The normal distribution is the most common type of distribution. It is encountered in the analysis of measurement errors, the control of technological processes and regimes, as well as in the analysis and prediction of various phenomena in biology, medicine and other fields of knowledge.
The term "normal distribution" is used in a conditional sense as generally accepted in the literature, although not entirely successful. Thus, the assertion that a certain attribute obeys the normal distribution law does not mean at all the existence of any unshakable norms that supposedly underlie the phenomenon, the reflection of which is the attribute in question, and the submission to other distribution laws does not mean some kind of abnormality of this phenomenon.
The main feature of the normal distribution is that it is the limit to which other distributions approach. The normal distribution was first discovered by Moivre in 1733. Only continuous random variables obey the normal law. The density of the normal distribution law has the form .
The mathematical expectation for the normal distribution law is . The dispersion is .
Basic properties of the normal distribution.
1. The distribution density function is defined on the entire real axis Oh , that is, each value X corresponds to a well-defined value of the function.
2. For all values X (both positive and negative) the density function takes positive values, that is, the normal curve is located above the axis Oh .
3. Limit of the density function with an unlimited increase X equals zero, .
4. The density function of the normal distribution at the point has a maximum.
5. The graph of the density function is symmetrical about a straight line.
6. The distribution curve has two inflection points with coordinates and .
7. The mode and median of the normal distribution coincide with the mathematical expectation A .
8. The shape of the normal curve does not change when the parameter is changed A .
9. The coefficients of skewness and kurtosis of the normal distribution are equal to zero.
The importance of calculating these coefficients for empirical distribution series is obvious, since they characterize the skewness and steepness of the given series compared to the normal one.
The probability of falling into the interval is found by the formula , where is an odd tabulated function.
Let's determine the probability that a normally distributed random variable deviates from its mathematical expectation by a value less than , that is, we find the probability of the inequality , or the probability of double inequality . Substituting into the formula, we get
Expressing the deviation of a random variable X in fractions of the standard deviation, that is, putting in the last equality, we get .
Then for , we get
when we get ,
when we receive .
It follows from the last inequality that practically the scattering of a normally distributed random variable lies in the section . The probability that a random variable will not fall into this area is very small, namely, it is equal to 0.0027, that is, this event can occur only in three cases out of 1000. Such events can be considered almost impossible. Based on the above reasoning, three sigma rule, which is formulated as follows: if a random variable has a normal distribution, then the deviation of this value from the mathematical expectation in absolute value does not exceed three times the standard deviation.
Example 28 . A part made by an automatic machine is considered fit if the deviation of its controlled size from the design one does not exceed 10 mm. Random deviations of the controlled size from the design size are subject to the normal distribution law with standard deviation mm and mathematical expectation. What percentage of good parts does the machine produce?
Solution. Consider a random variable X - deviation of the size from the design. The part will be recognized as fit if the random variable belongs to the interval . The probability of manufacturing a suitable part is found by the formula . Therefore, the percentage of good parts produced by the machine is 95.44%.
Binomial distribution
Binomial is the probability distribution of occurrence m number of events in P independent tests, in each of which the probability of occurrence of an event is constant and equal to R . The probability of the possible number of occurrences of an event is calculated by the Bernoulli formula: ,
Where . Permanent P And R , included in this expression, the parameters of the binomial law. The binomial distribution describes the probability distribution of a discrete random variable.
Basic numerical characteristics of the binomial distribution. The mathematical expectation is . The dispersion is . The skewness and kurtosis coefficients are equal to and . With an unlimited increase in the number of trials A And E tend to zero, therefore, we can assume that the binomial distribution converges to the normal one with increasing number of trials.
Example 29 . Independent tests are performed with the same probability of occurrence of the event A in every test. Find the probability of an event occurring A in one trial if the variance in the number of appearances across three trials is 0.63.
Solution. For the binomial distribution . Substitute the values, we get from here or then and .
Poisson distribution
Law of distribution of rare phenomena
The Poisson distribution describes the number of events m , occurring in equal time intervals, provided that the events occur independently of each other with a constant average intensity. At the same time, the number of trials P is large, and the probability of an event occurring in each trial R small. Therefore, the Poisson distribution is called the law of rare phenomena or the simplest flow. The parameter of the Poisson distribution is the value characterizing the intensity of occurrence of events in P tests. Poisson distribution formula.
The Poisson distribution well describes the number of claims for the payment of insurance sums per year, the number of calls received by the telephone exchange in a certain time, the number of element failures during reliability testing, the number of defective products, and so on.
Basic numerical characteristics for the Poisson distribution. The mathematical expectation is equal to the variance and is equal to A . That is . This is a distinctive feature of this distribution. The skewness and kurtosis coefficients are respectively equal to .
Example 30 . The average number of payments of sums insured per day is two. Find the probability that in five days you will have to pay: 1) 6 sums insured; 2) less than six amounts; 3) not less than six.distribution.
This distribution is often observed when studying the service life of various devices, the uptime of individual elements, parts of the system and the system as a whole, when considering random time intervals between the occurrence of two successive rare events.
The density of the exponential distribution is determined by the parameter , which is called failure rate. This term is associated with a specific area of application - the theory of reliability.
The expression for the integral function of the exponential distribution can be found using the properties of the differential function:
Mathematical expectation of the exponential distribution, variance, standard deviation. Thus, it is typical for this distribution that the standard deviation is numerically equal to the mathematical expectation. For any value of the parameter, the skewness and kurtosis coefficients are constant values.
Example 31 . The average operating time of the TV before the first failure is 500 hours. Find the probability that a TV set chosen at random will operate without breakdowns for more than 1000 hours.
Solution. Since the average time to first failure is 500, then . We find the desired probability by the formula .
In many problems related to normally distributed random variables, it is necessary to determine the probability that a random variable , obeying the normal law with parameters , falls into the interval from to . To calculate this probability, we use the general formula
where is the distribution function of the quantity .
Let us find the distribution function of a random variable distributed according to the normal law with parameters . The distribution density of the value is:
.
(6.3.2)
From here we find the distribution function
. (6.3.3)
Let us make the change of variable in the integral (6.3.3)
and bring it to the form:
(6.3.4)
The integral (6.3.4) is not expressed in terms of elementary functions, but it can be calculated in terms of a special function that expresses a definite integral of the expression or (the so-called probability integral), for which tables are compiled. There are many varieties of such functions, for example:
;
etc. Which of these functions to use is a matter of taste. We will choose as such a function
. (6.3.5)
It is easy to see that this function is nothing but the distribution function for a normally distributed random variable with parameters .
We agree to call the function a normal distribution function. The appendix (Table 1) shows tables of function values.
Let us express the distribution function (6.3.3) of the quantity with parameters and in terms of the normal distribution function . Obviously,
.
(6.3.6)
Now let's find the probability of hitting a random variable on the segment from to . According to formula (6.3.1)
Thus, we have expressed the probability that a random variable , distributed according to the normal law with any parameters, will fall on the plot in terms of the standard distribution function , corresponding to the simplest normal law with parameters 0.1. Note that the function arguments in formula (6.3.7) have a very simple meaning: there is a distance from the right end of the section to the center of dispersion, expressed in standard deviations; - the same distance for the left end of the section, and this distance is considered positive if the end is located to the right of the dispersion center, and negative if to the left.
Like any distribution function, the function has the following properties:
3. - non-decreasing function.
In addition, from the symmetry of the normal distribution with parameters about the origin, it follows that
Using this property, in fact, it would be possible to limit the function tables to only positive values of the argument, but in order to avoid an unnecessary operation (subtraction from one), Table 1 of the appendix provides values for both positive and negative arguments.
In practice, one often encounters the problem of calculating the probability that a normally distributed random variable will fall into an area that is symmetrical about the center of dispersion. Consider such a section of length (Fig. 6.3.1). Let us calculate the probability of hitting this site using the formula (6.3.7):
Taking into account the property (6.3.8) of the function and giving the left side of the formula (6.3.9) a more compact form, we obtain a formula for the probability of a random variable distributed according to the normal law falling into a section symmetric with respect to the scattering center:
.
(6.3.10)
Let's solve the following problem. Let us set aside successive segments of length from the scattering center (Fig. 6.3.2) and calculate the probability that a random variable will fall into each of them. Since the curve of the normal law is symmetrical, it is enough to postpone such segments only in one direction.
According to the formula (6.3.7) we find:
(6.3.11)
As can be seen from these data, the probabilities of hitting each of the following segments (fifth, sixth, etc.) with an accuracy of 0.001 are equal to zero.
Rounding the probabilities of hitting the segments to 0.01 (up to 1%), we get three numbers that are easy to remember:
0,34; 0,14; 0,02.
The sum of these three values is 0.5. This means that for a normally distributed random variable, all dispersions (up to fractions of a percent) fit into the section .
This allows, knowing the standard deviation and the mathematical expectation of a random variable, to approximately indicate the range of its practically possible values. This method of estimating the range of possible values of a random variable is known in mathematical statistics as the “rule of three sigma”. The rule of three sigma also implies an approximate method for determining the standard deviation of a random variable: they take the maximum practically possible deviation from the average and divide it by three. Of course, this rough method can only be recommended if there are no other, more accurate ways to determine .
Example 1. A random variable , distributed according to the normal law, is an error in measuring a certain distance. When measuring, a systematic error is allowed in the direction of overestimation by 1.2 (m); the standard deviation of the measurement error is 0.8 (m). Find the probability that the deviation of the measured value from the true value does not exceed 1.6 (m) in absolute value.
Solution. The measurement error is a random variable obeying the normal law with parameters and . We need to find the probability that this quantity falls on the interval from to . By formula (6.3.7) we have:
Using the function tables (Appendix, Table 1), we find:
;
,
Example 2. Find the same probability as in the previous example, but on the condition that there is no systematic error.
Solution. By formula (6.3.10), assuming , we find:
.
Example 3. At a target that looks like a strip (freeway), the width of which is 20 m, shooting is carried out in a direction perpendicular to the freeway. Aiming is carried out along the center line of the highway. The standard deviation in the firing direction is equal to m. There is a systematic error in the firing direction: the undershoot is 3 m. Find the probability of hitting the freeway with one shot.
(real, strictly positive)
Normal distribution, also called Gaussian distribution or Gauss - Laplace- probability distribution , which in the one-dimensional case is given by the probability density function , coinciding with the Gaussian function :
f (x) = 1 σ 2 π e − (x − μ) 2 2 σ 2 , (\displaystyle f(x)=(\frac (1)(\sigma (\sqrt (2\pi ))))\ ;e^(-(\frac ((x-\mu)^(2))(2\sigma ^(2)))),)where the parameter μ is the mathematical expectation (mean value), median and mode of the distribution, and the parameter σ is the standard deviation ( σ ² - variance) 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 mean μ = 0 and standard deviation σ = 1 .
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The importance of the normal distribution in many fields of science (for example, in 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 variables, 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 normal. This law of probability theory has as a consequence the wide 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 dispersions σ 1 2 (\displaystyle \sigma _(1)^(2)) And σ 2 2 (\displaystyle \sigma _(2)^(2)) respectively, then X 1 + X 2 (\displaystyle X_(1)+X_(2)) also has a normal distribution with expected value μ 1 + μ 2 (\displaystyle \mu _(1)+\mu _(2)) and dispersion σ 1 2 + σ 2 2 . (\displaystyle \sigma _(1)^(2)+\sigma _(2)^(2).) This implies 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 Pseudo-Random Variables
The simplest approximate modeling methods are based on the central limit theorem. Namely, if we add several independent identically distributed quantities with a finite variance , then the sum will be distributed approximately Fine. For example, if you add 100 independent standard evenly distributed random variables, then the distribution of the sum will be approximately normal.
For software generation of normally distributed pseudo-random variables, it is preferable to use the Box - Muller transformation. It allows you to generate one normally distributed value based on one uniformly distributed one.
Normal distribution in nature and applications
The normal distribution is often found in nature. For example, the following random variables are well modeled by the normal distribution:
- shooting deflection.
- measurement errors (however, the errors of some measuring instruments have non-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, such as binomial and Poisson. Many non-deterministic physical processes are modeled by this distribution.
Relationship with other distributions
- The normal distribution is a type XI Pearson 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) are independent standard normal random variables), then it will have a Cauchy distribution.
- If z 1 , … , z k (\displaystyle z_(1),\ldots ,z_(k)) are jointly independent standard normal random variables, i.e. 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 a 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
In practice, most random variables, which are affected by a large number of random factors, obey the normal law of probability distribution. Therefore, in various applications of probability theory, this law is of particular importance.
A random variable $X$ obeys the normal probability distribution law if its probability distribution density has the following form
$$f\left(x\right)=((1)\over (\sigma \sqrt(2\pi )))e^(-(((\left(x-a\right))^2)\over ( 2(\sigma )^2)))$$
Schematically, the graph of the function $f\left(x\right)$ is shown in the figure and has the name "Gaussian curve". To the right of this graphic is the German 10 Mark banknote, which was in use even before the introduction of the euro. If you look closely, then on this banknote you can see the Gaussian curve and its discoverer, the greatest mathematician Carl Friedrich Gauss.
Let's go back to our density function $f\left(x\right)$ and give some explanation about the distribution parameters $a,\ (\sigma )^2$. The parameter $a$ characterizes the center of dispersion of the values of the random variable, that is, it has the meaning of the mathematical expectation. When the parameter $a$ changes and the parameter $(\sigma )^2$ remains unchanged, we can observe the shift of the graph of the function $f\left(x\right)$ along the abscissa axis, while the density graph itself does not change its shape.
The parameter $(\sigma )^2$ is the variance and characterizes the shape of the density curve $f\left(x\right)$. When changing the parameter $(\sigma )^2$ with the parameter $a$ unchanged, we can observe how the density graph changes its shape, shrinking or stretching, while not shifting along the abscissa.
Probability of a normally distributed random variable falling into a given interval
As is known, the probability that a random variable $X$ falls into the interval $\left(\alpha ;\ \beta \right)$ can be calculated $P\left(\alpha< X < \beta \right)=\int^{\beta }_{\alpha }{f\left(x\right)dx}$. Для нормального распределения случайной величины $X$ с параметрами $a,\ \sigma $ справедлива следующая формула:
$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right)$$
Here the function $\Phi \left(x\right)=((1)\over (\sqrt(2\pi )))\int^x_0(e^(-t^2/2)dt)$ is the Laplace function . The values of this function are taken from . The following properties of the function $\Phi \left(x\right)$ can be noted.
1 . $\Phi \left(-x\right)=-\Phi \left(x\right)$, i.e. the function $\Phi \left(x\right)$ is odd.
2 . $\Phi \left(x\right)$ is a monotonically increasing function.
3 . $(\mathop(lim)_(x\to +\infty ) \Phi \left(x\right)\ )=0.5$, $(\mathop(lim)_(x\to -\infty ) \ Phi \left(x\right)\ )=-0.5$.
To calculate the values of the $\Phi \left(x\right)$ function, you can also use the $f_x$ function wizard of the Excel package: $\Phi \left(x\right)=NORMDIST\left(x;0;1;1\right )-0.5$. For example, let's calculate the values of the function $\Phi \left(x\right)$ for $x=2$.
The probability that a normally distributed random variable $X\in N\left(a;\ (\sigma )^2\right)$ falls into an interval symmetric with respect to the expectation $a$ can be calculated by the formula
$$P\left(\left|X-a\right|< \delta \right)=2\Phi \left({{\delta }\over {\sigma }}\right).$$
Three sigma rule. It is practically certain that a normally distributed random variable $X$ falls into the interval $\left(a-3\sigma ;a+3\sigma \right)$.
Example 1 . The random variable $X$ is subject to the normal probability distribution law with parameters $a=2,\ \sigma =3$. Find the probability that $X$ falls into the interval $\left(0,5;1\right)$ and the probability that the inequality $\left|X-a\right|< 0,2$.
Using the formula
$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right),$$
find $P\left(0,5;1\right)=\Phi \left(((1-2)\over (3))\right)-\Phi \left(((0,5-2)\ over (3))\right)=\Phi \left(-0.33\right)-\Phi \left(-0.5\right)=\Phi \left(0.5\right)-\Phi \ left(0.33\right)=0.191-0.129=$0.062.
$$P\left(\left|X-a\right|< 0,2\right)=2\Phi \left({{\delta }\over {\sigma }}\right)=2\Phi \left({{0,2}\over {3}}\right)=2\Phi \left(0,07\right)=2\cdot 0,028=0,056.$$
Example 2 . Suppose that during the year the price of shares of a certain company is a random variable distributed according to the normal law with a mathematical expectation equal to 50 conventional monetary units and a standard deviation equal to 10. What is the probability that on a randomly chosen day of the period under discussion, the price for the share will be:
a) more than 70 conventional monetary units?
b) below 50 per share?
c) between 45 and 58 conventional monetary units per share?
Let the random variable $X$ be the price of shares of some company. By condition $X$ is subject to normal distribution with parameters $a=50$ - mathematical expectation, $\sigma =10$ - standard deviation. Probability $P\left(\alpha< X < \beta \right)$ попадания $X$ в интервал $\left(\alpha ,\ \beta \right)$ будем находить по формуле:
$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right).$$
$$a)\ P\left(X>70\right)=\Phi \left(((\infty -50)\over (10))\right)-\Phi \left(((70-50)\ over (10))\right)=0.5-\Phi \left(2\right)=0.5-0.4772=0.0228.$$
$$b)\ P\left(X< 50\right)=\Phi \left({{50-50}\over {10}}\right)-\Phi \left({{-\infty -50}\over {10}}\right)=\Phi \left(0\right)+0,5=0+0,5=0,5.$$
$$c)\ P\left(45< X < 58\right)=\Phi \left({{58-50}\over {10}}\right)-\Phi \left({{45-50}\over {10}}\right)=\Phi \left(0,8\right)-\Phi \left(-0,5\right)=\Phi \left(0,8\right)+\Phi \left(0,5\right)=$$
The law of normal distribution of probabilities of a continuous random variable occupies a special place among various theoretical laws, since it is the main one in many practical studies. He describes most of the random phenomena associated with production processes.
Random phenomena obeying the normal distribution law include measurement errors of production parameters, the distribution of technological manufacturing errors, the height and weight of most biological objects, etc.
normal call the law of probability distribution of a continuous random variable, which is described by a differential function
a - mathematical expectation of a random variable;
The standard deviation of the normal distribution.
The graph of the differential function of the normal distribution is called the normal curve (Gaussian curve) (Fig. 7).
Rice. 7 Gaussian curve
Properties of a normal curve (Gaussian curve):
1. the curve is symmetrical about the straight line x = a;
2. the normal curve is located above the X axis, i.e., for all values of X, the function f(x) is always positive;
3. The ox axis is the horizontal asymptote of the graph, because
4. for x = a, the function f(x) has a maximum equal to
,at points A and B at and the curve has inflection points whose ordinates are equal.
At the same time, the probability that the absolute value of the deviation of a normally distributed random variable from its mathematical expectation will not exceed the standard deviation is equal to 0.6826.
at points E and G, for and , the value of the function f(x) is equal to
and the probability that the absolute value of the deviation of a normally distributed random variable from its mathematical expectation will not exceed twice the standard deviation is 0.9544.
Asymptotically approaching the abscissa axis, the Gaussian curve at points C and D, at and , comes very close to the abscissa axis. At these points, the value of the function f(x) is very small
and the probability that the absolute value of the deviation of a normally distributed random variable from its mathematical expectation will not exceed three times the standard deviation is 0.9973. This property of the Gaussian curve is called " three sigma rule".
If a random variable is normally distributed, then the absolute value of its deviation from the mathematical expectation does not exceed three times the standard deviation.
Changing the value of the parameter a (the mathematical expectation of a random variable) does not change the shape of the normal curve, but only leads to its shift along the X axis: to the right if a increases, and to the left if a decreases.
When a=0, the normal curve is symmetrical about the y-axis.
Changing the value of the parameter (standard deviation) changes the shape of the normal curve: with increasing ordinates of the normal curve decrease, the curve is stretched along the X axis and pressed against it. When decreasing, the ordinates of the normal curve increase, the curve shrinks along the X-axis and becomes more "peaked".
At the same time, for any values of and, the area bounded by the normal curve and the X axis remains equal to one (i.e., the probability that a random variable distributed normally will take on a value bounded by the normal curve on the X axis is equal to 1).
Normal distribution with arbitrary parameters and , i.e., described by a differential function
called general normal distribution.
The normal distribution with parameters and is called normalized distribution(Fig. 8). In a normalized distribution, the differential distribution function is:
Rice. 8 Normalized curve
The integral function of the general normal distribution has the form:
Let a random variable X be distributed according to the normal law in the interval (c, d). Then the probability that X takes a value belonging to the interval (c, d) is equal to
Example. The random variable X is distributed according to the normal law. The mathematical expectation and standard deviation of this random variable are a=30 and . Find the probability that X takes a value in the interval (10, 50).
By condition: . Then
Using ready-made Laplace tables (see Appendix 3), we have.
