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 A and V 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. Durations of time (in hours) of failure-free operation of 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 A equals , in the second, etc., the event A appears exactly once, is equal to the coefficient at in the expansion of the generating function in powers of . Let's 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 A 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 lot;

-- 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, the random variable X has a triangular distribution (Simpson distribution) on the interval [ a, b]. Numerical characteristics X:

Hence, . 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 different number contracts (out of four) under 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 the law of distribution of 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 the numerical characteristics of the quantity :

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 expectation and variance of the number of phone calls he made before hitting desired number, if he dials the last digit at random, but 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 average standard deviation 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:

.

Hence:

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 lot;

-- 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.

Unlike a discrete random variable, continuous random variables cannot be specified in the form of a table of its distribution law, since it is impossible to list and write out all its values ​​in a certain sequence. One of possible ways setting a continuous random variable is to use the distribution function.

DEFINITION. The distribution function is a function that determines the probability that a random variable will take on a value that is depicted on the real axis by a point lying to the left of the dot x, i.e.

Sometimes, instead of the term "Distribution function", the term "Integral function" is used.

Distribution function properties:

1. The value of the distribution function belongs to the segment: 0F(x)1
2. F(x) is a non-decreasing function, i.e. F(x 2)F(x 1) if x 2 >x 1

Corollary 1. The probability that a random variable will take a value contained in the interval (a, b) is equal to the increment of the distribution function on this interval:

P(aX

Example 9. A random variable X is given by a distribution function:

Find the probability that, as a result of the test, X will take a value belonging to the interval (0; 2): P(0

Solution: Since on the interval (0;2) by condition, F(x)=x/4+1/4, then F(2)-F(0)=(2/4+1/4)-(0 /4+1/4)=1/2. So P(0

Corollary 2. The probability that a continuous random variable X takes one definite value is equal to zero.

Corollary 3. If the possible values ​​of a random variable belong to the interval (a;b), then: 1) F(x)=0 for xa; 2) F(x)=1 for xb.
The following limit relations are valid:

The graph of the distribution function is located in the strip bounded by straight lines y=0, y=1 (the first property). As x increases in the interval (a;b), which contains all possible values ​​of the random variable, the graph "rises up". For xa, the ordinates of the graph are equal to zero; at xb, the ordinates of the graph are equal to one:

Picture 1

Example 10. A discrete random variable X is given by a distribution table:

X	1	4	8
P	0.3	0.1	0.6

Find the distribution function and build its graph.
Solution: The distribution function can be analytically written as follows:

Figure-2

DEFINITION: The probability distribution density of a continuous random variable X is the function f (x) - the first derivative of the distribution function F (x): f (x) \u003d F "(x)

It follows from this definition that the distribution function is the antiderivative of the distribution density.

Theorem. The probability that a continuous random variable X will take a value belonging to the interval (a; b) is equal to a certain integral of the distribution density, taken in the range from a to b:

(8)

Probability density properties:

1. The probability density is a non-negative function: f(x)0.
2. The definite integral from -∞ to +∞ of the probability distribution density of a continuous random variable is equal to 1: f(x)dx=1.
3. The definite integral from -∞ to x of the probability distribution density of a continuous random variable is equal to the distribution function of this variable: f(x)dx=F(x)

Example 11. Given the probability distribution density of a random variable X

Find the probability that, as a result of the test, X will take a value belonging to the interval (0.5; 1).

Solution: Desired probability:

Let us extend the definition of the numerical characteristics of discrete quantities to continuous quantities. Let a continuous random variable X be given by the distribution density f(x).

DEFINITION. The mathematical expectation of a continuous random variable X, the possible values ​​of which belong to the segment , is called a definite integral:

M(x)=xf(x)dx (9)

If the possible values ​​belong to the entire x-axis, then:

M(x)=xf(x)dx (10)

The mode M 0 (X) of a continuous random variable X is its possible value, which corresponds to the local maximum of the distribution density.

The median M e (X) of a continuous random variable X is its possible value, which is determined by the equality:

P(X e (X))=P(X>M e (X))

DEFINITION. The dispersion of a continuous random variable is the mathematical expectation of the square of its deviation. If the possible values ​​of X belong to the segment , then:

D(x)=2 f(x)dx (11)
or
D(x)=x 2 f(x)dx- 2 (11*)

If the possible values ​​belong to the entire x-axis, then.

9. Continuous random variable, its numerical characteristics

A continuous random variable can be specified using two functions. Integral probability distribution function of a random variable X is called the function defined by the equality
.

The integral function provides a general way of specifying both discrete and continuous random variables. In the case of a continuous random variable . All events: have the same probability equal to the increment of the integral function on this interval, i.e. For example, for a discrete random variable given in example 26, we have:

Thus, the graph of the integral function of the function under consideration is the union of two rays and three segments parallel to the Ox axis.

Example 27. A continuous random variable X is given by the integral probability distribution function

.

Construct a graph of the integral function and find the probability that, as a result of the test, the random variable X will take a value in the interval (0.5; 1.5).

Solution. On the interval
the graph is the straight line y \u003d 0. On the interval from 0 to 2 - the parabola given by the equation
. On the interval
the graph is the straight line y = 1.

The probability that the random variable X as a result of the test will take a value in the interval (0.5; 1.5) is found by the formula .

In this way, .

Properties of the integral probability distribution function:

The distribution law of a continuous random variable is conveniently specified using another function, namely, probability density functions
.

The probability that the value taken by the random variable X falls within the interval
, is determined by the equality
.

The graph of the function is called distribution curve. Geometrically, the probability of a random variable X falling into the interval is equal to the area of ​​the corresponding curvilinear trapezoid, bounded by the distribution curve, the Ox axis and straight lines
.

Probability density function properties:

9.1. Numerical characteristics of continuous random variables

Expected value(average value) of a continuous random variable X is defined by the equality
.

M(X) is denoted by a. The mathematical expectation of a continuous random variable has properties similar to those of a discrete variable:

dispersion discrete random variable X is the mathematical expectation of the squared deviation of the random variable from its mathematical expectation, i.e. . For a continuous random variable, the variance is given by
.

The dispersion has the following properties:

The last property is very convenient to apply to finding the variance of a continuous random variable.

The concept of standard deviation is introduced similarly. RMS continuous random variable X is called the square root of the variance, i.e.
.

Example 28. A continuous random variable X is given by a probability density function
in the interval (10;12), outside this interval the value of the function is 0. Find 1) the value of the parameter a, 2) mathematical expectation M(X), variance
, standard deviation, 3) integral function
and build graphs of the integral and differential functions.

one). To find the parameter a use the formula
. We get . In this way,
.

2). To find the mathematical expectation, we use the formula: , whence it follows that
.

We will find the dispersion using the formula:
, i.e. .

Let's find the standard deviation by the formula: , from where we get that
.

3). The integral function is expressed in terms of the probability density function as follows:
. Hence,
at
, = 0 for
and = 1 at
.

Graphs of these functions are presented in fig. 4. and fig. 5.

Fig.4 Fig.5.

9.2. Uniform probability distribution of a continuous random variable

Probability distribution of a continuous random variable X evenly on the interval if its probability density is constant on this interval and equals zero outside this interval, i.e. . It is easy to show that in this case
.

If the interval
is contained in the interval, then
.

Example 29. An instant signal event must occur between 1 p.m. and 5 p.m. The signal waiting time is a random variable X. Find the probability that the signal will be fixed between two and three o'clock in the afternoon.

Solution. The random variable X has a uniform distribution, and by the formula we find that the probability that the signal will be between 2 and 3 o'clock in the afternoon is equal to
.

In educational and other literature, it is often denoted in the literature through
.

9.3. Normal probability distribution of a continuous random variable

The probability distribution of a continuous random variable is called normal if its probability distribution law is determined by the probability density
. For such quantities a- expected value,
- standard deviation.

Theorem. Probability of a normally distributed continuous random variable falling into a given interval
is determined by the formula
, where
is the Laplace function.

A consequence of this theorem is the three sigma rule, i.e. it is almost certain that a normally distributed, continuous random variable X takes its values ​​in the interval
. This rule is derived from the formula
, which is a particular case of the formulated theorem.

Example 30. The life of the TV is a random variable X, subject to the normal distribution law, with a warranty period of 15 years and a standard deviation of 3 years. Find the probability that the TV will last from 10 to 20 years.

Solution. According to the condition of the problem, the mathematical expectation a= 15, standard deviation .

Let's find . Thus, the probability of a TV operating from 10 to 20 years is more than 0.9.

9.4 Chebyshev's inequality

Takes place Chebyshev's lemma. If a random variable X takes only non-negative values ​​and has a mathematical expectation, then for any positive v
.

Taking into account that , as the sum of the probabilities of opposite events, we get that
.

Chebyshev's theorem. If the random variable X has a finite variance
and mathematical expectation M(X), then for any positive the inequality
.

Whence it follows that
.

Example 31. A batch of parts has been made. The average length of the parts is 100 cm, and the standard deviation is 0.4 cm. Estimate from below the probability that the length of a part taken at random will be at least 99 cm. and not more than 101 cm.

Solution. dispersion. The mathematical expectation is 100. Therefore, to estimate the probability of the considered event from below
we apply the Chebyshev inequality, in which
, then
.

10. Elements of mathematical statistics

Statistical population name a set of homogeneous objects or phenomena. Number P elements of this set is called the volume of the set. Observed values feature X is called options. If the options are in ascending order, then discrete variation series. In the case of grouping, the option by intervals is obtained interval variation series. Under frequency t feature values ​​understand the number of members of the population with a given variant.

The ratio of the frequency to the size of the statistical population is called relative frequency sign:
.

The ratio between the variants of the variational series and their frequencies is called statistical distribution of the sample. A graphical representation of a statistical distribution can be polygon frequencies.

Example 32. By interviewing 25 first-year students, the following data on their age was obtained:
. Compile a statistical distribution of students by age, find the range of variation, build a frequency polygon and compile a series of distributions of relative frequencies.

Solution. Using the data obtained during the survey, we will compose the statistical distribution of the sample

The range of the variation sample is 23 - 17 = 6. To build a frequency polygon, build points with coordinates
and connect them in series.

The distribution series of relative frequencies has the form:

10.1. Numerical characteristics of the variation series

Let the sample be given by the frequency distribution series of feature X:

The sum of all frequencies is P.

Arithmetic mean of the sample call the quantity
.

dispersion or a measure of the dispersion of the values ​​of the attribute X in relation to its arithmetic mean is the value
. The standard deviation is called the square root of the dispersion, i.e. .

The ratio of the standard deviation to the arithmetic mean of the sample, expressed as a percentage, is called coefficient of variation:
.

Empirical relative frequency distribution function call a function that determines for each value the relative frequency of an event
, i.e.
, where - number of options, smaller X, a P- sample size.

Example 33. In the conditions of example 32, find numerical characteristics
.

Solution. Find the arithmetic mean of the sample using the formula , then .

The variance of the attribute X is found by the formula: , i.e. . The standard deviation of the sample is
. The coefficient of variation is
.

10.2. Probability estimation by relative frequency. Confidence interval

Let it be held P independent trials, in each of which the probability of occurrence of event A is constant and equal to R. In this case, the probability that the relative frequency will differ from the probability of the occurrence of event A in each test in absolute value by no more than , is approximately equal to twice the value of the Laplace integral function:
.

interval estimation call such an assessment, which is determined by two numbers that are the ends of the interval covering the estimated parameter of the statistical population.

Confidence intervalis called the interval, which with a given confidence probability covers the estimated parameter of the statistical population. Considering the formula in which we replace the unknown quantity R to its approximate value obtained from the sample data, we get:
. This formula is used to estimate the probability by relative frequency. Numbers
and
called the lower and respectively the upper trust boundaries, - marginal error for a given confidence level
.

Example 34. The factory floor produces electric light bulbs. When checking 625 lamps, 40 were defective. Find, with a confidence probability of 0.95, the boundaries in which the percentage of defective light bulbs produced by the factory shop is concluded.

Solution. According to the task. We use the formula
. According to table 2 of the appendix, we find the value of the argument, pi in which the value of the integral Laplace function is 0.475. We get that
. In this way, . Therefore, it can be said with a probability of 0.95 that the share of defects produced by the workshop is high, namely, it varies from 6.2% to 6.6%.

10.3. Estimation of parameters in statistics

Let the quantitative attribute X of the entire study population (general population) have a normal distribution.

If the standard deviation is known, then the confidence interval covering the mathematical expectation a

, where P is the sample size, - sample arithmetic mean, t is the argument of the integral Laplace function, for which
. At the same time, the number
is called the estimation accuracy.

If the standard deviation is unknown, then according to the sample data, it is possible to construct a random variable that has a Student's distribution with P– 1 degrees of freedom, which is determined by only one parameter P and does not depend on the unknown a and . Student's distribution even for small samples
gives quite satisfactory estimates. Then the confidence interval covering the mathematical expectation a of this feature with a given confidence probability , is found from the condition

, where S is the corrected root mean square, - Student's coefficient, is found according to the data
from table 3 of the appendix.

The confidence interval covering the standard deviation of this feature with a confidence probability , is found by the formulas: and , where
is in the table of values q according to .

10.4. Statistical methods for studying dependencies between random variables

The correlation dependence of Y on X is the functional dependence of the conditional average from X. The equation
represents the regression equation of Y on X, and
- regression equation X on Y.

Correlation dependence can be linear and curvilinear. In the case of a linear correlation dependence, the regression straight line equation has the form:
, where slope a straight regression line Y on X is called the sample regression coefficient Y on X and is denoted
.

For small samples, the data is not grouped, the parameters
are found by the least squares method from the system of normal equations:

, where P is the number of observations of the values ​​of pairs of interrelated quantities.

Sample linear correlation coefficient shows the tightness of the relationship between Y and X. The correlation coefficient is found by the formula
, moreover
, namely:

The sample equation of the straight line regression Y on X has the form:

.

With a large number of observations of signs X and Y, a correlation table is compiled with two inputs, with the same value X observed times, same value at observed times, the same pair
observed once.

Example 35. A table of observations of signs X and Y is given.

Find the sample equation of the straight line regression Y on X.

Solution. The relationship between the studied traits can be expressed by the equation of a straight line of regression Y on X: . To calculate the coefficients of the equation, we will compile a calculation table:

observation number

To find the distribution function of a discrete random variable you need to use this calculator. 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.

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.

Knowing that

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.

Numerical characteristics of continuous random variables. Let a continuous random variable X be given by the distribution function f(x)

Let a continuous random variable X be given by the distribution function f(x). Let us assume that all possible values ​​of the random variable belong to the interval [ a,b].

Definition. mathematical expectation continuous random variable X, the possible values ​​of which belong to the segment , is called a definite integral

If the possible values ​​of a random variable are considered on the entire number axis, then the mathematical expectation is found by the formula:

In this case, of course, it is assumed that the improper integral converges.

Definition. dispersion continuous random variable is called the mathematical expectation of the square of its deviation.

By analogy with the variance of a discrete random variable, the following formula is used for the practical calculation of the variance:

Definition. Standard deviation is called the square root of the variance.

Definition. Fashion M 0 of a discrete random variable is called its most probable value. For a continuous random variable, the mode is the value of the random variable at which the distribution density has a maximum.

If the distribution polygon for a discrete random variable or the distribution curve for a continuous random variable has two or more maxima, then such a distribution is called bimodal or multimodal. If a distribution has a minimum but no maximum, then it is called antimodal.

Definition. Median M D of a random variable X is its value, relative to which it is equally likely to obtain a larger or smaller value of the random variable.

Geometrically, the median is the abscissa of the point at which the area bounded by the distribution curve is divided in half. Note that if the distribution is unimodal, then the mode and median coincide with the mathematical expectation.

Definition. Starting moment order k random variable X is called the mathematical expectation of X k.

For a discrete random variable: .

.

The initial moment of the first order is equal to the mathematical expectation.

Definition. Central moment order k random variable X is called the mathematical expectation of the value

For a discrete random variable: .

For a continuous random variable: .

The first order central moment is always zero, and the second order central moment is equal to the dispersion. The central moment of the third order characterizes the asymmetry of the distribution.

Definition. The ratio of the central moment of the third order to the standard deviation in the third degree is called asymmetry coefficient.

Definition. To characterize the sharpness and flatness of the distribution, a quantity called kurtosis.

In addition to the quantities considered, the so-called absolute moments are also used:

Absolute starting moment: . Absolute central moment: . The absolute central moment of the first order is called arithmetic mean deviation.

Example. For the example considered above, determine the mathematical expectation and variance of the random variable X.

Example. An urn contains 6 white and 4 black balls. A ball is removed from it five times in a row, and each time the ball taken out is returned back and the balls are mixed. Taking the number of extracted white balls as a random variable X, draw up the law of distribution of this quantity, determine its mathematical expectation and variance.

Because balls in each experiment are returned back and mixed, then the trials can be considered independent (the result of the previous experiment does not affect the probability of occurrence or non-occurrence of an event in another experiment).

Thus, the probability of a white ball appearing in each experiment is constant and equal to

Thus, as a result of five successive trials, the white ball may not appear at all, appear once, twice, three, four or five times. To draw up a distribution law, you need to find the probabilities of each of these events.

1) The white ball did not appear at all:

2) The white ball appeared once:

3) The white ball will appear twice: .