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 the values \u200b\u200benclosed in the interval (2.5; 3.6).

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

Example 2.2. At what values \u200b\u200bof the parameters AND and AT function F(x) \u003d A + Be - x can be a distribution function for non-negative values \u200b\u200bof a random variable X.

Decision: Since all possible values \u200b\u200bof the random variable Xbelong to an interval, then in order for the function to be a distribution function for X, the property should be executed:

.

Answer: .

Example 2.3. Random variable X is given by the distribution function

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

Decision:The 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 with one throw is 0.3. Draw up the distribution law of the number of hits with three throws.

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

X:

Example 2.5. The two shooters fire one shot at the target. The probability of hitting it by the first shooter is 0.5, the second - 0.4. Draw up the distribution law for the number of hits on the target.

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

Let's compose the law of probability distribution of SV X:

Example 2.6. 3 elements are tested, working independently of each other. The time durations (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: 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 will fail.

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

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

Let's compose the generating function:

The coefficient at is equal to the probability that the event ANDwill 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) of a random variable X:

Find the distribution function F (x).

Decision: 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. Draw up the distribution law for the number of failed elements in one experiment.

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

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

Example 2.9. There are 4 standard parts in a batch of 6 parts. Three parts were chosen at random. Draw up the distribution law of the number of standard parts among the selected ones.

Decision:Random value X - the number of standard parts among the selected ones - can take 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 batch;

– number of selected parts;

-- number of standard parts selected.

.

.

.

Example 2.10. The random variable has a distribution density

and not known, but, and. Find and.

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

Consequently, ... Solving 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, the insurance company pays out the sums insured for 10% of contracts in connection with the occurrence of an insured event. Calculate the mathematical expectation and variance of the number of such contracts among the randomly selected four.

Decision:The mathematical expectation and variance can be found by the formulas:

.

Possible values \u200b\u200bof 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), under which the sums insured were paid:

.

The series of distribution of SV (the number of contracts with the occurrence of an insured event) is as follows:

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

Answer:,.

Example 2.12. Of the five roses, two are white. Draw up the distribution law of a random variable expressing the number of white roses among two simultaneously taken.

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

where -- number of roses;

-- number of white roses;

– the number of roses taken at the same time;

-- the number of white roses taken.

.

.

.

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

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

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

where -- number of assembled units;

-- the number of units requiring additional lubrication;

– number of selected units;

-- number of units requiring additional lubrication among the selected ones.

.

.

.

.

.

.

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

Example 2.14. Of the 10 hours 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 in need of cleaning, examines them one by one and, having found such a watch, stops further viewing. Find the expected value and variance of the number of hours viewed.

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

.

.

.

.

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

Now let's calculate the numerical characteristics of the quantity:

Answer:,.

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

Decision: A random variable can take the following values:. Since the subscriber does not dial the dialed digit in the future, the probabilities of these values \u200b\u200bare equal.

Let's compose a series of distribution 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 reliability tests for each device in the series is p... Determine the mathematical expectation of the number of devices that failed, if they were tested N devices.

Decision:Discrete random variable X is the number of failed devices in N independent tests, in each of which the probability of failure is 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.

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

We find:.

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

Decision: In this case, all conducted experiments 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 by the Bernoulli formula:

Answer: .

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

Decision: The problem can be solved in two ways.

1) Possible values \u200b\u200bof CB 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 the mathematical expectation, we determine the probability:

Find the variance of the RV X:

.

2) You can use the formula:

.

Answer: .

Example 2.20. Mathematical expectation and standard deviation of a normally distributed random variable X 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).

Decision: The probability of hitting a normal random variable X to the section from to is expressed through the Laplace function:

Example 2.21. Given a function:

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

Decision:For a function to be the density of distribution of some random variable, it must be non-negative, and it must satisfy the property:

.

Consequently:

Let's calculate the mathematical expectation by the formula:

.

Let's calculate the variance by the formula:

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

Decision: The distribution law of a discrete random variable X - the number of occurrences of an event in independent tests, in each of which the probability of occurrence of an event is equal, is called binomial. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of occurrence of 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 for three shots.

Decision: Since three independent tests are carried out, and the probability of occurrence of the event A (hit) is the same in each test, 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 probability of occurrence and non-occurrence of an event in one trial:

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

Average number of customers who came in 5 minutes: . .

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

Decision: In this example, the expected value is , and the failure rate is.

Then the required probability:

Example 2.30. A group of 15 students holds a meeting in a room 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 seventh place in the row?

Decision:

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 batch;

– number of selected parts;

-- the number of non-standard parts selected.

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 \u200b\u200bin a certain sequence. One of the possible ways to define a continuous random variable is to use a distribution function.

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

Sometimes the term “Cumulative function” is used instead of the term “Distribution function”.

Distribution function properties:

1. The values \u200b\u200bof the distribution function belong 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\u003e x 1

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

P (aX

Example 9. Random variable X is given by the 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) \u003d x / 4 + 1/4, then F (2) -F (0) \u003d (2/4 + 1/4) - (0 / 4 + 1/4) \u003d 1/2. So, P (0

Corollary 2. The probability that a continuous random variable X will take one definite value is zero.

Corollary 3. If the possible values \u200b\u200bof the random variable belong to the interval (a; b), then: 1) F (x) \u003d 0 for xa; 2) F (x) \u003d 1 for xb.
The following limit relations are valid:

The graph of the distribution function is located in a strip bounded by straight lines y \u003d 0, y \u003d 1 (first property). As x increases in the interval (a; b), which contains all possible values \u200b\u200bof the random variable, the graph "rises up". At 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. Discrete random variable X is given by the distribution table:

X	1	4	8
P	0.3	0.1	0.6

Find the distribution function and plot it.
Solution: The distribution function analytically can be written as follows:

Figure-2

DEFINITION: The density of the probability distribution 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 for 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 definite 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 1: f (x) dx \u003d 1.
3. The definite integral from -∞ to x of the probability density of a continuous random variable is equal to the distribution function of this quantity: f (x) dx \u003d F (x)

Example 11. The density of the probability distribution of a random variable X is given

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

Solution: Seeking 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 \u200b\u200bof which belong to the segment, is a definite integral:

M (x) \u003d xf (x) dx (9)

If the possible values \u200b\u200bbelong to the entire Ox axis, then:

M (x) \u003d xf (x) dx (10)

The mode M 0 (X) of a continuous random variable X is called 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 called its possible value, which is determined by the equality:

P (X e (X)) \u003d P (X\u003e M e (X))

DEFINITION. The variance of a continuous random variable is the mathematical expectation of the square of its deviation. If the possible values \u200b\u200bof X belong to the segment, then:

D (x) \u003d 2 f (x) dx (11)
or
D (x) \u003d x 2 f (x) dx- 2 (11 *)

If the possible values \u200b\u200bbelong to the entire x-axis, then.

9. Continuous random variable, its numerical characteristics

A continuous random variable can be set using two functions. The integral function of the probability distribution of a random variable X is the function defined by the equality
.

The integral function provides a general way to define 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 in this interval, ie .. For example, for a discrete random variable specified 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 cumulative probability distribution function

.

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

Decision. On the interval
the graph is the straight line y \u003d 0. In the interval from 0 to 2 there is a parabola given by the equation
... On the interval
the graph is the straight line y \u003d 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 cumulative probability distribution function:

It is convenient to define the distribution law of a continuous random variable using another function, namely, probability density functions
.

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

The function graph is called distribution curve... Geometrically, the probability of a random variable X hitting the gap is equal to the area of \u200b\u200bthe 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(mean value) of a continuous random variable X is determined by the equality
.

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

Dispersiona discrete random variable X is the mathematical expectation of the square of the deviation of the random variable from its mathematical expectation, i.e. ... For a continuous random variable, the variance is determined by the formula
.

The dispersion has the following properties:

The last property is very convenient to use to find the variance of a continuous random variable.

The concept of the standard deviation is introduced in a similar way. Standard deviation of continuousa random variable X is called the square root of the variance, i.e.
.

Example 28... Continuous random variable X is given by the 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 and, 2) mathematical expectation M (X), variance
, standard deviation, 3) integral function
and build graphs of integral and differential functions.

1). To find the parameter and 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 variance by the formula:
, i.e. ...

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

3). The integral function is expressed in terms of the probability density function as follows:
... Consequently,
at
, \u003d 0 for
and \u003d 1 for
.

The graphs of these functions are shown in Fig. 4. and fig. five.

Fig. 4 Fig. 5.

9.2. Uniform probability distribution of a continuous random variable

Probability distribution of a continuous random variable X evenly on an interval if its probability density is constant on this interval and equal to 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 event consisting of an instant signal must occur between 1 pm and 5 pm. 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.

Decision. 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
.

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 and - expected value,
is the standard deviation.

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

The consequence of this theorem is the three sigma rule, i.e. it is practically certain that a normally distributed, continuous random variable X takes on its values \u200b\u200bin the interval
... This rule is deduced from the formula
, which is a special case of the stated theorem.

Example 30.The life of the TV set 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 10 to 20 years.

Decision. By the condition of the problem, the mathematical expectation and \u003d 15, standard deviation.

Find ... Thus, the probability of TV operation from 10 to 20 years is more than 0.9.

9.4 Chebyshev's inequality

Occurs chebyshev's lemma... If a random variable X takes only non-negative values \u200b\u200band has a mathematical expectation, then for any positive at
.

Considering that, as the sum of the probabilities of opposite events, we obtain that
.

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

Whence it follows that
.

Example 31.A batch of parts has been manufactured. 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 randomly taken part will be at least 99cm. and no more than 101cm.

Decision. Dispersion. The mathematical expectation is equal to 100. Therefore, to estimate from below the probability of the event under consideration
we apply the Chebyshev inequality, in which
then
.

10. Elements of mathematical statistics

Statistical populationname a lot of homogeneous objects or phenomena. Number p elements of this set is called the volume of the population. Observed values feature X is called options... If the options are arranged in ascending order, then discrete variation series... In the case of grouping the variant by intervals, it turns out interval variation series... Under frequency tattribute values \u200b\u200bunderstand the number of members of the population with a given variant.

The ratio of frequency to volume of the statistical population is called relative frequency signs:
.

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

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

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

The sampling range of variation is 23 - 17 \u003d 6. To build a frequency polygon, points with coordinates
and connect them in series.

The distribution of relative frequencies is as follows:

10.1 Numerical characteristics of the variation series

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

The sum of all frequencies is p.

The arithmetic mean of the sample call the value
.

Dispersion or the measure of the dispersion of the values \u200b\u200bof the attribute X in relation to its arithmetic mean is called the value
... The mean square deviation is called the square root of the variance, i.e. ...

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

An empirical distribution function of relative frequenciesis a function that determines for each value the relative frequency of the event
, i.e.
where - number of options, less x, and p - sample size.

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

Decision. Let's find the arithmetic mean of the sample by the formula, then.

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

10.2. Estimation of probability by relative frequency. Confidence interval

Let it be carried out p independent tests, 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 occurrence of event A in each test in absolute value by no more than by, is approximately equal to twice the value of the integral Laplace function:
.

Interval assessmentsuch an estimate is called, which is determined by two numbers that are the ends of the interval covering the estimated parameter of the statistical population.

Confidence interval is called the interval, which with a given confidence level 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 of relative frequency. Numbers
and
called the lower and, respectively, the upper confidence limits, - the limiting error for a given confidence level
.

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

Decision. According to the problem statement. We use the formula
... According to table 2 of the appendix, we find the value of the argument pi of which the value of the Laplace integral function is 0.475. We get that
... In this way, . Therefore, we can say with a probability of 0.95 that the share of rejects produced by the shop is high, namely, it varies within the range from 6.2% to 6.6%.

10.3. Parameter estimation in statistics

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

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

where p - sample size, - sample arithmetic mean, t Is the argument of the Laplace integral function for which
... Moreover, 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 pand does not depend on unknowns andand. Student's distribution even for small samples
gives quite satisfactory marks. Then the confidence interval covering the mathematical expectation and of this feature with a given confidence probability is found from the condition

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

The confidence interval covering the standard deviation of this feature with a confidence level is found by the formulas: and, where
located 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 called the functional dependence of the conditional mean from x. The equation
represents the regression equation Y on X, and
- the equation of regression X to Y.

The correlation dependence can be linear and curvilinear. In the case of a linear correlation dependence, the equation of a straight line regression has the form:
where the slope andstraight line regression 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 method of least squares from the system of normal equations:

where p - the number of observations of values \u200b\u200bof pairs of interrelated quantities.

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

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

.

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

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

Find the sample equation of a straight line of regression Y by X.

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

Observation number

To find the distribution function of a discrete random variable, you must use this calculator. Exercise 1... The distribution density of a continuous random variable X has the form:
To find:
a) parameter A;
b) the 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).

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

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

Assignment 4... The probability density of some random variable is given as follows: f (x) \u003d A / x 4 (x \u003d 1; + ∞)
Find coefficient A, distribution function F (x), mathematical expectation and variance, and also the probability that a random variable will take a value in the interval. Plot f (x) and F (x).

Task... The distribution function of some continuous random variable is defined as follows:

Determine the parameters a and b, find an expression for the probability density f (x), 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).

Let us find the distribution density function as a derivative of the distribution function.

Knowing that

find the parameter a:


or 3a \u003d 1, whence a \u003d 1/3
The parameter b is found from the following properties:
F (4) \u003d a * 4 + b \u003d 1
1/3 * 4 + b \u003d 1 whence b \u003d -1/3
Therefore, the distribution function has the form: F (x) \u003d (x-1) / 3

Expected value.


Dispersion.

1 / 9 4 3 - (1 / 9 1 3) - (5 / 2) 2 = 3 / 4
Let us find the probability that the random variable will take 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 density of the probability distribution f (x) of a continuous random variable X is given. Required:

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

f (x) \u003d A * sqrt (x), 1 ≤ x ≤ 4.
Decision:

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

Let us find the parameter A from the condition:



or
14/3 * A-1 \u003d 0
Where from,
A \u003d 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 \u200b\u200bof the random variable belong to the segment [ a, b].

Definition. Mathematical expectationa continuous random variable X, the possible values \u200b\u200bof which belong to an interval, is called a definite integral

If the possible values \u200b\u200bof a random variable are considered on the entire numerical 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 a continuous random variable is called the mathematical expectation of the square of its deviation.

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

Definition. Mean square deviationcalled the square root of the variance.

Definition. FashionM 0 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 does not have a maximum, then it is called anti-modal.

Definition. Median M D of a random variable X is called its value relative to which it is equally probable 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 halved. Note that if the distribution is unimodal, then the mode and median coincide with the mathematical expectation.

Definition. The starting pointorder kof a random variable X is called the mathematical expectation of the quantity X k.

For a discrete random variable:.

.

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

Definition. Central pointorder 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 variance. The central moment of the third order characterizes the distribution asymmetry.

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

Definition. To characterize the peakedness 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 point:. Absolute Center Point: ... The absolute central moment of the first order is called arithmetic mean.

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

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

Because the balls in each experiment are returned back and mixed, then the tests 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 consecutive trials, the white ball may not appear at all, it may appear once, two, three, four or five times. To draw up the distribution law, it is necessary 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 two times: .