In many problems related to normally distributed random variables, it is necessary to determine the probability of a random variable , subject to a normal law with parameters, falling on the segment from to . To calculate this probability we use the general formula

where is the distribution function of the quantity .

Let's find the distribution function of a random variable distributed according to a normal law with parameters. The distribution density of the value is equal to:

From here we find the distribution function

. (6.3.3)

Let us make a change of variable in the integral (6.3.3)

and let's put it in this form:

(6.3.4)

Integral (6.3.4) is not expressed through elementary functions, but it can be calculated through a special function expressing definite integral from the expression or (the so-called probability integral) for which the 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 more than a distribution function for a normally distributed random variable with parameters .

Let us agree to call the function a normal distribution function. The appendix (Table 1) contains tables of function values.

Let us express the distribution function (6.3.3) of the quantity with parameters and through the normal distribution function. Obviously,

Now let's find the probability of a random variable falling on the section from to . According to formula (6.3.1)

Thus, we expressed the probability of a random variable distributed according to a normal law with any parameters getting into a section through the standard distribution function corresponding to the simplest normal law with parameters 0.1. Note that the arguments of the function in formula (6.3.7) have a very simple meaning: there is the distance from the right end of the section to the center of scattering, 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 center of dispersion, 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 relative to the origin, it follows that

Using this property, strictly speaking, it would be possible to limit the function tables to only positive argument values, but in order to avoid an unnecessary operation (subtraction from one), Appendix Table 1 provides values ​​for both positive and negative arguments.

In practice, we often encounter the problem of calculating the probability of a normally distributed random variable falling into an area that is symmetrical with respect to the center of scattering. Let's consider such a section of length (Fig. 6.3.1). Let's calculate the probability of hitting this area using formula (6.3.7):

Taking into account the property (6.3.8) of the function and giving the left side of 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 an area symmetrical with respect to the center of scattering:

. (6.3.10)

Let's solve the following problem. Let us plot successive segments of length from the center of dispersion (Fig. 6.3.2) and calculate the probability of a random variable falling into each of them. Since the normal curve is symmetrical, it is enough to plot such segments only in one direction.

Using 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 getting into segments to 0.01 (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 dispersion (with an accuracy of fractions of a percent) fits within the area .

This allows, knowing the standard deviation and mathematical expectation of a random variable, to roughly 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 called the "three sigma rule". The rule of three sigma also implies an approximate method for determining the standard deviation of a random variable: take the maximum practically possible deviation from the mean and divide it by three. Of course, this rough technique can only be recommended if there are no other, more accurate methods for determining.

Example 1. A random variable distributed according to a normal law represents 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 will not exceed 1.6 (m) in absolute value.

Solution. The measurement error is a random variable subject to the normal law with parameters and . We need to find the probability of this quantity falling on the section from to . According to 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 provided that there is no systematic error.

Solution. Using formula (6.3.10), assuming , we find:

Example 3. A target that looks like a strip (motorway), the width of which is 20 m, is fired in a direction perpendicular to the highway. Aiming is carried out along the center line of the highway. The standard deviation in the shooting direction is equal to m. There is a systematic error in the shooting direction: the undershoot is 3 m. Find the probability of hitting a highway with one shot.

The most famous and frequently used law in probability theory is the normal distribution law or Gauss's law .

main feature the normal distribution law is that it is ultimate law for other distribution laws.

Note that for a normal distribution the integral function has the form:

.

Let us now show that the probabilistic meaning of the parameters is as follows: A is the mathematical expectation, - the standard deviation (i.e.) of the normal distribution:

a) by definition of the mathematical expectation of a continuous random variable, we have

Really

,

since under the integral sign there is odd function, and the limits of integration are symmetrical about the origin;

- Poisson integral .

So, the mathematical expectation of a normal distribution is equal to the parameter A .

b) by definition of the variance of a continuous random variable and, taking into account that , we can write

.

Integrating by parts, putting , let's find

Hence .

So, the standard deviation of the normal distribution is equal to the parameter.

In case and normal distribution called the normalized (or standard normal) distribution. Then, obviously, the normalized density (differential) and the normalized integral distribution function will be written respectively in the form:

(The function, as you know, is called the Laplace function (see LECTURE 5) or the probability integral. Both functions, that is , tabulated and their values ​​are recorded in the corresponding tables).

Properties of normal distribution (properties of normal curve):

1. Obviously, a function on the entire number line.

2. , that is, the normal curve is located above the axis Oh .

3. , that is, the axis Oh serves as the horizontal asymptote of the graph.

4. A normal curve is symmetrical about a straight line x = a (accordingly, the graph of the function is symmetrical about the axis OU ).

Therefore, we can write: .

5. .

6. It is easy to show that the points And are inflection points of the normal curve (prove it yourself).

7.It's obvious that

but since , That . Besides , therefore, all odd moments are equal to zero.

For even moments we can write:

8. .

9. .

10. , Where .

11. For negative values ​​of the random variable: , where .

13. The probability of a random variable falling into a section symmetrical with respect to the center of the distribution is equal to:

EXAMPLE 3. Show that a normally distributed random variable X deviates from mathematical expectation M(X) no more than .

Solution. For normal distribution: .

In other words, the probability that the absolute value of the deviation will exceed triple the standard deviation is very small, namely equal to 0.0027. This means that only in 0.27% of cases this can happen. Such events, based on the principle of the impossibility of unlikely events, can be considered practically impossible.

So, an event with a probability of 0.9973 can be considered practically reliable, that is, the random variable deviates from the mathematical expectation by no more than .

EXAMPLE 4. Knowing the characteristics of the normal distribution of a random variable X - tensile strength of steel: kg/mm ​​2 and kg/mm ​​2, find the probability of obtaining steel with a tensile strength from 31 kg/mm ​​2 to 35 kg/mm ​​2.

Solution.

3. Exponential distribution (exponential distribution law)

Exponential is the probability distribution of a continuous random variable. X , which is described by a differential function (distribution density)

where is a constant positive value.

The exponential distribution is defined one parameter. This feature of the exponential distribution indicates its advantage compared to distributions that depend on a larger number of parameters. Usually the parameters are unknown and their estimates (approximate values) have to be found; Of course, it is easier to evaluate one parameter than two, or three, etc.

It is easy to write the integral exponential distribution function:

We defined the exponential distribution using a differential function; it is clear that it can be determined using the integral function.

Comment: Consider a continuous random variable T - length of time of non-failure operation of the product. Its accepted values ​​are denoted by t , . Cumulative distribution function defines probability of failure products over a period of time t . Consequently, the probability of failure-free operation during the same time, duration t , that is, the probability of the opposite event is equal to

Normal distribution ( normal distribution) - plays an important role in data analysis.

Sometimes instead of the term normal distribution use the term Gaussian distribution in honor of K. Gauss (older terms that are practically not used nowadays: Gauss’s law, Gauss-Laplace distribution).

Univariate normal distribution

A normal distribution has a density::

In this formula, the fixed parameters are average, - standard deviation.

Density graphs for various parameters are given.

The characteristic function of the normal distribution has the form:

Differentiating the characteristic function and setting t = 0, we obtain moments of any order.

The normal distribution density curve is symmetrical with respect to and has a single maximum at this point, equal to

The standard deviation parameter varies from 0 to ∞.

Average varies from -∞ to +∞.

As the parameter increases, the curve spreads along the axis X, as it approaches 0, it shrinks around the average value (the parameter characterizes the spread, scattering).

When it changes the curve shifts along the axis X(see graphs).

By varying the parameters and , we obtain various models random variables, arising in telephony.

A typical application of the normal law in the analysis of, for example, telecommunications data is modeling signals, describing noise, interference, errors, and traffic.

Univariate Normal Distribution Plots

Figure 1. Density plot of normal distribution: mean is 0, standard deviation is 1

Figure 2. Density plot of the standard normal distribution with regions containing 68% and 95% of all observations

Figure 3. Density graphs of normal distributions with zero mean and different deviations (=0.5, =1, =2)

Figure 4 Graphs of two normal distributions N(-2,2) and N(3,2).

Notice that the center of the distribution has shifted when changing the parameter.

Comment

In a programme STATISTICA The designation N(3,2) refers to the normal or Gaussian law with the parameters: mean = 3 and standard deviation =2.

In the literature, sometimes the second parameter is interpreted as dispersion, i.e. square standard deviation.

Calculating Normal Distribution Percentage Points Using a Probability Calculator STATISTICA

Using a probability calculator STATISTICA You can calculate various characteristics of distributions without resorting to the cumbersome tables used in old books.

Step 1. Let's launch Analysis / Probability calculator / Distributions.

In the distribution section, select normal.

Figure 5. Launching the probability distribution calculator

Step 2. We indicate the parameters that interest us.

For example, we want to calculate the 95% quantile of a normal distribution with a mean of 0 and a standard deviation of 1.

Let's indicate these parameters in the fields of the calculator (see the calculator fields mean and standard deviation).

Let's enter the parameter p=0.95.

Checkbox “Reverse f.r.” will appear automatically. Check the “Schedule” box.

Click the “Calculate” button in the upper right corner.

Figure 6. Setting parameters

Step 3. In the Z field we get the result: the quantile value is 1.64 (see the next window).

Figure 7. Viewing the result of the calculator

Figure 8. Density plots and distribution functions. Straight line x=1.644485

Figure 9. Graphs of the normal distribution function. Vertical dotted lines - x=-1.5, x=-1, x=-0.5, x=0

Figure 10. Graphs of the normal distribution function. Vertical dotted lines - x=0.5, x=1, x=1.5, x=2

Estimation of normal distribution parameters

Normal distribution values ​​can be calculated using interactive calculator.

Bivariate normal distribution

The one-dimensional normal distribution naturally generalizes to two-dimensional normal distribution.

For example, if you consider a signal at only one point, then a one-dimensional distribution is enough for you, at two points - two-dimensional, at three points - three-dimensional, etc.

The general formula for the bivariate normal distribution is:

Where is the pairwise correlation between X 1 And X 2;

X 1 respectively;

Mean and standard deviation of a variable X 2 respectively.

If random variables X 1 And X 2 are independent, then the correlation is 0, = 0, respectively, the middle term in the exponent vanishes, and we have:

f(x 1 ,x 2) = f(x 1)*f(x 2)

For independent quantities, the two-dimensional density decomposes into the product of two one-dimensional densities.

Density plots of bivariate normal distributions

Figure 11. Density plot of a bivariate normal distribution (zero vector of means, unit covariance matrix)

Figure 12. Section of the density graph of a two-dimensional normal distribution with a plane z=0.05

Figure 13. Density plot of a two-dimensional normal distribution (zero vector of expected value, covariance matrix with 1 on the main diagonal and 0.5 on the side diagonal)

Figure 14. Section of the density graph of a two-dimensional normal distribution (zero vector of mathematical expectation, covariance matrix with 1 on the main diagonal and 0.5 on the side diagonal) by plane z= 0.05

Figure 15. Density plot of a two-dimensional normal distribution (zero vector of expected value, covariance matrix with 1 on the main diagonal and -0.5 on the side diagonal)

Figure 16. Section of the density graph of a two-dimensional normal distribution (zero vector of mathematical expectation, covariance matrix with 1 on the main diagonal and -0.5 on the side diagonal) by plane z=0.05

Figure 17. Sections of density graphs of a two-dimensional normal distribution with a plane z=0.05

To better understand the bivariate normal distribution, try solving the following problem.

Task. Look at the graph of the bivariate normal distribution. Think about it, can it be represented as a rotation of the graph of a one-dimensional normal distribution? When should you use the deformation technique?

In practice, most random variables affected by a large number of random factors are subject to the normal probability distribution law. Therefore, in various applications of probability theory, this law is of particular importance.

The random variable $X$ obeys the normal probability distribution law if its probability distribution density has the following form

$$f\left(x\right)=((1)\over (\sigma \sqrt(2\pi )))e^(-(((\left(x-a\right))^2)\over ( 2(\sigma )^2)))$$

The graph of the function $f\left(x\right)$ is shown schematically in the figure and is called “Gaussian curve”. To the right of this graph is the German 10 mark banknote, which was used before the introduction of the euro. If you look closely, you can see on this banknote the Gaussian curve and its discoverer, the greatest mathematician Carl Friedrich Gauss.

Let's return to our density function $f\left(x\right)$ and give some explanations regarding the distribution parameters $a,\ (\sigma )^2$. The parameter $a$ characterizes the center of dispersion of the values ​​of a random variable, that is, it has the meaning of a mathematical expectation. When the parameter $a$ changes and the parameter $(\sigma )^2$ remains unchanged, we can observe a shift in the graph of the function $f\left(x\right)$ along the abscissa, while the density graph itself does not change its shape.

The parameter $(\sigma )^2$ is the variance and characterizes the shape of the density graph 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, compressing or stretching, without moving along the abscissa axis.

Probability of a normally distributed random variable falling into a given interval

As is known, the probability of a random variable $X$ falling 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 . $\Phi \left(-x\right)=-\Phi \left(x\right)$, that is, 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 function $\Phi \left(x\right)$, you can also use the function $f_x$ wizard in Excel: $\Phi \left(x\right)=NORMDIST\left(x;0;1;1\right )-0.5$. For example, let's calculate the values ​​of the function $\Phi \left(x\right)$ for $x=2$.

The probability of a normally distributed random variable $X\in N\left(a;\ (\sigma )^2\right)$ falling into an interval symmetric with respect to the mathematical expectation $a$ can be calculated using the formula

$$P\left(\left|X-a\right|< \delta \right)=2\Phi \left({{\delta }\over {\sigma }}\right).$$

Three sigma rule. It is almost certain that a normally distributed random variable $X$ will fall 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 of $X$ falling into the interval $\left(0.5;1\right)$ and the probability of satisfying the inequality $\left|X-a\right|< 0,2$.

Using formula

$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right),$$

we 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 selected day of the period under discussion the price for the promotion 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 a 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:

$$P\left(\alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right).$$

$$а)\ P\left(X>70\right)=\Phi \left(((\infty -50)\over (10))\right)-\Phi \left(((70-50)\ over (10))\right)=0.5-\Phi \left(2\right)=0.5-0.4772=0.0228.$$

$$b)\P\left(X< 50\right)=\Phi \left({{50-50}\over {10}}\right)-\Phi \left({{-\infty -50}\over {10}}\right)=\Phi \left(0\right)+0,5=0+0,5=0,5.$$

$$in)\ P\left(45< X < 58\right)=\Phi \left({{58-50}\over {10}}\right)-\Phi \left({{45-50}\over {10}}\right)=\Phi \left(0,8\right)-\Phi \left(-0,5\right)=\Phi \left(0,8\right)+\Phi \left(0,5\right)=$$

Normal probability distribution law

Without exaggeration, it can be called a philosophical law. Observing various objects and processes in the world around us, we often come across the fact that something is not enough, and that there is a norm:


Here is a basic view density functions normal probability distribution, and I welcome you to this interesting lesson.

What examples can you give? There are simply darkness of them. This is, for example, the height, weight of people (and not only), their physical strength, mental abilities, etc. There is a "main mass" (for one reason or another) and there are deviations in both directions.

These are different characteristics of inanimate objects (same size, weight). This is a random duration of processes, for example, the time of a hundred-meter race or the transformation of resin into amber. From physics, I remembered air molecules: some of them are slow, some are fast, but most move at “standard” speeds.

Next, we deviate from the center by one more standard deviation and calculate the height:

Marking points on the drawing (green color) and we see that this is quite enough.

At the final stage, we carefully draw a graph, and especially carefully reflect it convex/concave! Well, you probably realized a long time ago that the x-axis is horizontal asymptote, and it is absolutely forbidden to “climb” behind it!

When filing a solution electronically, it’s easy to create a graph in Excel, and unexpectedly for myself, I even recorded a short video on this topic. But first, let's talk about how the shape of the normal curve changes depending on the values ​​of and.

When increasing or decreasing "a" (with constant “sigma”) the graph retains its shape and moves right/left respectively. So, for example, when the function takes the form and our graph “moves” 3 units to the left - exactly to the origin of coordinates:


A normally distributed quantity with zero mathematical expectation received a completely natural name - centered; its density function is even, and the graph is symmetrical about the ordinate.

In case of change of "sigma" (with constant “a”), the graph “stays the same” but changes shape. When enlarged, it becomes lower and elongated, like an octopus stretching its tentacles. And, conversely, when decreasing the graph becomes narrower and taller- it turns out to be a “surprised octopus”. Yes, when decrease“sigma” twice: the previous graph narrows and stretches up twice:

Everything is in full accordance with geometric transformations of graphs.

A normal distribution with a unit sigma value is called normalized, and if it is also centered(our case), then such a distribution is called standard. It has even more simple function density, which has already been encountered in Laplace's local theorem: . The standard distribution has found wide application in practice, and very soon we will finally understand its purpose.

Well, now let's watch the movie:

Yes, absolutely right - somehow undeservedly it remained in the shadows probability distribution function. Let's remember her definition:
– the probability that a random variable will take a value LESS than the variable that “runs through” all real values ​​to “plus” infinity.

Inside the integral, a different letter is usually used so that there are no “overlaps” with the notation, because here each value is associated with improper integral, which is equal to some number from the interval .

Almost all values ​​cannot be calculated accurately, but as we have just seen, with modern computing power this is not difficult. Thus, for the standard distribution function, the corresponding Excel function generally contains one argument:

=NORMSDIST(z)

One, two - and you're done:

The drawing clearly shows the implementation of all distribution function properties, and from the technical nuances here you should pay attention to horizontal asymptotes and the inflection point.

Now let's remember one of the key tasks of the topic, namely, find out how to find the probability that a normal random variable will take the value from the interval. Geometrically, this probability is equal to area between the normal curve and the x-axis in the corresponding section:

but every time I try to get an approximate value is unreasonable, and therefore it is more rational to use "easy" formula:
.

! Also remembers , What

Here you can use Excel again, but there are a couple of significant “buts”: firstly, it is not always at hand, and secondly, “ready-made” values ​​will most likely raise questions from the teacher. Why?

I have talked about this many times before: at one time (and not very long ago) a regular calculator was a luxury, and in educational literature The “manual” method of solving the problem under consideration is still preserved. Its essence is to standardize values ​​“alpha” and “beta”, that is, reduce the solution to the standard distribution:

Note : the function is easy to obtain from the general caseusing linear replacements. Then also:

and from the replacement carried out exactly follows the formula for the transition from the values ​​of an arbitrary distribution to the corresponding values ​​of the standard distribution.

Why is this necessary? The fact is that the values ​​were meticulously calculated by our ancestors and compiled into a special table, which is in many books on terwer. But even more often there is a table of values, which we have already dealt with in Laplace's integral theorem:

If we have at our disposal a table of values ​​of the Laplace function , then we solve through it:

Fractional values ​​are traditionally rounded to 4 decimal places, as is done in the standard table. And for control there is Point 5 layout.

I remind you that, and to avoid confusion always control, a table of WHAT function is in front of your eyes.

Answer is required to be given as a percentage, so the calculated probability must be multiplied by 100 and the result provided with a meaningful comment:

– with a flight from 5 to 70 m, approximately 15.87% of shells will fall

We train on our own:

Example 3

The diameter of factory-made bearings is a random variable, normally distributed with a mathematical expectation of 1.5 cm and a standard deviation of 0.04 cm. Find the probability that the size of a randomly selected bearing ranges from 1.4 to 1.6 cm.

In the sample solution and below, I will use the Laplace function as the most common option. By the way, note that according to the wording, the ends of the interval can be included in the consideration here. However, this is not critical.

And already in this example we encountered a special case - when the interval is symmetrical with respect to the mathematical expectation. In such a situation, it can be written in the form and, using the oddity of the Laplace function, simplify the working formula:

The delta parameter is called deviation from the mathematical expectation, and the double inequality can be “packaged” using module:

– the probability that the value of a random variable will deviate from the mathematical expectation by less than .

It’s good that the solution fits in one line :)
– the probability that the diameter of a randomly taken bearing differs from 1.5 cm by no more than 0.1 cm.

The result of this task turned out to be close to unity, but I would like even greater reliability - namely, to find out the boundaries within which the diameter is located almost everyone bearings. Is there any criterion for this? Exists! The question posed is answered by the so-called

three sigma rule

Its essence is that practically reliable is the fact that a normally distributed random variable will take a value from the interval .

Indeed, the probability of deviation from the expected value is less than:
or 99.73%

In terms of bearings, these are 9973 pieces with a diameter from 1.38 to 1.62 cm and only 27 “substandard” copies.

IN practical research The three sigma rule is usually applied in the opposite direction: if statistically It was found that almost all values random variable under study fall within an interval of 6 standard deviations, then there are compelling reasons to believe that this value is distributed according to a normal law. Verification is carried out using theory statistical hypotheses.

We continue to solve the harsh Soviet problems:

Example 4

The random value of the weighing error is distributed according to the normal law with zero mathematical expectation and a standard deviation of 3 grams. Find the probability that the next weighing will be carried out with an error not exceeding 5 grams in absolute value.

Solution very simple. By condition, we immediately note that at the next weighing (something or someone) we will almost 100% get the result with an accuracy of 9 grams. But the problem involves a narrower deviation and according to the formula:

– the probability that the next weighing will be carried out with an error not exceeding 5 grams.

Answer:

The solved problem is fundamentally different from a seemingly similar one. Example 3 lesson about uniform distribution. There was an error rounding measurement results, here we are talking about the random error of the measurements themselves. Such errors arise due to the technical characteristics of the device itself. (the range of acceptable errors is usually indicated in his passport), and also through the fault of the experimenter - when we, for example, “by eye” take readings from the needle of the same scales.

Among others, there are also so-called systematic measurement errors. It's already non-random errors that occur due to incorrect setup or operation of the device. For example, unregulated floor scales can steadily “add” kilograms, and the seller systematically weighs down customers. Or it can be calculated not systematically. However, in any case, such an error will not be random, and its expectation is different from zero.

…I’m urgently developing a sales training course =)

Let’s solve the inverse problem ourselves:

Example 5

The diameter of the roller is a random normally distributed random variable, its standard deviation is equal to mm. Find the length of the interval, symmetrical with respect to the mathematical expectation, into which the length of the roller diameter is likely to fall.

Point 5* design layout to help. Please note that the mathematical expectation is not known here, but this does not in the least prevent us from solving the problem.

AND exam task, which I highly recommend for consolidating the material:

Example 6

A normally distributed random variable is specified by its parameters (mathematical expectation) and (standard deviation). Required:

a) write down the probability density and schematically depict its graph;
b) find the probability that it will take a value from the interval ;
c) find the probability that the absolute value will deviate from no more than ;
d) using the “three sigma” rule, find the values ​​of the random variable.

Such problems are offered everywhere, and over the years of practice I have solved hundreds and hundreds of them. Be sure to practice drawing a drawing by hand and using paper tables;)

Well, I’ll look at an example of increased complexity:

Example 7

The probability distribution density of a random variable has the form . Find, mathematical expectation, variance, distribution function, build density graphs and distribution functions, find.

Solution: First of all, let us note that the condition does not say anything about the nature of the random variable. The presence of an exponent in itself does not mean anything: it may turn out, for example, indicative or even arbitrary continuous distribution. And therefore the “normality” of the distribution still needs to be justified:

Since the function determined at any real value , and it can be reduced to the form , then the random variable is distributed according to the normal law.

Here we go. For this select a complete square and organize three-story fraction:


Be sure to perform a check, returning the indicator to its original form:

, which is what we wanted to see.

Thus:
- By rule of operations with powers"pinch off" And here you can immediately write down the obvious numerical characteristics:

Now let's find the value of the parameter. Since the normal distribution multiplier has the form and , then:
, from where we express and substitute into our function:
, after which we will once again go through the recording with our eyes and make sure that the resulting function has the form .

Let's build a density graph:

and distribution function graph :

If you don’t have Excel or even a regular calculator at hand, then the last graph can easily be built manually! At a point, the distribution function takes on a value and is found here