The probability that the SV deviation X from her M.O. a in absolute value will be less than a given positive number, equal to

If we put in this equality, then we get

s w: space \u003d "720" /\u003e"> ,

That is, the normally distributed SV X deviates from his M.O. a, as a rule, by less than 3. This is the so-called 3 sigma rule, which is often used in mathematical statistics.

Function of one random variable. The mathematical expectation of a function of one SV. (Tetr)

If each possible value of the random variable X corresponds to one possible value of a random variable Y then Y called function of a random argument X: Y \u003d φ (X ).

Let us find out how to find the distribution law of a function according to the known distribution law of the argument.

1) Let the argument X Is a discrete random variable, with different values X correspond different meanings Y ... Then the probabilities of the corresponding values X and Y are equal .

2) If different values X may correspond identical values Y , then the probabilities of the argument values \u200b\u200bfor which the function takes the same value are added.

3) If X - continuous random variable, Y \u003d φ (X ), φ (x ) Is a monotone and differentiable function, and ψ (at ) Is the function inverse to φ (x ).

The mathematical expectation of a function of one random argument.

Let be Y \u003d φ (X ) - random argument function X , and you need to find it expected valueknowing the distribution law X .

1) If X Is a discrete random variable, then

2) If X Is a continuous random variable, then M (Y ) can be searched in different ways. If the distribution density is known g (y ), then

21. Function of two random arguments. Distribution of the function Z \u003d X + Y for discrete independent SV X and Y. (tetr)

If each pair of possible values \u200b\u200bof random variables X and Y corresponds to one possible value of a random variable Z, then Z is called a function of two random arguments X and Y and is written Z \u003d φ (X, Y). If X and Y are discrete independent random variables, then in order to find the distribution of the function Z \u003d X + Y, it is necessary to find all possible values \u200b\u200bof Z, for which it is enough to add each possible value of X with all possible values \u200b\u200bof Y; the probabilities of the found possible values \u200b\u200bof Z are equal to the products of the probabilities of the added values \u200b\u200bof X and Y. If X and Y are continuous independent random variables, then the distribution density g (z) of the sum Z \u003d X + Y (provided that the distribution density of at least one of the arguments is given in the interval (- oo, oo) by one formula) can be found by the formula, or by an equivalent formula, where f1 and f2 are the density of the distribution of arguments; if the possible values \u200b\u200bof the arguments are non-negative, then the distribution density g (z) of the quantity Z \u003d X + Y is found by the formula, or by the equivalent formula. In the case when both densities f1 (x) and f2 (y) are given on finite intervals, to find the density g (z) of the quantity Z \u003d X + Y, it is advisable to first find the distribution function G (z) and then differentiate it with respect to z : g (z) \u003d G '(z). If X and Y are independent random variables given by the corresponding distribution densities f1 (x) and f2 (y), then the probability of a random point (X, Y) falling into the domain D is double integral over this area from the product of distribution densities: P [(X, Y) cD] \u003d ... Discrete independent random variables X and Y are given by distributions:

R 0.3 0.7 R 0.6 0.4

Find the distribution of the random variable Z \u003d X + K. Solution. In order to compose the distribution of the value Z \u003d X + Y, it is necessary to find all possible values \u200b\u200bof Z and their probabilities. Possible Z values \u200b\u200bare the sums of each possible X value with all possible Y values: Z 1 \u003d 1 + 2 \u003d 3; z 2 \u003d 1 + 4 \u003d 5; z 3 \u003d 3 + 2 \u003d 5; z4 \u003d 3 + 4 \u003d 7. Find the probabilities of these possible values. For Z \u003d 3, it is sufficient that the value of X takes the value x1 \u003d l and the value of the K-value y1 \u003d 2. The probabilities of these possible values, as follows from these distribution laws, are 0.3 and 0.6, respectively. Since the arguments X and Y are independent, then the events X \u003d 1 and Y \u003d 2 are independent and, therefore, the probability of their joint occurrence (i.e., the probability of the event Z \u003d 3) according to the multiplication theorem is 0.3 * 0.6 \u003d 0 ,18. Similarly, we find:

I B \u003d! - f4 \u003d 5) \u003d 0.3 0.4 \u003d 0.12;

P (Z \u003d 34-2 \u003d 5) \u003d 0.7 0.6 \u003d 0.42;

P (Z \u003d 3rd \u003d 7) \u003d 0.7-0.4 \u003d 0.28. Let us write the desired distribution by adding the probabilities of inconsistent events Z \u003d z 2 \u003d 5, Z \u003d z 3 \u003d 5 (0.12 + 0.42 \u003d 0.54):

Z 3 5 7; P 0.18 0.54 0.28. Control: 0.18 + 0.54 + 0.28 \u003d 1.

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

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

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

№	Index	Normal distribution law	Note
	Definition	Normal is called the probability distribution of a continuous random variable X, whose density has the form
			where m x is the mathematical expectation of a random variable X, σ x is the standard deviation
2	Distribution function
	Probability hitting the interval (a; b)		- integral Laplace function
	Probability the fact that the absolute value of the deviation is less than the positive number δ		for m x \u003d 0

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

A task.

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

Decision. and) We find the probability density of a random variable X, distributed according to the normal law:

provided that m x \u003d 20, σ \u003d 0.2.

b) For the normal distribution of a random variable, the probability of falling into the interval (19.7; 20.3) is determined:
F ((20.3-20) / 0.2) - F ((19.7-20) / 0.2) \u003d F (0.3 / 0.2) - F (-0.3 / 0, 2) \u003d 2F (0.3 / 0.2) \u003d 2F (1.5) \u003d 2 * 0.4332 \u003d 0.8664.
We found the value Ф (1.5) \u003d 0.4332 in the applications, in the table of values \u200b\u200bof the Laplace integral function Φ (x) ( table 2 )

in) We find the probability that the absolute value of the deviation is less than a positive number 0.1:
R (| X-20 |< 0,1) = 2Ф(0,1/0,2) = 2Ф(0,5) = 2*0,1915 = 0,383.
We found the value Ф (0.5) \u003d 0.1915 in applications, in the table of values \u200b\u200bof the integral Laplace function Φ (x) ( table 2 )

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

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

Using the application ( table 2 ), we find δ / σ \u003d 0.74. Hence δ \u003d 0.74 * σ \u003d 0.74 * 0.2 \u003d 0.148 mm.

e) Since the sought interval is symmetric about the mean value m x \u003d 20, then it can be defined as the set of X values \u200b\u200bsatisfying the inequality 20 - δ< X < 20 + δ или |x − 20| < δ .

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

Using the application ( table 2 ), we find δ / σ \u003d 1.96. Hence δ \u003d 1.96 * σ \u003d 1.96 * 0.2 \u003d 0.392.
The sought interval : (20 - 0.392; 20 + 0.392) or (19.608; 20.392).

It is said that SV X has even distribution in the section from a to b, if its density f (x) is constant in this section, that is

.

For example, a measurement is made using a device with coarse divisions; the nearest integer is taken as an approximate value of the measured value. SV X - the measurement error is evenly distributed over the site, since none of the values \u200b\u200bof the random variable is in any way preferable to others.

Exponential (exponential) is the probability distribution of a continuous random variable, which is described by the density

where is a constant positive value.

An example of a continuous random variable distributed according to an exponential law is the time between the occurrence of two consecutive events of the simplest flow.

Often the duration of the uptime of the elements has an exponential distribution, the distribution function of which is
determines the probability of an element failure in a time of duration t.

- failure rate (average number of failures per unit of time).

Normal law distribution (sometimes called gauss's law) plays an extremely important role in the theory of probability and ranks among other distribution laws special position... The distribution density of the normal law has the form

,

where m is the mathematical expectation,

- standard deviation X.

The probability that the normally distributed SV X will take a value belonging to the interval is calculated by the formula: ,

where Ф (X) - laplace function... Its values \u200b\u200bare determined from the table in the appendix of the textbook on probability theory.

The probability that the deviation of a normally distributed random variable X from the mathematical expectation in absolute value is less than a given positive number is calculated by the formula

.

EXAMPLES OF SOLVING PROBLEMS

EXAMPLE 13.2.41. The value of one division of the ammeter scale is 0.1 A. Readings are rounded to the nearest whole division. Find the probability that an error exceeding 0.02 A.

Decision. The round-off error can be thought of as CB X, which is evenly distributed between two adjacent divisions. Density of uniform distribution, where (b-a) is the length of the interval containing the possible values \u200b\u200bof X. In the problem under consideration, this length is 0.1. therefore ... So, .

The counting error will exceed 0.02 if it is enclosed in the interval (0.02; 0.08). According to the formula we have

EXAMPLE 13.2.42. The duration of the uptime of the element has an exponential distribution. Find the probability that over a time duration of hours:

a) the element will fail;

b) the element will not fail.

Decision. a) The function determines the probability of failure of an element for a time duration t, therefore, substituting, we obtain the probability of failure:.

b) The events "the element will fail" and "the element will not fail" are opposite, therefore the probability that the element will not fail.

EXAMPLE 13.2.43. Random variable X is distributed normally with parameters. Find the probability that SV X deviates from its mathematical expectation m by more than.

This probability is very small, that is, such an event can be considered almost impossible (you can make a mistake in about three cases out of 1000). This is the “rule of three sigma”: if a random variable is normally distributed, then the absolute value of its deviation from the mathematical expectation does not exceed three times the standard deviation.

EXAMPLE 13.2.44. The mathematical expectation and standard deviation of a normally distributed random variable are, respectively, 10 and 2. Find the probability that, as a result of the test, X will take the value included in the interval (12, 14).

Solution: For a normally distributed quantity

.

Substituting, we get

We find from the table.

Seeking probability.

Examples and tasks for independent solution

Solve problems using formulas for calculating the probability of continuous random variables and their characteristics

3.2.9.1. Find the mathematical expectation, variance and standard deviation of a random variable X, uniformly distributed in the interval (a, b).

Resp.:

3.2.9.2. Subway trains run regularly at intervals of 2 minutes. The passenger enters the platform at a random time. Find the distribution density of SV T - the time during which he will have to wait for the train; ... Find the probability that you will have to wait no more than half a minute.

Resp.:

3.2.9.3. The minute hand of the electric clock moves in leaps and bounds at the end of each minute. Find the probability that at a given instant the clock will show a time that differs from the true one by no more than 20 s.

Resp.:2/3

3.2.9.4. Random variable X is distributed evenly over the site (a, b). Find the probability that, as a result of the experiment, it deviates from its mathematical expectation by more than.

Resp.:0

3.2.9.5. Random variables X and Y are independent and evenly distributed: X - in the interval (a, b), Y - in the interval (c, d). Find the expected value of the XY product.

Resp.:

3.2.9.6. Find the mathematical expectation, variance, and standard deviation of an exponentially distributed random variable.

Resp.:

3.2.9.7. Write the density and distribution function of the exponential law if parameter.

Resp.: ,

3.2.9.8. The random variable has an exponential distribution with a parameter. To find .

Resp.:0,233

3.2.9.9. The uptime of the element is distributed according to the exponential law, where t is the time, h. Find the probability that the element will work 100 h without failure.

Resp.:0,37

3.2.9.10. Three elements are tested that work independently of one another. The duration of the uptime of the elements is distributed according to the exponential law: for the first element ; for the second ; for the third element ... Find the probability that in the time interval (0; 5) hours will fail: a) only one element; b) only two elements; c) all three elements.

Resp.: a) 0.292; b) 0.466; c) 0.19

3.2.9.11. Prove that if a continuous random variable is exponentially distributed, then the probability that X will take a value less than the mathematical expectation M (X) does not depend on the value of the parameter; b) find the probability that X\u003e M (X).

Resp.:

3.2.9.12. The mathematical expectation and standard deviation of the normally distributed random variable are, respectively, 20 and 5. Find the probability that, as a result of the test, X will take the value enclosed in the interval (15; 25).

Resp.: 0,6826

3.2.9.13. Some substance is weighed without systematic errors. Random weighing errors are subject to the normal law with a standard deviation r. Find the probability that a) weighing will be performed with an error not exceeding 10 r in absolute value; b) of three independent weighings, the error of at least one will not exceed 4d in absolute value.

Resp.:

3.2.9.14. Random variable X is distributed normally with mathematical expectation and standard deviation. Find an interval symmetric with respect to the mathematical expectation, in which the value X will fall with a probability of 0.9973 as a result of the test.

Resp.:(-5,25)

3.2.9.15. The factory produces balls for bearings, the nominal diameter of which is 10 mm, and the actual diameter is random and distributed according to the normal law with mm and mm. During the inspection, all balls that do not pass through a round hole with a diameter of 10.7 mm and all that pass through a round hole with a diameter of 9.3 mm are rejected. Find the percentage of balls that will be rejected.

Resp.:8,02%

3.2.9.16. The machine is stamping parts. The length of the part X is controlled, which is distributed normally with the design length (mathematical expectation) equal to 50 mm. In fact, the length of the manufactured parts is not less than 32 and not more than 68 mm. Find the probability that the length of the part taken at random: a) is greater than 55 mm; b) less than 40 mm.

Hint: From equality pre-find.

Resp.: a) 0.0823; b) 0.0027

3.2.9.17. Chocolate boxes are packed automatically; their average weight is 1.06 kg. Find the variance if 5% of the boxes weigh less than 1 kg. It is assumed that the mass of the boxes is distributed according to the normal law.

Resp.:0,00133

3.2.9.18. A bomber flying along the bridge, which is 30 meters long and 8 meters wide, dropped bombs. The random variables X and Y (the distance from the vertical and horizontal axes of symmetry of the bridge to the place where the bomb fell) are independent and distributed normally with standard deviations of 6 and 4 m, respectively, and mathematical expectations, equal to zero... Find: a) the probability of one thrown bomb hitting the bridge; b) the probability of the destruction of the bridge if two bombs are dropped, and it is known that one hit is enough to destroy the bridge.

Resp.:

3.2.9.19. In a normally distributed population, 11% of X values \u200b\u200bare less than 0.5 and 8% of X values \u200b\u200bare greater than 5.8. Find the parameters m and the given distribution. \u003e
Examples of problem solving\u003e

> > Distributions of continuous random variables

Normal probability distribution

Without exaggeration, it can be called a philosophical law. Observing various objects and processes of the surrounding world, we are often faced with the fact that something is small, and that there is a norm:


Before you is a principled view density functions normal probability distribution, and I welcome you to this exciting lesson.

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

These are different characteristics of inanimate objects (the same size, weight). This is a random duration of processes, for example, the time of a 100-meter run or the transformation of resin into amber. From physics I remembered air molecules: among them there are slow ones, there are fast ones, but most of them move with "standard" speeds.

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

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

At the final stage, carefully draw the schedule, and especially neat reflect it bulge / concavity! Well, you probably realized a long time ago that the abscissa axis is horizontal asymptote, and it is categorically impossible to "climb" behind it!

With the electronic registration of the solution, the schedule 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 \u200b\u200bof and.

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


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

If sigma changes (with constant "a"), the graph "stays in place" but changes shape. As it increases, it becomes lower and elongated, like an octopus stretching out its tentacles. Conversely, when decreasing, the graph getting narrower and taller- it turns out "surprised octopus". So, for decreasing "Sigma" twice: the previous chart narrows and stretches upwards twice:

Everything is in full accordance with geometric transformations of graphs.

A normal distribution with a single sigma value 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 was already encountered in local Laplace theorem: ... The standard distribution has found wide application in practice, and very soon we will finally understand its purpose.

Well, now let's watch a movie:

Yes, quite right - somehow undeservedly remained in the shadows probability distribution function... We remember her definition:
- the probability that a random variable will take on a value LESS than a variable that "runs" all real values \u200b\u200bto "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 assigned improper integral which is equal to some the number from the interval.

Almost all of the values \u200b\u200bdefy accurate calculation, but as we just saw, with modern computing power, this is not difficult. So, for the function standard distribution, the corresponding Excel function generally contains one argument:

\u003d 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 us 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 area between the normal curve and the abscissa axis in the corresponding area:

but every time to grind out an approximate value unreasonable, and therefore it is more rational to use "Light" formula:
.

! Also remembers , what

Here you can use Excel again, but there are a couple of weighty "buts": firstly, it is not always at hand, and secondly, the "ready-made" values \u200b\u200bare likely to raise questions from the teacher. Why?

I have talked about this many times before: at one time (and not very long ago) an ordinary calculator was a luxury, and in the educational literature there is still a “manual” way of solving the problem in question. Its essence is to standardize values \u200b\u200b"alpha" and "beta", that is, to reduce the solution to the standard distribution:

Note : the function is easy to obtain from the general case using linear replacements ... Then and:

and from the replacement just follows the formula transition from values \u200b\u200bof an arbitrary distribution - to the corresponding values \u200b\u200bof the standard distribution.

Why is this needed? The fact is that the values \u200b\u200bwere meticulously calculated by our ancestors and compiled in a special table, which is in many books on the Terver. But even more often there is a table of values, with which we have already dealt in integral Laplace theorem:

If we have at our disposal a table of values \u200b\u200bof the Laplace function , then we solve through it:

Fractional values \u200b\u200bare traditionally rounded to 4 decimal places, as is done in a typical table. And for control there is Item 5 layout.

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

Answer it is required to give 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 shells will fall

We train ourselves:

Example 3

The diameter of bearings manufactured at the factory is a random value distributed normally with a mathematical expectation of 1.5 cm and a standard deviation of 0.04 cm.Find the probability that the size of a bearing taken at random ranges from 1.4 to 1.6 cm.

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

And already in this example we encountered a special case - when the interval is symmetric 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, to simplify the working formula:

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

- 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 :)
- 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 to have even greater reliability - namely, to find out the boundaries in which the diameter is almost everyone bearings. Is there any criterion for this? Exist! The question posed is answered by the so-called

the 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 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 investigated random variable fit into an interval of 6 standard deviations in length, then there are good reasons to believe that this value is distributed according to the normal law. Verification is carried out using theory statistical hypotheses.

We continue to solve severe 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 condition, and 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 has 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 Example 3 lesson about even distribution... There was an error rounding off measurement results, here we are talking about the random error of the measurements themselves. Such errors arise due to technical characteristics the device itself (the range of acceptable errors is usually 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 the so-called systematic measurement errors. This is already nonrandom errors that arise due to incorrect setup or operation of the device. So, for example, unadjusted floor scales can steadily "add" the kilogram, and the seller systematically weigh the buyers. Or not systematically, because you can short-circuit. However, in any case, such an error will not be accidental, and its expectation is nonzero.

... I am urgently developing a training course for salespeople \u003d)

We independently solve the inverse problem:

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, symmetric with respect to the mathematical expectation, in which the length of the bead diameter is likely to fall.

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 assignment, which I highly recommend to consolidate 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 deviates in absolute value from no more than;
d) applying the "three sigma" rule, find the values \u200b\u200bof the random variable.

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

Well, I will analyze an example of increased complexity:

Example 7

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

Decision: First of all, let us note 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 may turn out, for example, indicative or generally arbitrary continuous distribution... And therefore, the "normality" of the distribution still needs to be justified:

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

We give. For this select a full square and organize three-story shot:


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

what we wanted to see.

In this way:
- by rule of action 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 factor of the normal distribution has the form and, then:
, from where 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 build a density graph:

and the distribution function graph :

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

Example 1.Mathematical expectation of a normally distributed continuous RV X M(X) \u003d 6, and the standard deviation s ( X) = 2.

Find: 1) the probability of hitting the values \u200b\u200bof SV X in the interval (2; 9);

3) an interval symmetrical about a X with probability g \u003d 0.9642.

Decision... 1) Let us find the probability of hitting the values \u200b\u200bof SV X in the interval (2; 9).

Laplace function values taken from the table. The property of oddness of the function Ф (- X) \u003d - Ф ( X).

2) Determine the probability

Because a = M(X) \u003d 6 and s \u003d s ( X) \u003d 2, then

3) Find the interval symmetric with respect to a, in which the values \u200b\u200bof SV X with probability g \u003d 0.9642.

From the table of values \u200b\u200bof the Laplace function, we find that is, d \u003d 4.2. Then the interval is –4.2< X – 6 < 4,2 и
1,8 < X < 10,2.

Example 2.Random value T(hours) - the device uptime has an exponential distribution. Find the probability that the device will work without repair for at least 600 hours if the mean time of failure-free operation of devices of this type is 400 hours.

Decision. M(T) \u003d 400 hours, therefore, according to the formula (1.46) Since for the exponential distribution then
0,2233.

Example 3.Random value X distributed uniformly on the segment [ a, b]. Find the probability of hitting a random variable X per segment
, entirely contained within the segment [ a, b].

Decision... Let's use the formula where the probability density

.

In this way

Example 4.Electric trains run strictly on schedule at intervals
20 minutes. Find the probability that a passenger arriving at the platform will wait for the next electric train for more than 10 minutes, as well as the average waiting time.

Decision. X - the waiting time (min.) Of an electric train can be considered a uniformly distributed random variable with a density:

and this is the average waiting time for an electric train.

Example 5.The machine produces bushings. The bushing is considered good if the deviation X its diameter from the design size is less than 1mm in absolute value. Assuming that the random variable X distributed normally with standard deviation s \u003d 0.5 mm and mathematical expectation a \u003d 0, find how many suitable bushings there will be among 100 manufactured ones, as well as the probability that the deviation from the design size will be at least 0.4 mm and not more than 0.8 mm.

Decision... Let's use the formula () for d \u003d 1, s \u003d 0.5 and a = 0.

It follows that approximately 95 out of 100 bushings will be usable.

To find the probability that the deviation from the design size will be at least 0.4 mm and not more than 0.8 mm, we use the formula (1.54)

at a \u003d 0, s \u003d 0.5, a \u003d 0.4, b \u003d 0.8.

The values \u200b\u200bof the function Ф ( x) we find from the table.

Job options

OPTION 1

X (CB X) is given by the distribution series:

x i
p i	0,1	0,2	0,3	0,1	0,3

F(x M(X), variance D(XX), fashion M 0 (X); 3) probability P(8≤ X < 30). Построить многоугольник распределения и график F(x).

Problem 2. Each of the shooters shoots at the target once. The probability that the first, second and third shooters will hit the target in one shot are respectively equal to 0.8; 0.6 and 0.9. For
CB X- the total number of hits on the target under the specified conditions, draw up a distribution series and find F(x), M(X), s ( X) and D(X).

Problem 3. The probability of occurrence of some event AND in each experiment is 0.6. It is required: 1) to construct a series of discrete distribution CB X - the number of occurrences of the event AND in four independent experiments; 2) estimate the probability that in a series of 80 independent experiments this event will appear at least 60 times.

Problem 4. Discrete CB X given by a distribution series:

x i	–2	–1
p i	0,05	0,10	0,15	?	0,15	0,20	0,10

Find distribution series CB Y = –2X 2 + 3, M(Y) and D(Y).

Task 5. Continuous CB X

Find: a) distribution density f(x); b) M(x); in) d) the probability that in three independent tests CB X takes exactly two values \u200b\u200bthat belong to the interval

Task 6. The function is set

A CB X... To find F(x), M(X) and D(X). Build a graph F(x).

Task 7. Set M(X) \u003d 14 and s ( X CB X... To find:

1) probability ;

2) probability ;

3) symmetrical with respect to a CB X with probability g \u003d 0.8385.

Problem 8. The stopwatch scale has a division value of 0.2 s. The time is counted down to the nearest whole division, rounded to the nearest side. The reading error under these conditions can be considered a uniformly distributed random variable.

Find the probability of making this stopwatch countdown with an error of a) less than 0.05 s; b) not less than 0.01 s and not more than 0.05 s.

OPTION 2

Problem 1. Discrete random variable X (CB X) is given by the distribution series:

x i	–2	–1
p i	0,2	0,2	0,1	0,3	0,2

Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), variance D(X), the standard deviation s ( X), fashion M 0 (X); 3) probability P(–2 ≤ X < 5). Построить многоугольник распределения и график F(x).

Problem 2. There are 100 tickets in the lottery, of which 10 are winning. Someone buys 4 tickets. For SV X - the number of winning tickets among those that will be bought, make a distribution series and find F(x), M(X), s ( X).

Task 3. Reports are drawn up independently of one another. The probability of making a mistake in each report is 0.3. Required: 1) build a distribution series CB X - the number of reports with errors among the four generated; calculate M(X), D(X) and s ( X); 2) estimate the probability that when compiling 50 reports, there will be 20 reports with errors.

Problem 4. It is known that the discrete CB Xcan take only two values x 1 \u003d –2 and x 2 \u003d 3 and its mathematical expectation M(X) \u003d 1.5. Make distribution series CB X and CB Z \u003d Find F(z) and s ( Z).

Task 5. Continuous CB X given by the distribution function

f(x); 2) M(x) and D(X);
3) 4) the probability that in three independent trials CB X will take a value belonging to the interval (1; 4) exactly once.

Task 6. The function is set

Determine parameter value A, at which this function defines the density of the probability distribution of some continuous CB X... To find F(x), M(X), D(X). Build a graph F(x).

Task 7. Set M(X) \u003d 12 and s ( X CB X... To find:

1) probability ;

2) probability ;

3) symmetrical with respect to a the interval in which the values \u200b\u200bfall CB X with probability g \u003d 0.4515.

Problem 8. A random error in measuring a certain part is subject to the normal law with the parameter s \u003d 20 mm. Find the probability that: a) the part is measured with an error not exceeding 22 mm in absolute value; b) in none of the two measurements performed, the error will not exceed 22 mm in absolute value.

OPTION 3

Problem 1. Discrete random variable X (CB X) is given by the distribution series:

x i
p i	0,3	0,1	0,1	0,4	0,1

Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), variance D(X), the standard deviation s ( X), fashion M 0 (X); 3) probability P(1 ≤ X < 7). Построить многоугольник распределения и график F(x).

Problem 2. Of the three athletes who entered the national youth team of the country in the high jump competition, one can pass qualified starts with a probability of 0.9, the second with a probability of 0.8 and the third with a probability of 0.6. For CB X - the number of athletes of the national team who will advance to the next round of competitions, draw up a distribution series and find M(X), s ( X).

Task 3. A series of independent shots is fired at the target. The probability of hitting the target with each shot is 0.8. Required: 1) build a distribution series CB X - the number of hits with three shots; 2) estimate the probability that at 100 shots there will be at least 90 hits.

Problem 4. Discrete random variable X (CB X) is given by the distribution series:

x i	–3	–2	–1
p i	0,1	0,2	0,3	0,2	?

Find the series and distribution function CB Y = 2X + 1, M(Y) and D(Y).

Task 5. Continuous CB X given by the distribution function

Find: 1) distribution density f(x); 2) M(x) and D(X);
3) P(–2,3 < X <1,5);4) вероятность того, что в трех независимых испытаниях CB X takes exactly two values \u200b\u200bthat belong to the interval (–2.3; 1.5).

Task 6. The function is set

Determine parameter value A, at which this function defines the density of the probability distribution of some continuous CB X... To find F(x), and M(X). Build a graph F(x).

Task 7. Set M(X) \u003d 13 and s ( X CB X... To find:

1) probability ;

2) probability ;

3) symmetrical with respect to a the interval in which the values \u200b\u200bfall CB X with probability g \u003d 0.9973.

Problem 8. It is known that the TV repair time is a random variable X, distributed according to the exponential law, while the average TV repair time is two weeks. Find the probability that the repair of the TV brought to the workshop will take: a) less than 10 days; b) from 9 to 12 days.

OPTION 4

Problem 1. Discrete random variable X (CB X) is given by the distribution series:

x i	–10	–5
p i	0,1	0,1	0,4	0,1	0,3

Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), variance D(X), the standard deviation s ( X), fashion M 0 (X); 3) probability P(–10≤ X < 1). Построить многоугольник распределения и график F(x).

Problem 2. The attendant has 5 different keys to different rooms. Taking out a key at random, he tries to open the door of one of the rooms. For discrete CB X - the number of attempts to open the door (the verified key is not used a second time) compile a distribution series and find F(x) and M(X).

Problem 3. The probability of manufacturing a part with specified accuracy parameters from a standard workpiece for each part is 0.8.

Required: 1) build a distribution series CB X- the number of parts with specified accuracy characteristics that will be manufactured from five standard blanks; 2) estimate the probability that 70 parts with specified accuracy characteristics will be produced from 90 blanks.

CB X and Y:

x i
p i	?	0,5	0,2

y i
p i	0,6	?

Make a distribution series CB Z = Y – X... To find M(Z) and D(Z).

Task 5. Continuous CB X given by the distribution function

Find: 1) distribution density f(x); 2) M(x); 3) CB X takes exactly three times values \u200b\u200bthat belong to the interval

Task 6. The function is set

Determine parameter value A, at which this function defines the density of the probability distribution of some continuous CB X... To find F(x), M(X) and D(X). Build a graph F(x).

Task 7. Set M(X) \u003d 16 and s ( X) \u003d 2 normally distributed continuous CB X... To find:

1) probability ;

2) probability ;

3) symmetrical with respect to a the interval in which the values \u200b\u200bfall CB X with probability g \u003d 0.9281.

Problem 8. The height of an adult male is SV Xdistributed according to the normal law with parameters and \u003d 175 cm and s \u003d 10 cm. Find the probability that the height of a randomly selected man will be: a) less than 180 cm; b) not less than 170 cm and not more than 175 cm.

OPTION 5

Problem 1. Discrete random variable X (CB X) is given by the distribution series:

x i
p i	0,2	0,1	0,4	0,2	0,1

Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), variance D(X), the standard deviation s ( X), fashion M 0 (X); 3) probability P(40≤ X < 80). Построить многоугольник распределения и график F(x).

Problem 2. The target consists of a circle and two concentric rings. Hitting circle gives 6 points, ring 2 gives 4 points, and hitting ring 3 gives two points. The probabilities of hitting the circle and rings 2 and 3 are, respectively, 0.2; 0.3 and 0.5. For discrete SV X - the sum of points knocked out as a result of three hits, make a distribution series and find F(x), M(X), s ( X).

Task 3. The automatic line consists of n independently working machines of the same type. The probability that the machine will require adjustment during a shift for each machine is 0.3. Required: 1) build a distribution series CB X- the number of machines that require adjustment during the shift, if n \u003d 4; 2) estimate the probability that 20 machines will require adjustment per shift, if n = 100.

Problem 4. Joint distribution of discrete CB X and Y given by the table:

Y X
	0,20	0,15	0,10
	0,30	0,20	0,05

Draw up the distribution law CB Z = Y + X... To find M(Z) and D(Z).

Task 5. Continuous CB X given by the distribution function

Find: 1) distribution density f(x); 2) M(x) and D(X);
3) P(3 < X < 9); 4) вероятность того, что в четырех независимых испытаниях CB X takes exactly two values \u200b\u200bthat belong to the interval (3; 9).

Task 6. The function is set

Determine parameter value A, at which this function defines the density of the probability distribution of some continuous CB X... To find F(x), M(X). Build a graph F(x).

Task 7. Set M(X) \u003d 10 and s ( X) \u003d 4 normally distributed continuous CB X... To find:

1) probability ;

2) probability ;

3) symmetrical with respect to a the interval in which the values \u200b\u200bfall CB X with probability g \u003d 0.5161.

Problem 8. The minute hand of the electric clock moves in leaps and bounds at the end of each minute. Random value X - the difference between the time shown on the scoreboard and the real time is evenly distributed. Find the probability that at some point in time the clock will indicate a time that differs from the true one: a) not less than 10 s and not more than 25 s; b) not less than 25 s.

OPTION 6

Problem 1. Discrete random variable X (CB X) is given by the distribution series:

x i	–5	–3	–1
p i	0,2	0,2	0,1	0,4	0,1

Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), variance D(X), the standard deviation s ( X), fashion M 0 (X); 3) probability P(– 3≤ X < 1). Построить многоугольник распределения и график F(x).

Problem 2. There are 12 students in a group, 5 of whom live in a hostel. 4 students are selected at random from the list. For SV X - the number of students living in the hostel among those who will be selected, draw up a distribution series and find F(x), M(X) and D(X).

Task 3. When manufacturing parts of the same type on outdated equipment, each part can be defective with a probability of 0.1. Plot distribution series CB X< 3);
4) the probability that in four independent tests CB X takes exactly two values \u200b\u200bthat belong to the interval (1; 3).

Task 6. The function is set

Determine parameter value A, at which this function defines the density of the probability distribution of some continuous CB X... To find F(x), M(X) and D(X). Build a graph F(x).

Task 7. Set M(X) \u003d 11 and s ( X) \u003d 3 normally distributed continuous CB X... To find:

1) probability ;

2) probability ;

3) symmetrical with respect to a the interval in which the values \u200b\u200bfall CB X with probability g \u003d 0.9973.

Task 8. The uptime of a TV of this brand is a random variable distributed according to the normal law with parameters and \u003d 12 years and s \u003d 2 years. Find the probability that the TV will work without repair: a) from 9 to 12 years;
b) at least 10 years.

OPTION 7

Problem 1. Discrete random variable X (CB X) is given by the distribution series:

x i
p i	0,2	0,1	0,2	0,3	0,2

Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), variance D(X), the standard deviation s ( X), fashion M 0 (X); 3) probability P(2≤ X < 10). Построить многоугольник распределения и график F(x).

Task 2. A worker maintains 4 independently operating machines. The probability that within an hour the machine does not require the attention of the worker for the first machine is 0.7; for the second - 0.75; for the third - 0.8; for the fourth - 0.9. For discrete SV X - the number of machines that do not require the attention of the worker for an hour, draw up a distribution series and find F(x), M(X) and D(X).

Task 3. There is n independently working machines. Plot distribution series CB X- the number of machines operating at a given time, if n \u003d 6, and the probability that the machine is running at a given time is 0.9; calculate M(X) and D(X). Evaluate the likelihood that an enterprise that has n\u003d 180 and the probability of work for each machine is 0.98, the number of machines currently working will be at least 170.

Problem 4. The laws of distribution of independent discrete CB X and Y:

x i
p i	0,3	?	0,5

y i	–2	–1
p i	?	0,4

Make a distribution series CB Z = XY + 2. Find M(Z) and D(Z).