Find the normal distribution of a random variable. Normal distribution of continuous random variables

Normal law of probability distribution

Without exaggeration, it can be called a philosophical law. Observing various objects and processes of the world around us, we often encounter 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 most interesting lesson.

What examples can be given? They are just darkness. This, for example, is the height, weight of people (and not only), their physical strength, mental abilities, etc. There is a "mass" (in one way or another) and there are deviations in both directions.

These are different characteristics of inanimate objects (the same dimensions, 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, air molecules came to mind: among them there are slow ones, there are fast ones, but most of them 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 convexity / concavity! Well, you probably realized a long time ago that the abscissa axis is horizontal asymptote, and it is absolutely impossible to “climb” for it!

With the electronic design of the solution, the graph is easy to build 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 unchanged "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:

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

In the event of a change in "sigma" (with constant "a"), the graph "remains in place", but changes shape. When enlarged, it becomes lower and elongated, like an octopus stretching its tentacles. And vice versa, when decreasing the graph becomes narrower and taller- it turns out "surprised octopus." Yes, at decrease"sigma" two times: the previous chart narrows and stretches up twice:

Everything is in full accordance with geometric transformations of graphs.

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

Now let's watch a movie:

Yes, quite right - somehow undeservedly we have remained in the shadows probability distribution function. We remember her definition:
- the probability that a random variable will take a value LESS than the variable , which "runs" all real values \u200b\u200bto "plus" infinity.

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

Almost all values cannot be calculated exactly, but as we have just seen, with modern computing power, this is not difficult. So, for the function of the standard distribution, 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 an inflection point.

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

but each time grind out 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, most likely, will raise questions from the teacher. Why?

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

Note : the function is easy to obtain from the general caseusing a linear substitutions. Then and:

and from the replacement just follows the formula transition from the values of an arbitrary distribution to the corresponding values of the standard distribution.

Why is this needed? The fact is that the values were scrupulously calculated by our ancestors and summarized in a special table, which is in many books on terver. But even more common is the table of values, which we have already dealt with in Laplace 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 Item 5 layout.

I remind you that , and to avoid confusion always be in control, table of WHAT function before your eyes.

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

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

We train on our own:

Example 3

The diameter of bearings manufactured at the factory is a random variable normally distributed with an expectation of 1.5 cm and a standard deviation of 0.04 cm. Find the probability that the size of a randomly taken 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, here you can include the ends of the interval in the consideration. However, this is not critical.

And already in this example, we met 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 oddness of the Laplace function, simplify the working formula:

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

is the probability that the value of a random variable deviates from the mathematical expectation by less than .

Well, the solution that fits in one line :)
is the probability that the diameter of a bearing taken at random 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 more reliability - namely, to find out the boundaries in which the diameter is almost everyone bearings. Is there any criterion for this? Exist! The question 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 expectation is less than:
or 99.73%

In terms of "bearings" - these are 9973 pieces with a diameter of 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 found that almost all values random variable under study fit into an interval of 6 standard deviations, then there are good reasons to believe that this value is distributed according to the normal law. Verification is carried out using the theory statistical hypotheses.

We continue to solve the harsh Soviet tasks:

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.

Decision very simple. By the condition, and 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 in the problem there is 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:

A 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 permissible errors, as a rule, is indicated in his passport), and also through the fault of the experimenter - when, for example, "by eye" we take readings from the arrow of the same scales.

Among others, there are also so-called systematic measurement errors. It's already nonrandom errors that occur due to incorrect setup or operation of the device. So, for example, unadjusted floor scales can consistently "add" a kilogram, and the seller systematically underweight buyers. Or not systematically because you can shortchange. However, in any case, such an error will not be random, and its expectation is different from zero.

…I am urgently developing a sales training course =)

Let's solve the problem on our own:

Example 5

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

Item 5* design layout to help. Please note that the mathematical expectation is not known here, but this does not in the least interfere with solving the problem.

And the exam task, which I highly recommend to consolidate the material:

Example 6

A normally distributed random variable is given 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 modulo deviates from no more than ;
d) applying the rule of "three sigma", find the values of the random variable .

Such problems are offered everywhere, and over the years of practice I have been able to solve hundreds and hundreds of them. Be sure to practice hand drawing and using paper spreadsheets ;)

Well, I will analyze an example of increased complexity:

Example 7

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

Decision: first of all, let's pay attention that the condition does not say anything about the nature of the random variable. By itself, the presence of the exhibitor does not mean anything: it can be, for example, demonstrative or generally arbitrary continuous distribution. And therefore, the “normality” of the distribution still needs to be substantiated:

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.

We present. For this select a full 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:
- on power rule"pinching 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 which we express and substitute into our function:
, after which we will once again go over the record with our eyes and make sure that the resulting function has the form .

Let's plot the density:

and the plot of the distribution function :

If there is no Excel and even a regular calculator at hand, then the last chart is easily built manually! At the point, the distribution function takes on the value and here is

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

The normal distribution is also called Gauss law.

The normal distribution law is central to the theory of probability. This is due to the fact that this law manifests itself in all cases when a random variable is the result 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 the standard deviation of the random variable X.

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

The normal distribution density plot is called normal curve or Gaussian curve.

A normal curve has the following properties:

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

2) For all X the distribution function takes only positive values.

3) The OX axis is the horizontal asymptote of the probability density graph, since with an 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 function has a maximum equal to
.

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

(x - a) enters the squared distribution density function.

6) To find the inflection points of the graph, we 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.

At these points, the value of the function is
.

Let's build a graph of 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 a> 0, then the graph will shift in the positive direction if a < 0 – в отрицательном.

At a= 0 and  = 1 the curve is called normalized. Normalized Curve Equation:

Laplace function

Find the probability that a random variable distributed according to the normal law falls into a given interval.

Denote

Because integral
is not expressed in terms of elementary functions, then the function

which is called Laplace function or probability integral.

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

On fig. 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 the Laplace function, which is related to the Laplace function by the relation:

On fig. 7 shows a plot of the normalized Laplace function.

P three sigma rule

When considering the normal distribution, an important special case is distinguished, known as three sigma rule.

Let's 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 accept  = 3, then we obtain using the tables of values of the Laplace function:

Those. the probability that a random variable deviates from its mathematical expectation by an amount greater than three times the standard deviation is practically zero.

This rule is called three sigma rule.

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

Lecture conclusion:

In the lecture, we considered the laws of distribution of continuous quantities. In preparation for the next lecture and practical exercises, you should independently supplement your lecture notes with an in-depth study of the recommended literature and solving the proposed problems.

Brief theory

Normal is the probability distribution of a continuous random variable , whose density has the form:

where is the mathematical expectation , is the standard deviation .

The probability that it will take a value belonging to the interval:

where is the Laplace function:

The probability that the absolute value of the deviation is less than a positive number:

In particular, for , the following equality holds:

When solving problems that practice puts forward, one has to deal with various distributions of continuous random variables.

In addition to the normal distribution, the main distribution laws for continuous random variables are:

Problem solution example

The part is made on the machine. Its length is a random variable distributed according to the normal law with parameters , . Find the probability that the length of the part will be between 22 and 24.2 cm. What deviation of the length of the part from can be guaranteed with a probability of 0.92; 0.98? Within what limits, symmetrical with respect to , will practically all dimensions of the parts lie?

join the VK group.

Decision:

The probability that a random variable distributed according to the normal law will be in the interval:

We get:

The probability that a random variable, distributed according to the normal law, deviates from the mean by no more than :

By condition

If you do not need help now, but may need it in the future, then in order not to lose contact,

There will also be tasks for an independent solution, to which you can see the answers.

Normal distribution: theoretical foundations

Examples of random variables distributed according to the normal law are the height of a person, the mass of the caught fish of the same species. Normal distribution means the following : there are values of human height, the mass of fish of the same species, which are intuitively perceived as "normal" (and in fact - averaged), and they are much more common in a sufficiently large sample than those that differ up or down.

The normal probability distribution of a continuous random variable (sometimes the Gaussian distribution) can be called bell-shaped due to the fact that the density function of this distribution, which is symmetric about the mean, is very similar to the cut of a bell (red curve in the figure above).

The probability of meeting certain values in the sample is equal to the area of \u200b\u200bthe figure under the curve, and in the case of a normal distribution, we see that under the top of the "bell", which corresponds to values tending to the average, the area, and therefore the probability, is greater than under the edges. Thus, we get the same thing that has already been said: the probability of meeting a person of "normal" height, catching a fish of "normal" weight is higher than for values that differ up or down. In very many cases of practice, measurement errors are distributed according to a law close to normal.

Let's stop again at the figure at the beginning of the lesson, which shows the density function of the normal distribution. The graph of this function was obtained by calculating some data sample in the software package STATISTICS. On it, the histogram columns represent intervals of sample values whose distribution is close (or, as they say in statistics, do not differ significantly from) to the normal distribution density function graph itself, which is a red curve. The graph shows that this curve is indeed bell-shaped.

The normal distribution is valuable in many ways because knowing only the mean of a continuous random variable and the standard deviation, you can calculate any probability associated with that variable.

The normal distribution has the added benefit of being one of the easiest to use statistical criteria used to test statistical hypotheses - Student's t-test- can be used only in the case when the sample data obey the normal distribution law.

The density function of the normal distribution of a continuous random variable can be found using the formula:

where x- value of the variable, - mean value, - standard deviation, e\u003d 2.71828 ... - the base of the natural logarithm, \u003d 3.1416 ...

Properties of the normal distribution density function

Changes in the mean move the bell curve in the direction of the axis Ox. If it increases, the curve moves to the right, if it decreases, then to the left.

If the standard deviation changes, then the height of the curve vertex changes. When the standard deviation increases, the top of the curve is higher, when it decreases, it is lower.

The probability that the value of a normally distributed random variable will fall within a given interval

Already in this paragraph, we will begin to solve practical problems, the meaning of which is indicated in the title. Let us analyze what possibilities the theory provides for solving problems. The starting concept for calculating the probability of a normally distributed random variable falling into a given interval is the integral function of the normal distribution.

Integral normal distribution function:

However, it is problematic to obtain tables for every possible combination of mean and standard deviation. Therefore, one of the simple ways to calculate the probability of a normally distributed random variable falling into a given interval is to use probability tables for a standardized normal distribution.

A normal distribution is called a standardized or normalized distribution., whose mean value is , and the standard deviation is .

Density function of the standardized normal distribution:

Cumulative function of the standardized normal distribution:

The figure below shows the integral function of the standardized normal distribution, the graph of which was obtained by calculating some data sample in the software package STATISTICS. The graph itself is a red curve, and the sample values are approaching it.

To enlarge the picture, you can click on it with the left mouse button.

Standardizing a random variable means moving from the original units used in the task to standardized units. Standardization is performed according to the formula

In practice, all possible values of a random variable are often not known, so the values of the mean and standard deviation cannot be accurately determined. They are replaced by the arithmetic mean of the observations and the standard deviation s. Value z expresses the deviations of the values of a random variable from the arithmetic mean when measuring standard deviations.

Open interval

The probability table for the standardized normal distribution, which is available in almost any book on statistics, contains the probabilities that a random variable having a standard normal distribution Z takes on a value less than a certain number z. That is, it will fall into the open interval from minus infinity to z. For example, the probability that the value Z less than 1.5 is equal to 0.93319.

Example 1 The company manufactures parts that have a normally distributed lifetime with a mean of 1000 and a standard deviation of 200 hours.

For a randomly selected part, calculate the probability that its service life will be at least 900 hours.

Decision. Let's introduce the first notation:

The desired probability.

The values of the random variable are in the open interval. But we can calculate the probability that a random variable will take a value less than a given value, and according to the condition of the problem, it is required to find an equal or greater value than a given one. This is the other part of the space under the bell curve. Therefore, in order to find the desired probability, it is necessary to subtract from one the mentioned probability that the random variable will take a value less than the specified 900:

Now the random variable needs to be standardized.

We continue to introduce the notation:

z = (X ≤ 900) ;

x= 900 - given value of a random variable;

μ = 1000 - average value;

σ = 200 - standard deviation.

Based on these data, we obtain the conditions of the problem:

According to the tables of a standardized random variable (interval boundary) z= −0.5 corresponds to the probability 0.30854. Subtract it from unity and get what is required in the condition of the problem:

So, the probability that the life of the part will be at least 900 hours is 69%.

This probability can be obtained using the MS Excel function NORM.DIST (the value of the integral value is 1):

P(X≥900) = 1 - P(X≤900) = 1 - NORM.DIST(900; 1000; 200; 1) = 1 - 0.3085 = 0.6915.

About calculations in MS Excel - in one of the subsequent paragraphs of this lesson.

Example 2 In a certain city, the average annual family income is a normally distributed random variable with a mean value of 300,000 and a standard deviation of 50,000. It is known that the income of 40% of families is less than the value A. Find value A.

Decision. In this problem, 40% is nothing more than the probability that a random variable will take a value from an open interval that is less than a certain value, indicated by the letter A.

To find the value A, we first compose the integral function:

According to the task

μ = 300000 - average value;

σ = 50000 - standard deviation;

x = A is the value to be found.

Making up equality

According to the statistical tables, we find that the probability of 0.40 corresponds to the value of the interval boundary z = −0,25 .

Therefore, we make the equality

and find its solution:

A = 287300 .

Answer: income of 40% of families is less than 287300.

Closed interval

In many problems, it is required to find the probability that a normally distributed random variable takes a value in the interval from z 1 to z 2. That is, it will fall into the closed interval. To solve such problems, it is necessary to find in the table the probabilities corresponding to the boundaries of the interval, and then find the difference between these probabilities. This requires subtracting the smaller value from the larger one. Examples for solving these common problems are as follows, and it is proposed to solve them yourself, and then you can see the correct solutions and answers.

Example 3 The profit of an enterprise for a certain period is a random variable subject to the normal distribution law with an average value of 0.5 million c.u. and a standard deviation of 0.354. Determine, with an accuracy of two decimal places, the probability that the profit of the enterprise will be from 0.4 to 0.6 c.u.

Example 4 The length of the manufactured part is a random variable distributed according to the normal law with parameters μ =10 and σ =0.071 . Find, with an accuracy of two decimal places, the probability of marriage if the allowable dimensions of the part should be 10 ± 0.05.

Hint: in this problem, in addition to finding the probability of a random variable falling into a closed interval (the probability of obtaining a non-defective part), one more action is required.

allows you to determine the probability that the standardized value Z not less -z and no more +z, where z- an arbitrarily chosen value of a standardized random variable.

An Approximate Method for Checking the Normality of a Distribution

An approximate method for checking the normality of the distribution of sample values is based on the following property of a normal distribution: skewness β 1 and kurtosis coefficient β 2 zero.

Asymmetry coefficient β 1 numerically characterizes the symmetry of the empirical distribution with respect to the mean. If the skewness is equal to zero, then the arithmetric mean, median and mode are equal: and the distribution density curve is symmetrical about the mean. If the coefficient of asymmetry is less than zero (β 1 < 0 ), then the arithmetic mean is less than the median, and the median, in turn, is less than the mode () and the curve is shifted to the right (compared to the normal distribution). If the coefficient of asymmetry is greater than zero (β 1 > 0 ), then the arithmetic mean is greater than the median, and the median, in turn, is greater than the mode () and the curve is shifted to the left (compared to the normal distribution).

Kurtosis coefficient β 2 characterizes the concentration of the empirical distribution around the arithmetic mean in the direction of the axis Oy and the degree of peaking of the distribution density curve. If the kurtosis coefficient is greater than zero, then the curve is more elongated (compared to the normal distribution) along the axis Oy(the graph is more pointed). If the kurtosis coefficient is less than zero, then the curve is more flattened (compared to a normal distribution) along the axis Oy(the graph is more obtuse).

The skewness coefficient can be calculated using the MS Excel function SKRS. If you are checking one array of data, then you need to enter a range of data in one "Number" box.

The coefficient of kurtosis can be calculated using the MS Excel function kurtosis. When checking one data array, it is also enough to enter the data range in one "Number" box.

So, as we already know, with a normal distribution, the skewness and kurtosis coefficients are equal to zero. But what if we got skewness coefficients equal to -0.14, 0.22, 0.43, and kurtosis coefficients equal to 0.17, -0.31, 0.55? The question is quite fair, since in practice we are dealing only with approximate, selective values of asymmetry and kurtosis, which are subject to some inevitable, uncontrollable scatter. Therefore, it is impossible to require strict equality of these coefficients to zero, they should only be sufficiently close to zero. But what does enough mean?

It is required to compare the received empirical values with admissible values. To do this, you need to check the following inequalities (compare the values of the coefficients modulo with the critical values - the boundaries of the hypothesis testing area).

For the asymmetry coefficient β 1 .