^(2)\right )). And vice versa, if Y ∼ N (μ , σ 2) (\displaystyle Y\sim \mathrm (N) \left(\mu ,\sigma ^(2)\right)), That X = exp ⁡ (Y) ∼ L o g N (μ , σ 2) (\displaystyle X=\exp \left(Y\right)\sim \mathrm (LogN) \left(\mu ,\sigma ^(2) \right)).
  • The ratio of the squares of two standard normal random variables has

  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- waiting time for the train.

  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 %.

  Definition. Normal is the probability distribution of a continuous random variable, which is described by the probability density

  

  The normal distribution law is also called Gauss's law.

  The normal distribution law occupies a central place in probability theory. This is due to the fact that this law manifests itself in all cases where a random variable is the result of the action of a large number of different factors. All other distribution laws approach the normal law.

  It can be easily shown that the parameters And , included in the distribution density are, respectively, the mathematical expectation and standard deviation of the random variable X.

  Let's find the distribution function F(x) .

  

  The density graph of a normal distribution is called normal curve or Gaussian curve.

  A normal curve has the following properties:

  1) The function is defined on the entire number line.

  2) In front of everyone X the distribution function takes only positive values.

  3) The OX axis is the horizontal asymptote of the probability density graph, because with unlimited increase in the absolute value of the argument X, the value of the function tends to zero.

  4) Find the extremum of the function.

  

  Because at y’ > 0 at x < m And y’ < 0 at x > m, then at the point x = t the function has a maximum equal to
  .

  5) The function is symmetrical with respect to a straight line x = a, because difference

  (x – a) is included in the squared distribution density function.

  6) To find the inflection points of the graph, we will find the second derivative of the density function.

  

  At x = m+  and x = m-  the second derivative is equal to zero, and when passing through these points it changes sign, i.e. at these points the function has an inflection point.

  At these points the function value is equal to
  .

  Let's plot the distribution density function (Fig. 5).

  

  Graphs were built for T=0 and three possible values ​​of the standard deviation  = 1,  = 2 and  = 7. As you can see, as the value of the standard deviation increases, the graph becomes flatter, and the maximum value decreases.

  If A> 0, then the graph will shift in a positive direction if A < 0 – в отрицательном.

  At A= 0 and  = 1 the curve is called normalized. Normalized curve equation:

  

  Laplace function

  Let's find the probability of a random variable distributed according to a normal law falling into a given interval.

  

  Let's denote
  

  Because integral
  is not expressed through elementary functions, then the function is introduced into consideration

  ,

  which is called Laplace function or probability integral.

  The values ​​of this function for various values X calculated and presented in special tables.

  In Fig. Figure 6 shows a graph of the Laplace function.

  

  The Laplace function has the following properties:

  1) F(0) = 0;

  2) F(-x) = - F(x);

  3) F( ) = 1.

  The Laplace function is also called error function and denote erf x.

  Still in use normalized Laplace function, which is related to the Laplace function by the relation:

  

  In Fig. Figure 7 shows a graph of the normalized Laplace function.

  

  P three sigma rule

  When considering the normal distribution law, an important special case stands out, known as three sigma rule.

  Let us write down the probability that the deviation of a normally distributed random variable from the mathematical expectation is less than a given value :

  If we take  = 3, then using tables of values ​​of the Laplace function we obtain:

  Those. the probability that a random variable will deviate from its mathematical expectation by an amount greater than triple the standard deviation is practically zero.

  This rule is called three sigma rule.

  In practice, it is believed that if the three-sigma rule is satisfied for any random variable, then this random variable has a normal distribution.

  Conclusion of the lecture:

  In the lecture, we examined the laws of distribution of continuous quantities. In preparation for the subsequent lecture and practical classes, you must independently supplement your lecture notes when studying the recommended literature in depth and solving the proposed problems.

  As mentioned earlier, examples of probability distributions continuous random variable X are:

  • uniform distribution
  • exponential distribution probabilities of a continuous random variable;
  • normal probability distribution of a continuous random variable.

  Let us give the concept of a normal distribution law, the distribution function of such a law, and the procedure for calculating the probability of a random variable X falling into a certain interval.

  №	Index	Normal distribution law	Note
  	Definition	Called normal probability distribution of a continuous random variable X, whose density has the form
  			where m x is the mathematical expectation of the random variable X, σ x is the standard deviation
  2	Distribution function
  	Probability falling into the interval (a;b)		- Laplace integral function
  	Probability the fact that the absolute value of the deviation is less than a positive number δ		at m x = 0

  An example of solving a problem on the topic “Normal distribution law of a continuous random variable”

  Task.

  The length X of a certain part is a random variable distributed according to the normal distribution law, and has an average value of 20 mm and a standard deviation of 0.2 mm.
  Necessary:
  a) write down the expression for the distribution density;
  b) find the probability that the length of the part will be between 19.7 and 20.3 mm;
  c) find the probability that the deviation does not exceed 0.1 mm;
  d) determine what percentage are parts whose deviation from the average value does not exceed 0.1 mm;
  e) find what the deviation should be set so that the percentage of parts whose deviation from the average does not exceed the specified value increases to 54%;
  f) find an interval symmetrical about the average value in which X will be located with probability 0.95.

  Solution. A) We find the probability density of a random variable X distributed according to a normal law:

  provided that m x =20, σ =0.2.

  b) For a normal distribution of a random variable, the probability of falling into the interval (19.7; 20.3) is determined by:
  Ф((20.3-20)/0.2) – Ф((19.7-20)/0.2) = Ф(0.3/0.2) – Ф(-0.3/0, 2) = 2Ф(0.3/0.2) = 2Ф(1.5) = 2*0.4332 = 0.8664.
  We found the value Ф(1.5) = 0.4332 in the appendices, in the table of values ​​of the Laplace integral function Φ(x) ( table 2 )

  V) We find the probability that the absolute value of the deviation is less than a positive number 0.1:
  R(|X-20|< 0,1) = 2Ф(0,1/0,2) = 2Ф(0,5) = 2*0,1915 = 0,383.
  We found the value Ф(0.5) = 0.1915 in the appendices, in the table of values ​​of the Laplace integral function Φ(x) ( table 2 )

  G) Since the probability of a deviation less than 0.1 mm is 0.383, it follows that on average 38.3 parts out of 100 will have such a deviation, i.e. 38.3%.

  d) Since the percentage of parts whose deviation from the average does not exceed the specified value has increased to 54%, then P(|X-20|< δ) = 0,54. Отсюда следует, что 2Ф(δ/σ) = 0,54, а значит Ф(δ/σ) = 0,27.

  Using the application ( table 2 ), we find δ/σ = 0.74. Hence δ = 0.74*σ = 0.74*0.2 = 0.148 mm.

  e) Since the required interval is symmetrical with respect to the average value m x = 20, it can be defined as the set of values ​​of X satisfying the inequality 20 − δ< X < 20 + δ или |x − 20| < δ .

  According to the condition, the probability of finding X in the desired interval is 0.95, which means P(|x − 20|< δ)= 0,95. С другой стороны P(|x − 20| < δ) = 2Ф(δ/σ), следовательно 2Ф(δ/σ) = 0,95, а значит Ф(δ/σ) = 0,475.

  Using the application ( table 2 ), we find δ/σ = 1.96. Hence δ = 1.96*σ = 1.96*0.2 = 0.392.
  Search interval : (20 – 0.392; 20 + 0.392) or (19.608; 20.392).