RANDOM VALUES

Example 2.1. Random value X given by the distribution function

Find the probability that as a result of the test X will take values ​​between (2.5; 3.6).

Solution: X in the interval (2.5; 3.6) can be determined in two ways:

Example 2.2. At what values ​​of the parameters BUT and AT function F(x) = A + Be - x can be a distribution function for non-negative values ​​of a random variable X.

Solution: Since all possible values ​​of the random variable X belong to the interval , then in order for the function to be a distribution function for X, the property should hold:

.

Answer: .

Example 2.3. The random variable X is given by the distribution function

Find the probability that, as a result of four independent trials, the value X exactly 3 times will take a value belonging to the interval (0.25; 0.75).

Solution: Probability of hitting a value X in the interval (0.25; 0.75) we find by the formula:

Example 2.4. The probability of the ball hitting the basket in one throw is 0.3. Draw up the law of distribution of the number of hits in three throws.

Solution: Random value X- the number of hits in the basket with three throws - can take on the values: 0, 1, 2, 3. The probabilities that X

X:

Example 2.5. Two shooters make one shot at the target. The probability of hitting it by the first shooter is 0.5, the second - 0.4. Write down the law of distribution of the number of hits on the target.

Solution: Find the law of distribution of a discrete random variable X- the number of hits on the target. Let the event be a hit on the target by the first shooter, and - hit by the second shooter, and - respectively, their misses.

Let us compose the law of probability distribution of SV X:

Example 2.6. 3 elements are tested, working independently of each other. Duration of time (in hours) uptime elements have distribution density functions: for the first: F 1 (t) =1-e- 0,1 t, for the second: F 2 (t) = 1-e- 0,2 t, for the third one: F 3 (t) =1-e- 0,3 t. Find the probability that in the time interval from 0 to 5 hours: only one element will fail; only two elements will fail; all three elements fail.

Solution: Let's use the definition of the generating function of probabilities:

The probability that in independent trials, in the first of which the probability of occurrence of an event BUT equals , in the second, etc., the event BUT appears exactly once, is equal to the coefficient at in the expansion of the generating function in powers of . Let us find the probabilities of failure and non-failure, respectively, of the first, second and third element in the time interval from 0 to 5 hours:

Let's create a generating function:

The coefficient at is equal to the probability that the event BUT will appear exactly three times, that is, the probability of failure of all three elements; the coefficient at is equal to the probability that exactly two elements will fail; coefficient at is equal to the probability that only one element will fail.

Example 2.7. Given a probability density f(x) random variable X:

Find the distribution function F(x).

Solution: We use the formula:

.

Thus, the distribution function has the form:

Example 2.8. The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Compile the law of distribution of the number of failed elements in one experiment.

Solution: Random value X- the number of elements that failed in one experiment - can take the values: 0, 1, 2, 3. Probabilities that X takes these values, we find by the Bernoulli formula:

Thus, we obtain the following law of the probability distribution of a random variable X:

Example 2.9. There are 4 standard parts in a lot of 6 parts. 3 items were randomly selected. Draw up the law of distribution of the number of standard parts among the selected ones.

Solution: Random value X- the number of standard parts among the selected ones - can take the values: 1, 2, 3 and has a hypergeometric distribution. The probabilities that X

where -- the number of parts in the batch;

-- the number of standard parts in the lot;

– number of selected parts;

-- the number of standard parts among those selected.

.

.

.

Example 2.10. The random variable has a distribution density

where and are not known, but , a and . Find and .

Solution: In this case random value X has a triangular distribution (Simpson distribution) on the interval [ a, b]. Numerical characteristics X:

Consequently, . Deciding this system, we get two pairs of values: . Since, according to the condition of the problem, we finally have: .

Answer: .

Example 2.11. On average, for 10% of contracts, the insurance company pays the sums insured in connection with the occurrence of an insured event. Calculate expected value and the variance in the number of such contracts among four randomly selected ones.

Solution: The mathematical expectation and variance can be found using the formulas:

.

Possible values ​​of SV (number of contracts (out of four) with the occurrence of an insured event): 0, 1, 2, 3, 4.

We use the Bernoulli formula to calculate the probabilities of a different number of contracts (out of four) for which the sums insured were paid:

.

The distribution series of CV (the number of contracts with the occurrence of an insured event) has the form:

	0,6561	0,2916	0,0486	0,0036	0,0001

Answer: , .

Example 2.12. Of the five roses, two are white. Write a distribution law for a random variable expressing the number of white roses among two taken at the same time.

Solution: In a sample of two roses, there may either be no white rose, or there may be one or two white roses. Therefore, the random variable X can take values: 0, 1, 2. The probabilities that X takes these values, we find by the formula:

where -- number of roses;

-- number of white roses;

– the number of simultaneously taken roses;

-- the number of white roses among those taken.

.

.

.

Then the law of distribution of a random variable will be as follows:

Example 2.13. Among the 15 assembled units, 6 need additional lubrication. Draw up a distribution law for the number of units in need of additional lubrication, among five randomly selected from the total number.

Solution: Random value X- the number of units that need additional lubrication among the five selected - can take the values: 0, 1, 2, 3, 4, 5 and has a hypergeometric distribution. The probabilities that X takes these values, we find by the formula:

where -- the number of assembled units;

-- number of units requiring additional lubrication;

– the number of selected aggregates;

-- the number of units that need additional lubrication among the selected ones.

.

.

.

.

.

.

Then the law of distribution of a random variable will be as follows:

Example 2.14. Of the 10 watches received for repair, 7 need a general cleaning of the mechanism. Watches are not sorted by type of repair. The master, wanting to find a watch that needs cleaning, examines them one by one and, having found such a watch, stops further viewing. Find the mathematical expectation and variance of the number of hours watched.

Solution: Random value X- the number of units that need additional lubrication among the five selected - can take the following values: 1, 2, 3, 4. The probabilities that X takes these values, we find by the formula:

.

.

.

.

Then the law of distribution of a random variable will be as follows:

Now let's calculate numerical characteristics values ​​:

Answer: , .

Example 2.15. The subscriber has forgotten the last digit of the phone number he needs, but remembers that it is odd. Find the mathematical expectation and variance of the number of dials he made before hitting the desired number, if he dials the last digit at random and does not dial the dialed digit in the future.

Solution: Random variable can take values: . Since the subscriber does not dial the dialed digit in the future, the probabilities of these values ​​are equal.

Let's compose a distribution series of a random variable:

	0,2

Let's calculate the mathematical expectation and variance of the number of dialing attempts:

Answer: , .

Example 2.16. The probability of failure during the reliability tests for each device of the series is equal to p. Determine the mathematical expectation of the number of devices that failed, if tested N appliances.

Solution: Discrete random variable X is the number of failed devices in N independent tests, in each of which the probability of failure is equal to p, distributed according to the binomial law. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of an event occurring in one trial:

Example 2.17. Discrete random variable X takes 3 possible values: with probability ; with probability and with probability . Find and knowing that M( X) = 8.

Solution: We use the definitions of mathematical expectation and the law of distribution of a discrete random variable:

We find: .

Example 2.18. The technical control department checks products for standardity. The probability that the item is standard is 0.9. Each batch contains 5 items. Find the mathematical expectation of a random variable X- the number of batches, each of which contains exactly 4 standard products, if 50 batches are subject to verification.

Solution: In this case, all experiments conducted are independent, and the probabilities that each batch contains exactly 4 standard products are the same, therefore, the mathematical expectation can be determined by the formula:

,

where is the number of parties;

The probability that a batch contains exactly 4 standard items.

We find the probability using the Bernoulli formula:

Answer: .

Example 2.19. Find the variance of a random variable X– number of occurrences of the event A in two independent trials, if the probabilities of the occurrence of an event in these trials are the same and it is known that M(X) = 0,9.

Solution: The problem can be solved in two ways.

1) Possible CB values X: 0, 1, 2. Using the Bernoulli formula, we determine the probabilities of these events:

, , .

Then the distribution law X looks like:

From the definition of mathematical expectation, we determine the probability:

Let's find the variance of SW X:

.

2) You can use the formula:

.

Answer: .

Example 2.20. Mathematical expectation and standard deviation of a normally distributed random variable X are 20 and 5, respectively. Find the probability that as a result of the test X will take the value contained in the interval (15; 25).

Solution: Probability of hitting a normal random variable X on the section from to is expressed in terms of the Laplace function:

Example 2.21. Given a function:

At what value of the parameter C this function is the distribution density of some continuous random variable X? Find the mathematical expectation and variance of a random variable X.

Solution: In order for a function to be the distribution density of some random variable , it must be non-negative, and it must satisfy the property:

.

Consequently:

Calculate the mathematical expectation using the formula:

.

Calculate the variance using the formula:

T is p. It is necessary to find the mathematical expectation and variance of this random variable.

Solution: The distribution law of a discrete random variable X - the number of occurrences of an event in independent trials, in each of which the probability of the occurrence of an event is , is called binomial. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of the occurrence of the event A in one trial:

.

Example 2.25. Three independent shots are fired at the target. The probability of hitting each shot is 0.25. Determine the standard deviation of the number of hits with three shots.

Solution: Since three independent trials are performed, and the probability of occurrence of the event A (hit) in each trial is the same, we will assume that the discrete random variable X - the number of hits on the target - is distributed according to the binomial law.

The variance of the binomial distribution is equal to the product of the number of trials and the probabilities of occurrence and non-occurrence of an event in one trial:

Example 2.26. The average number of clients visiting the insurance company in 10 minutes is three. Find the probability that at least one customer arrives in the next 5 minutes.

Average number of customers arriving in 5 minutes: . .

Example 2.29. The waiting time for an application in the processor queue obeys an exponential distribution law with an average value of 20 seconds. Find the probability that the next (arbitrary) request will wait for the processor for more than 35 seconds.

Solution: In this example, the expectation , and the failure rate is .

Then the desired probability is:

Example 2.30. A group of 15 students holds a meeting in a hall with 20 rows of 10 seats each. Each student takes a seat in the hall at random. What is the probability that no more than three people will be in the seventh place in the row?

Solution:

Example 2.31.

Then according to the classical definition of probability:

where -- the number of parts in the batch;

-- the number of non-standard parts in the lot;

– number of selected parts;

-- the number of non-standard parts among the selected ones.

Then the distribution law of the random variable will be as follows.

As is known, random variable is called a variable that can take on certain values ​​depending on the case. Random variables are denoted by capital letters of the Latin alphabet (X, Y, Z), and their values ​​- by the corresponding lowercase letters (x, y, z). Random variables are divided into discontinuous (discrete) and continuous.

Discrete random variable is called a random variable that takes only a finite or infinite (countable) set of values ​​with certain non-zero probabilities.

The distribution law of a discrete random variable is a function that connects the values ​​of a random variable with their corresponding probabilities. The distribution law can be specified in one of the following ways.

1 . The distribution law can be given by the table:

where λ>0, k = 0, 1, 2, … .

in) by using distribution function F(x) , which determines for each value x the probability that the random variable X takes a value less than x, i.e. F(x) = P(X< x).

Properties of the function F(x)

3 . The distribution law can be set graphically – distribution polygon (polygon) (see problem 3).

Note that in order to solve some problems, it is not necessary to know the distribution law. In some cases, it is enough to know one or more numbers that reflect the most important features of the distribution law. It can be a number that has the meaning of the "average value" of a random variable, or a number that shows the average size of the deviation of a random variable from its average value. Numbers of this kind are called numerical characteristics of a random variable.

Basic numerical characteristics of a discrete random variable :

• Mathematical expectation (mean value) of a discrete random variable M(X)=Σ x i p i.
  For binomial distribution M(X)=np, for Poisson distribution M(X)=λ
• Dispersion discrete random variable D(X)=M2 or D(X) = M(X 2) − 2. The difference X–M(X) is called the deviation of a random variable from its mathematical expectation.
  For binomial distribution D(X)=npq, for Poisson distribution D(X)=λ
• Standard deviation (standard deviation) σ(X)=√D(X).

Examples of solving problems on the topic "The law of distribution of a discrete random variable"

Task 1.

1000 lottery tickets have been issued: 5 of them win 500 rubles, 10 - 100 rubles, 20 - 50 rubles, 50 - 10 rubles. Determine the law of probability distribution of the random variable X - winnings per ticket.

Solution. According to the condition of the problem, the following values ​​of the random variable X are possible: 0, 10, 50, 100 and 500.

The number of tickets without winning is 1000 - (5+10+20+50) = 915, then P(X=0) = 915/1000 = 0.915.

Similarly, we find all other probabilities: P(X=0) = 50/1000=0.05, P(X=50) = 20/1000=0.02, P(X=100) = 10/1000=0.01 , P(X=500) = 5/1000=0.005. We present the resulting law in the form of a table:

Find the mathematical expectation of X: M(X) = 1*1/6 + 2*1/6 + 3*1/6 + 4*1/6 + 5*1/6 + 6*1/6 = (1+ 2+3+4+5+6)/6 = 21/6 = 3.5

Task 3.

The device consists of three independently operating elements. The probability of failure of each element in one experiment is 0.1. Draw up a distribution law for the number of failed elements in one experiment, build a distribution polygon. Find the distribution function F(x) and plot it. Find the mathematical expectation, variance and standard deviation of a discrete random variable.

Solution. 1. Discrete random variable X=(number of failed elements in one experiment) has the following possible values: x 1 =0 (none of the elements of the device failed), x 2 =1 (one element failed), x 3 =2 (two elements failed ) and x 4 \u003d 3 (three elements failed).

Failures of elements are independent of each other, the probabilities of failure of each element are equal to each other, therefore, it is applicable Bernoulli's formula . Given that, by condition, n=3, p=0.1, q=1-p=0.9, we determine the probabilities of the values:
P 3 (0) \u003d C 3 0 p 0 q 3-0 \u003d q 3 \u003d 0.9 3 \u003d 0.729;
P 3 (1) \u003d C 3 1 p 1 q 3-1 \u003d 3 * 0.1 * 0.9 2 \u003d 0.243;
P 3 (2) \u003d C 3 2 p 2 q 3-2 \u003d 3 * 0.1 2 * 0.9 \u003d 0.027;
P 3 (3) \u003d C 3 3 p 3 q 3-3 \u003d p 3 \u003d 0.1 3 \u003d 0.001;
Check: ∑p i = 0.729+0.243+0.027+0.001=1.

Thus, the desired binomial distribution law X has the form:

On the abscissa axis, we plot the possible values ​​x i, and on the ordinate axis, the corresponding probabilities р i . Let's construct points M 1 (0; 0.729), M 2 (1; 0.243), M 3 (2; 0.027), M 4 (3; 0.001). Connecting these points with line segments, we obtain the desired distribution polygon.

3. Find the distribution function F(x) = P(X

For x ≤ 0 we have F(x) = P(X<0) = 0;
for 0< x ≤1 имеем F(x) = Р(Х<1) = Р(Х = 0) = 0,729;
for 1< x ≤ 2 F(x) = Р(Х<2) = Р(Х=0) + Р(Х=1) =0,729+ 0,243 = 0,972;
for 2< x ≤ 3 F(x) = Р(Х<3) = Р(Х = 0) + Р(Х = 1) + Р(Х = 2) = 0,972+0,027 = 0,999;
for x > 3 it will be F(x) = 1, because the event is certain.

Graph of the function F(x)

4. For the binomial distribution X:
- mathematical expectation М(X) = np = 3*0.1 = 0.3;
- dispersion D(X) = npq = 3*0.1*0.9 = 0.27;
- standard deviation σ(X) = √D(X) = √0.27 ≈ 0.52.

Exercise 1. The distribution density of a continuous random variable X has the form:
Find:
a) parameter A ;
b) distribution function F(x) ;
c) the probability of hitting a random variable X in the interval ;
d) mathematical expectation MX and variance DX .
Plot the functions f(x) and F(x) .

Task 2. Find the variance of the random variable X given by the integral function.

Task 3. Find the mathematical expectation of a random variable X given a distribution function.

Task 4. The probability density of some random variable is given as follows: f(x) = A/x 4 (x = 1; +∞)
Find coefficient A , distribution function F(x) , mathematical expectation and variance, as well as the probability that a random variable takes a value in the interval . Plot f(x) and F(x) graphs.

A task. The distribution function of some continuous random variable is given as follows:

Determine the parameters a and b , find the expression for the probability density f(x) , the mathematical expectation and variance, as well as the probability that the random variable will take a value in the interval . Plot f(x) and F(x) graphs.

Let's find the distribution density function as a derivative of the distribution function.
F′=f(x)=a
Knowing that we will find the parameter a:

or 3a=1, whence a = 1/3
We find the parameter b from the following properties:
F(4) = a*4 + b = 1
1/3*4 + b = 1 whence b = -1/3
Therefore, the distribution function is: F(x) = (x-1)/3

Expected value.


Dispersion.

1 / 9 4 3 - (1 / 9 1 3) - (5 / 2) 2 = 3 / 4
Find the probability that a random variable takes a value in the interval
P(2< x< 3) = F(3) – F(2) = (1/3*3 - 1/3) - (1/3*2 - 1/3) = 1/3

Example #1. The probability distribution density f(x) of a continuous random variable X is given. Required:

1. Determine coefficient A .
2. find the distribution function F(x) .
3. schematically plot F(x) and f(x) .
4. find the mathematical expectation and variance of X .
5. find the probability that X takes a value from the interval (2;3).

f(x) = A*sqrt(x), 1 ≤ x ≤ 4.
Solution:

The random variable X is given by the distribution density f(x):

Find the parameter A from the condition:



or
14/3*A-1=0
Where,
A = 3 / 14

The distribution function can be found by the formula.

………………………………………………………

An - a random variable X has taken the value of An.

Obviously, the sum of events A1 A2, . , An is a certain event, since the random variable necessarily takes at least one of the values ​​x1, x2, xn.

Therefore, P (A1 È A2 È . È An) = 1.

In addition, the events A1, A2, ., An are incompatible, since a random variable in a single experiment can take only one of the values ​​x1, x2, ., xn. By the addition theorem for incompatible events, we obtain

P(A1)+P(A2)+ .+P(An)=1,

i.e. p1+p2+ . +pn = 1, or, in short,

Therefore, the sum of all numbers located in the second row of Table 1, which gives the distribution law of the random variable X, must be equal to one.

EXAMPLE 1. Let the random variable X be the number of points rolled when a die is rolled. Find the distribution law (in the form of a table).

Random variable X takes values

x1=1, x2=2, … , x6=6

with probabilities

p1= p2 = … = p6 =

The distribution law is given by the table:

table 2

EXAMPLE 2. Binomial distribution. Consider a random variable X - the number of occurrences of event A in a series of independent experiments, in each of which A occurs with probability p.

The random variable X can obviously take one of the following values:

0, 1, 2, ., k, ., n.

The probability of an event consisting in the fact that the random variable X will take on a value equal to k is determined by the Bernoulli formula:

Рn(k)= where q=1- р.

Such a distribution of a random variable is called the binomial distribution or Bernoulli distribution. The Bernoulli distribution is completely specified by two parameters: the number n of all trials and the probability p with which the event occurs in each individual trial.

The condition for the binomial distribution takes the form:

To prove the validity of this equality, it suffices in the identity

(q+px)n=

put x=1.

EXAMPLE 3. Poisson distribution. This is the name of the probability distribution of the form:

P(k)= .

It is determined by a single (positive) parameter a. If ξ is a random variable that has a Poisson distribution, then the corresponding parameter a - is the average value of this random variable:

a=Mξ=, where M is the mathematical expectation.

The random variable is:

EXAMPLE 4. exponential distribution.

If time is a random variable, let's denote it by τ, such that

where 0<λ=const, t ³ 0, причем, если t=0, то P(t)=0.

The mean value of the random variable t is:

The distribution density has the form:

4) Normal distribution

Let be independent, identically distributed random variables and let If the terms are small enough, and the number n is large enough, - if for n à ∞ the mathematical expectation of the random variable Мξ and the variance Dξ equal to Dξ=M(ξ–Мξ)2, are such that Мξ~а, Dξ~σ2, then

- normal or gaussian distribution

.

5) Geometric distribution. Let ξ denote the number of trials preceding the first "success". If we assume that each test lasts a unit of time, then we can consider ξ as the waiting time until the first "success". The distribution looks like:

Р(k)=p(1-p)k, (k=0, 1, 2) p>0

6) Hypergeometric distribution.

There are N - objects among which n - "special objects". Among all objects, k-objects are randomly selected. Find the probability that among the selected objects is equal to r - "special objects". The distribution looks like:

7) Pascal distribution.

Let x be the total number of "failures" preceding the arrival of the rth "success". The distribution looks like:

The distribution function has the form:

An equiprobable distribution implies that the random variable x can take any value on the interval with the same probability. In this case, the distribution density is calculated as

Plots of distribution density and distribution function are presented below.

Before explaining the concept of "white noise", it is necessary to give a number of definitions.

A random function is a function of a non-random argument t, which, for each fixed value of the argument, is a random variable. For example, if U is a random variable, then the function X(t)=t2U is random.

The section of a random function is a random variable corresponding to a fixed value of the argument of the random function. Thus, a random function can be considered as a set of random variables (X(t)), depending on the parameter t.

Distribution density probabilities X call the function f(x) is the first derivative of the distribution function F(x):

The concept of the probability distribution density of a random variable X for a discrete quantity is not applicable.

Probability density f(x) is called the differential distribution function:

Property 1. The distribution density is a non-negative value:

Property 2. The improper integral of the distribution density in the range from to is equal to one:

Example 1.25. Given the distribution function of a continuous random variable X:

f(x).

Solution: The distribution density is equal to the first derivative of the distribution function:

1. Given the distribution function of a continuous random variable X:

Find the distribution density.

2. The distribution function of a continuous random variable is given X:

Find the distribution density f(x).

1.3. Numerical characteristics of continuous random

quantities

Expected value continuous random variable X, the possible values ​​of which belong to the entire axis Oh, is determined by the equality:

It is assumed that the integral converges absolutely.

a,b), then:

f(x) is the distribution density of the random variable.

Dispersion continuous random variable X, the possible values ​​of which belong to the entire axis, is determined by the equality:

Special case. If the values ​​of the random variable belong to the interval ( a,b), then:

The probability that X will take values ​​belonging to the interval ( a,b), is determined by the equality:

.

Example 1.26. Continuous random variable X

Find the mathematical expectation, variance and probability of hitting a random variable X in the interval (0; 0.7).

Solution: The random variable is distributed over the interval (0,1). Let us define the distribution density of a continuous random variable X:

a) Mathematical expectation :

b) Dispersion

in)

Tasks for independent work:

1. Random variable X given by the distribution function:

M(x);

b) dispersion D(x);

X into the interval (2,3).

2. Random variable X

Find: a) mathematical expectation M(x);

b) dispersion D(x);

c) determine the probability of hitting a random variable X in the interval (1; 1.5).

3. Random value X is given by the integral distribution function:

Find: a) mathematical expectation M(x);

b) dispersion D(x);

c) determine the probability of hitting a random variable X in the interval.

1.4. Laws of distribution of a continuous random variable

1.4.1. Uniform distribution

Continuous random variable X has a uniform distribution on the interval [ a,b], if on this segment the density of the probability distribution of a random variable is constant, and outside it is equal to zero, i.e.:

Rice. four.

; ; .

Example 1.27. A bus of some route moves uniformly with an interval of 5 minutes. Find the probability that a uniformly distributed random variable X– the waiting time for the bus will be less than 3 minutes.

Solution: Random value X- uniformly distributed over the interval .

Probability Density: .

In order for the waiting time not to exceed 3 minutes, the passenger must arrive at the bus stop within 2 to 5 minutes after the departure of the previous bus, i.e. random value X must fall within the interval (2;5). That. desired probability:

Tasks for independent work:

1. a) find the mathematical expectation of a random variable X distributed uniformly in the interval (2; 8);

b) find the variance and standard deviation of a random variable X, distributed uniformly in the interval (2;8).

2. The minute hand of an electric clock jumps at the end of each minute. Find the probability that at a given moment the clock will show the time that differs from the true time by no more than 20 seconds.

1.4.2. The exponential (exponential) distribution

Continuous random variable X is exponentially distributed if its probability density has the form:

where is the parameter of the exponential distribution.

In this way

Rice. 5.

Numerical characteristics:

Example 1.28. Random value X- the operating time of the light bulb - has an exponential distribution. Determine the probability that the lamp will last at least 600 hours if the average lamp life is 400 hours.

Solution: According to the condition of the problem, the mathematical expectation of a random variable X equals 400 hours, so:

;

The desired probability , where

Finally:

Tasks for independent work:

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

2. Random variable X

Find the mathematical expectation and variance of a quantity X.

3. Random value X given by the probability distribution function:

Find the mathematical expectation and standard deviation of a random variable.

1.4.3. Normal distribution

normal is called the probability distribution of a continuous random variable X, whose density has the form:

where a– mathematical expectation, – standard deviation X.

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

, where

is the Laplace function.

A distribution that has ; , i.e. with a probability density called standard.

Rice. 6.

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

.

In particular, when a= 0 equality is true:

Example 1.29. Random value X distributed normally. Standard deviation . Find the probability that the deviation of a random variable from its mathematical expectation in absolute value will be less than 0.3.

Solution: .

Tasks for independent work:

1. Write the probability density of the normal distribution of a random variable X, knowing that M(x)= 3, D(x)= 16.

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

3. Random measurement errors are subject to the normal law with standard deviation mm and mathematical expectation a= 0. Find the probability that the error of at least one of 3 independent measurements does not exceed 4 mm in absolute value.

4. 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 the weighing will be carried out with an error not exceeding 10 g in absolute value.