Normal distribution in statistics. Normal distribution and its parameters. Univariate Normal Distribution Plots

Normal distribution is the most common type of distribution. One encounters it when analyzing measurement errors, monitoring technological processes and modes, as well as in the analysis and forecasting of various phenomena in biology , medicine and other areas of knowledge.

The term “normal distribution” is used in a conditional sense as generally accepted in the literature, although not entirely successful. Thus, the statement that a certain characteristic obeys a normal distribution law does not at all mean the presence of any unshakable norms that supposedly underlie the phenomenon of which the characteristic in question is a reflection, and 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. Normal distribution discovered for the first time 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 variance is equal to .

Basic properties of normal distribution.

1. The distribution density function is defined on the entire numerical axis Oh , that is, each value X corresponds to a very specific 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 unlimited increase X equal to zero, .

4. The normal distribution density function at a point has a maximum .

5. The graph of the density function is symmetrical about the 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 changing the parameter A .

9. Odds asymmetry And excess 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 this series in comparison with the normal one.

The probability of falling into the interval is found by the formula , Where odd tabulated function.

Let us determine the probability that a normally distributed random variable deviates from its mathematical expectation by an amount less than , that is, we will 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 when we get ,

when we get ,

when we receive .

From the last inequality it follows that practically the scattering of a normally distributed random variable is confined to the area . The probability that a random variable will not fall into this area is very small, namely 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 rule of three sigma, 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 produced by an automatic machine is considered suitable 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 are subject to the normal distribution law with a standard deviation of mm and mathematical expectation. What percentage of suitable parts does the machine produce?

Solution. Consider the random variable X - deviation of the size from the design one. The part will be considered valid if the random variable belongs to the interval. The probability of producing a suitable part can be found using the formula . Consequently, the percentage of suitable parts produced by the machine is 95.44%.

Binomial distribution

Binomial is the probability distribution of occurrence m number of events in n independent trials, in each of which the probability of an event occurring is constant and equal to r . The probability of the possible number of occurrences of an event is calculated using the Bernoulli formula: ,

Where . Permanent n And r , included in this expression, are 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 . Dispersion equal to . The coefficients of skewness and kurtosis are equal and . With an unlimited increase in the number of tests A And E tend to zero, therefore, we can assume that the binomial distribution converges to normal as the number of trials increases.

Example 29. Independent tests are carried out 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 of the number of occurrences across three trials is 0.63.

Solution. For binomial distribution . Let's substitute the values, we get from here or then and .

Poisson distribution

Law of distribution of rare phenomena

Poisson distribution describes the number of events m , occurring over equal periods of time, provided that events occur independently of each other with a constant average intensity. Moreover, the number of tests n is high, and the probability of the event occurring in each trial r small Therefore, the Poisson distribution is called the law of rare events or the simplest flow. The Poisson distribution parameter is the value characterizing the intensity of occurrence of events in n tests. Poisson distribution formula .

The Poisson distribution well describes the number of claims for payment of insurance amounts per year, the number of calls received at the telephone exchange in a certain time, the number of failures of elements during reliability tests, 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 distinctive feature this distribution. The coefficients of asymmetry and kurtosis are respectively equal.

Example 30. The average number of insurance payments per day is two. Find the probability that in five days you will have to pay: 1) 6 insurance amounts; 2) less than six amounts; 3) at least six. or exponential distribution.

This distribution is often observed when studying the service life of various devices, uptime individual elements, parts of the system and the system as a whole, when considering random time intervals between the occurrence of two consecutive 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 application area - reliability theory.

The expression for the integral function of the exponential distribution can be found using the properties of the differential function:

Expectation of exponential distribution, variance, standard deviation. Thus, it is characteristic of this distribution that the standard deviation is numerically equal to the mathematical expectation. For any value of the parameter, the coefficients of asymmetry and kurtosis are constants.

Example 31. The average operating time of a TV before the first failure is 500 hours. Find the probability that a randomly selected TV will operate without breakdowns for more than 1000 hours.

Solution. Since the average operating time to first failure is 500, then . We find the desired probability using the formula.

Definition 1

A random variable $X$ has a normal distribution (Gaussian distribution) if its distribution density is determined by the formula:

\[\varphi \left(x\right)=\frac(1)(\sqrt(2\pi )\sigma )e^(\frac(-((x-a))^2)(2(\sigma )^ 2))\]

Here $aϵR$ is the mathematical expectation, and $\sigma >0$ is the standard deviation.

Density of normal distribution.

Let us show that this function is indeed a distribution density. To do this, check the following condition:

Let's consider improper integral$\int\limits^(+\infty )_(-\infty )(\frac(1)(\sqrt(2\pi )\sigma )e^(\frac(-((x-a))^2)( 2(\sigma )^2))dx)$.

Let's make the replacement: $\frac(x-a)(\sigma )=t,\ x=\sigma t+a,\ dx=\sigma dt$.

Since $f\left(t\right)=e^(\frac(-t^2)(2))$ is an even function, then

The equality is satisfied, which means the function $\varphi \left(x\right)=\frac(1)(\sqrt(2\pi )\sigma )e^(\frac(-((x-a))^2)(2 (\sigma )^2))$ is indeed the distribution density of some random variable.

Let's consider some simple properties of the probability density function of the normal distribution $\varphi \left(x\right)$:

1. The graph of the probability density function of the normal distribution is symmetrical with respect to the straight line $x=a$.
2. The function $\varphi \left(x\right)$ reaches its maximum at $x=a$, and $\varphi \left(a\right)=\frac(1)(\sqrt(2\pi )\sigma ) e^(\frac(-((a-a))^2)(2(\sigma )^2))=\frac(1)(\sqrt(2\pi )\sigma )$
3. The function $\varphi \left(x\right)$ decreases as $x>a$ and increases as $x
4. The function $\varphi \left(x\right)$ has inflection points at $x=a+\sigma $ and $x=a-\sigma $.
5. The function $\varphi \left(x\right)$ asymptotically approaches the $Ox$ axis as $x\to \pm \infty $.
6. The schematic graph looks like this (Figure 1).

Figure 1. Fig. 1. Normal distribution density graph

Note that if $a=0$, then the graph of the function is symmetrical about the $Oy$ axis. Therefore, the function $\varphi \left(x\right)$ is even.

Normal probability distribution function.

To find the probability distribution function for a normal distribution, we use the following formula:

Hence,

Definition 2

The function $F(x)$ is called the standard normal distribution if $a=0,\ \sigma =1$, that is:

Here $Ф\left(x\right)=\frac(1)(\sqrt(2\pi ))\int\limits^x_0(e^(\frac(-t^2)(2))dt)$ - Laplace function.

Definition 3

Function $Ф\left(x\right)=\frac(1)(\sqrt(2\pi ))\int\limits^x_0(e^(\frac(-t^2)(2))dt)$ called the probability integral.

Numerical characteristics of normal distribution.

Mathematical expectation: $M\left(X\right)=a$.

Variance: $D\left(X\right)=(\sigma )^2$.

Mean square distribution: $\sigma \left(X\right)=\sigma $.

Example 1

An example of solving a problem on the concept of normal distribution.

Problem 1: The path length $X$ is a random continuous variable. $X$ is distributed according to the normal distribution law, the mean value of which is equal to $4$ kilometers, and the standard deviation is equal to $100$ meters.

1. Find the distribution density function $X$.
2. Draw a schematic graph of the distribution density.
3. Find the distribution function of the random variable $X$.
4. Find the variance.

1. To begin with, let’s imagine all the quantities in one dimension: 100m=0.1km

From Definition 1, we get:

\[\varphi \left(x\right)=\frac(1)(0.1\sqrt(2\pi ))e^(\frac(-((x-4))^2)(0.02 ))\]

(since $a=4\ km,\ \sigma =0.1\ km)$

1. Using the properties of the distribution density function, we have that the graph of the function $\varphi \left(x\right)$ is symmetrical with respect to the straight line $x=4$.

The function reaches its maximum at the point $\left(a,\frac(1)(\sqrt(2\pi )\sigma )\right)=(4,\ \frac(1)(0.1\sqrt(2\pi )))$

The schematic graph looks like:

Figure 2.

1. By definition of the distribution function $F\left(x\right)=\frac(1)(\sqrt(2\pi )\sigma )\int\limits^x_(-\infty )(e^(\frac(-( (t-a))^2)(2(\sigma )^2))dt)$, we have:

\

1. $D\left(X\right)=(\sigma )^2=0.01$.

The article shows in detail what it is normal law distribution of a random variable and how to use it when solving practical problems.

Normal distribution in statistics

The history of the law goes back 300 years. The first discoverer was Abraham de Moivre, who came up with the approximation back in 1733. Many years later, Carl Friedrich Gauss (1809) and Pierre-Simon Laplace (1812) derived mathematical functions.

Laplace also discovered a remarkable pattern and formulated central limit theorem (CPT), according to which the amount large quantity small and independent quantities has a normal distribution.

The normal law is not a fixed equation of the dependence of one variable on another. Only the nature of this dependence is recorded. The specific form of distribution is specified by special parameters. For example, y = ax + b is the equation of a straight line. However, where exactly it passes and at what angle is determined by the parameters A And b. Same with normal distribution. It is clear that this is a function that describes a tendency for values ​​to be highly concentrated around the center, but its exact shape is determined by special parameters.

The Gaussian normal distribution curve looks like this.

The normal distribution graph resembles a bell, which is why you might see the name bell curve. The graph has a “hump” in the middle and a sharp decrease in density at the edges. This is the essence of the normal distribution. The probability that a random variable will be near the center is much higher than that it will deviate greatly from the center.

The figure above shows two areas under the Gaussian curve: blue and green. Reasons, i.e. The intervals are equal for both sections. But the heights are noticeably different. The blue area is farther from the center and has a significantly lower height than the green area, which is located in the very center of the distribution. Consequently, the areas, that is, the probabilities of falling into the designated intervals, also differ.

The formula for normal distribution (density) is as follows.

The formula consists of two mathematical constants:

π – pi number 3.142;

e– natural logarithm base 2.718;

two changeable parameters that define the shape of a specific curve:

m– mathematical expectation (in various sources Other notations may be used, for example, µ or a);

σ 2– dispersion;

and the variable itself x, for which the probability density is calculated.

The specific form of the normal distribution depends on 2 parameters: ( m) And ( σ 2). Briefly indicated N(m, σ 2) or N(m, σ). Parameter m(mathematical expectation) determines the center of the distribution to which it corresponds maximum height graphics. Dispersion σ 2 characterizes the scope of variation, that is, the “smeariness” of the data.

The mathematical expectation parameter shifts the center of the distribution to the right or left without affecting the shape of the density curve itself.

But dispersion determines the sharpness of the curve. When the data has a small scatter, then all its mass is concentrated at the center. If the data has a large scatter, then it is “spread out” over a wide range.

The distribution density has no direct practical application. To calculate the probabilities, you need to integrate the density function.

The probability that a random variable will be less than a certain value x, is determined normal distribution function:

Using the mathematical properties of any continuous distribution, it is easy to calculate any other probabilities, since

P(a ≤ X< b) = Ф(b) – Ф(a)

Standard normal distribution

The normal distribution depends on the parameters of the mean and variance, which is why its properties are poorly visible. It would be nice to have some distribution standard that does not depend on the scale of the data. And it exists. Called standard normal distribution. In fact, this is an ordinary normal distribution, only with the parameters mathematical expectation 0 and variance 1, briefly written N(0, 1).

Any normal distribution can easily be converted into a standard distribution by normalizing:

Where z– a new variable that is used instead x;
m– mathematical expectation;
σ – standard deviation.

For sample data, estimates are taken:

Arithmetic mean and variance of the new variable z are now also 0 and 1 respectively. This can be easily verified using elementary algebraic transformations.

The name appears in the literature z-score. This is it – normalized data. Z-score can be directly compared with theoretical probabilities, because its scale coincides with the standard.

Let's now see what the density of the standard normal distribution looks like (for z-scores). Let me remind you that the Gaussian function has the form:

Let's substitute instead (x-m)/σ letter z, and instead σ – one, we get density function of the standard normal distribution:

Density chart:

The center, as expected, is at point 0. At the same point, the Gaussian function reaches its maximum, which corresponds to the random variable accepting its average value (i.e. x-m=0). The density at this point is 0.3989, which can be calculated even in your head, because e 0 =1 and all that remains is to calculate the ratio of 1 to the root of 2 pi.

Thus, the graph clearly shows that values ​​that have small deviations from the average appear more often than others, and those that are very far from the center occur much less frequently. The x-axis scale is measured in standard deviations, which allows you to get rid of units of measurement and obtain a universal structure of a normal distribution. The Gaussian curve for normalized data perfectly demonstrates other properties of the normal distribution. For example, that it is symmetrical about the ordinate axis. Most of all values ​​are concentrated within ±1σ from the arithmetic mean (we estimate by eye for now). Most of the data are within ±2σ. Almost all data are within ±3σ. The last property is widely known as three sigma rule for normal distribution.

The standard normal distribution function allows you to calculate probabilities.

It’s clear that no one counts manually. Everything is calculated and placed in special tables, which are at the end of any statistics textbook.

Normal distribution table

There are two types of normal distribution tables:

- table density;

- table functions(integral of density).

Table density rarely used. However, let's see how it looks. Let's say we need to get the density for z = 1, i.e. density of a value separated from the expectation by 1 sigma. Below is a piece of the table.

Depending on the organization of data we are looking for desired value by column and row names. In our example we take the line 1,0 and column 0 , because there are no hundredths. The value you are looking for is 0.2420 (the 0 before 2420 is omitted).

The Gaussian function is symmetrical about the ordinate. That's why φ(z)= φ(-z), i.e. density for 1 is identical to the density for -1 , which is clearly visible in the figure.

To avoid wasting paper, tables are printed only for positive values.

In practice, the values ​​are more often used functions standard normal distribution, that is, the probability for different z.

Such tables also contain only positive values. Therefore, to understand and find any you should know the required probabilities properties of the standard normal distribution.

Function Ф(z) symmetrical about its value 0.5 (and not the ordinate axis, like density). Hence the equality is true:

This fact is shown in the picture:

Function values Ф(-z) And Ф(z) divide the graph into 3 parts. Moreover, the upper and lower parts are equal (indicated by check marks). To complement the probability Ф(z) to 1, just add the missing value Ф(-z). You get the equality indicated just above.

If you need to find the probability of falling into the interval (0;z), that is, the probability of deviation from zero in a positive direction to a certain number of standard deviations, it is enough to subtract 0.5 from the value of the standard normal distribution function:

For clarity, you can look at the drawing.

On a Gaussian curve, this same situation looks like the area from center right to z.

Quite often, an analyst is interested in the probability of deviation in both directions from zero. And since the function is symmetrical about the center, the previous formula must be multiplied by 2:

Picture below.

Under the Gaussian curve it is central part, limited to the selected value –z left and z right.

These properties should be taken into account, because tabulated values ​​rarely correspond to the interval of interest.

To make the task easier, textbooks usually publish tables for functions of the form:

If you need the probability of deviation in both directions from zero, then, as we have just seen, the table value for this function is simply multiplied by 2.

Now let's look at specific examples. Below is a table of the standard normal distribution. Let's find the table values ​​for three z: 1.64, 1.96 and 3.

How to understand the meaning of these numbers? Let's start with z=1.64, for which the table value is 0,4495 . The easiest way to explain the meaning is in the figure.

That is, the probability that a standardized normally distributed random variable falls within the interval from 0 to 1,64 , is equal 0,4495 . When solving problems, you usually need to calculate the probability of deviation in both directions, so let’s multiply the value 0,4495 by 2 and we get approximately 0.9. The occupied area under the Gaussian curve is shown below.

Thus, 90% of all normally distributed values ​​fall within the interval ±1.64σ from the arithmetic mean. It was not by chance that I chose the meaning z=1.64, because the neighborhood around the arithmetic mean, occupying 90% of the entire area, is sometimes used to calculate confidence intervals. If the value being tested does not fall within the designated area, then its occurrence is unlikely (only 10%).

To test hypotheses, however, an interval covering 95% of all values ​​is more often used. Half the chance 0,95 - This 0,4750 (see the second highlighted value in the table).

For this probability z=1.96. Those. within almost ±2σ 95% of values ​​are from the average. Only 5% fall outside these limits.

Another interesting and frequently used table value corresponds to z=3, it is equal according to our table 0,4986 . Multiply by 2 and get 0,997 . So, within ±3σ Almost all values ​​are derived from the arithmetic mean.

This is what the 3 sigma rule looks like for a normal distribution in a diagram.

Using statistical tables you can get any probability. However, this method is very slow, inconvenient and very outdated. Today everything is done on the computer. Next, we move on to the practice of calculations in Excel.

Normal Distribution in Excel

Excel has several functions for calculating probabilities or inverses of a normal distribution.

NORMAL DIST function

Function NORM.ST.DIST. designed to calculate density ϕ(z) or probabilities Φ(z) according to normalized data ( z).

=NORM.ST.DIST(z;integral)

z– value of the standardized variable

integral– if 0, then the density is calculatedϕ(z) , if 1 is the value of the function Ф(z), i.e. probability P(Z

Let's calculate the density and function value for various z: -3, -2, -1, 0, 1, 2, 3(we will indicate them in cell A2).

To calculate the density, you will need the formula =NORM.ST.DIST(A2;0). In the diagram below, this is the red dot.

To calculate the value of the function =NORM.ST.DIST(A2;1). The diagram shows the shaded area under the normal curve.

In reality, it is more often necessary to calculate the probability that a random variable will not go beyond certain limits from the average (in standard deviations corresponding to the variable z), i.e. P(|Z| .

Let us determine the probability of a random variable falling within the limits ±1z, ±2z and ±3z from zero. Need a formula 2Ф(z)-1, in Excel =2*NORM.ST.DIST(A2;1)-1.

The diagram clearly shows the main basic properties of the normal distribution, including the three-sigma rule. Function NORM.ST.DIST. is an automatic table of normal distribution function values ​​in Excel.

There may also be an inverse problem: according to the available probability P(Z find the standardized value z,that is, a quantile of the standard normal distribution.

NORM.ST.REV function

NORM.ST.REV calculates the inverse of the standard normal distribution function. The syntax consists of one parameter:

=NORM.ST.REV(probability)

probability is a probability.

This formula is used as often as the previous one, because using the same tables you have to look not only for probabilities, but also for quantiles.

For example, when calculating confidence intervals, a confidence probability is specified, according to which it is necessary to calculate the value z.

Given that the confidence interval consists of an upper and lower limit and that the normal distribution is symmetrical around zero, it is enough to obtain the upper limit (positive deviation). The lower limit is taken with a negative sign. Let us denote the confidence probability as γ (gamma), then the upper limit of the confidence interval is calculated using the following formula.

Let's calculate the values ​​in Excel z(which corresponds to the deviation from the average in sigma) for several probabilities, including those that any statistician knows by heart: 90%, 95% and 99%. In cell B2 we indicate the formula: =NORM.ST.REV((1+A2)/2). By changing the value of the variable (probability in cell A2), we obtain different boundaries of the intervals.

The 95% confidence interval is 1.96, that is, almost 2 standard deviations. From here it is easy, even mentally, to estimate the possible spread of a normal random variable. In general, the 90%, 95% and 99% confidence intervals correspond to confidence intervals of ±1.64, ±1.96 and ±2.58σ.

In general, the NORM.ST.DIST and NORM.ST.REV functions allow you to perform any calculation related to the normal distribution. But to make it easier and reduce the number of steps, Excel has several other features. For example, you can use CONFIDENCE NORM to calculate confidence intervals for the mean. To check the arithmetic mean there is the formula Z.TEST.

Let's look at a couple more useful formulas with examples.

NORMAL DIST function

Function NORMAL DIST. different from NORM.ST.DIST. only because it is used to process data of any scale, and not just normalized ones. Normal distribution parameters are specified in the syntax.

=NORM.DIST(x,average,standard_deviation,integral)

average– mathematical expectation used as the first parameter of the normal distribution model

standard_off– standard deviation – the second parameter of the model

integral– if 0, then the density is calculated, if 1 – then the value of the function, i.e. P(X

For example, the density for the value 15, which was extracted from a normal sample with an expectation of 10, a standard deviation of 3, is calculated as follows:

If the last parameter is set to 1, then we get the probability that the normal random variable will be less than 15 for the given distribution parameters. Thus, probabilities can be calculated directly from the original data.

NORM.REV function

This is a quantile of the normal distribution, i.e. the value of the inverse function. The syntax is as follows.

=NORM.REV(probability,average,standard_deviation)

probability- probability

average– mathematical expectation

standard_off– standard deviation

The purpose is the same as NORM.ST.REV, only the function works with data of any scale.

An example is shown in the video at the end of the article.

Normal Distribution Modeling

Some problems require the generation of normal random numbers. There is no ready-made function for this. However, Excel has two functions that return random numbers: CASE BETWEEN And RAND. The first produces random, uniformly distributed integers within specified limits. The second function generates uniformly distributed random numbers between 0 and 1. To make an artificial sample with any given distribution, you need the function RAND.

Let's say that to conduct an experiment it is necessary to obtain a sample from a normally distributed population with an expectation of 10 and a standard deviation of 3. For one random value, we will write a formula in Excel.

NORM.INV(RAND();10;3)

Let's extend it to the required number of cells and the normal sample is ready.

To model standardized data, you should use NORM.ST.REV.

The process of converting uniform numbers to normal numbers can be shown in the following diagram. From the uniform probabilities that are generated by the RAND formula, horizontal lines are drawn to the graph of the normal distribution function. Then, from the points of intersection of the probabilities with the graph, projections are lowered onto the horizontal axis.