The random variable is given by the distribution function; find the constant. Expectation of a continuous random variable
RANDOM VARIABLES
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 contained in the interval (2.5; 3.6).
Solution: X in the interval (2.5; 3.6) can be determined in two ways:
Example 2.2. At what parameter values A And IN function F(x) = A + Be - x can be a distribution function for non-negative values 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 must be satisfied:
.
Answer: 
.
Example 2.3. The random variable X is specified by the distribution function
Find the probability that, as a result of four independent tests, 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 using the formula:
Example 2.4. The probability of the ball hitting the basket with one shot is 0.3. Draw up a distribution law for the number of hits with three throws.
Solution: Random value X– the number of hits in the basket with three shots – can take the following values: 0, 1, 2, 3. Probabilities that X
X:
Example 2.5. Two shooters each fire one shot at a target. The probability of the first shooter hitting it is 0.5, the second - 0.4. Draw up a distribution law for the number of hits on a target.
Solution: Let's find the law of distribution of a discrete random variable X– number of hits on the target. Let the event be the first shooter hitting the target, and let the second shooter hit the target, and be their misses, respectively.
Let's compose the law of probability distribution of SV X:
Example 2.6. Three elements are tested, operating independently of each other. The duration of time (in hours) of failure-free operation of elements has a distribution density function: 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.
Solution: Let's use the definition of the probability generating function:
The probability that in independent trials, in the first of which the probability of an event occurring A equal to , in the second, etc., event A appears exactly once, equal to the coefficient 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 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; the coefficient at is equal to the probability that only one element will fail.
Example 2.7. Given the probability density f(x)random variable X:
Find the distribution function F(x).
Solution: We use the formula:
.
Thus, the distribution function looks like:
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 a distribution law for the number of failed elements in one experiment.
Solution: Random value X– the number of elements that failed in one experiment – can take the following values: 0, 1, 2, 3. Probabilities that X takes these values, we find using Bernoulli’s formula:
Thus, we obtain the following law of probability distribution of a random variable X:
Example 2.9. In a batch of 6 parts there are 4 standard ones. 3 parts were selected at random. Draw up a distribution law for 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 following values: 1, 2, 3 and has a hypergeometric distribution. Probabilities that X
Where -- number of parts in the batch;
-- number of standard parts in a batch;
– number of selected parts;
-- number of standard parts among those selected.
.
.
.
Example 2.10. The random variable has a distribution density
and are not known, but , a and . Find and.
Solution: IN in this case random value 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 conditions of the problem, we finally have: 
.
Answer: .
Example 2.11. On average 10% of contracts Insurance Company pays insurance amounts in connection with the occurrence of an insured event. Calculate the mathematical expectation and dispersion of 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 Bernoulli's formula to calculate the probabilities various numbers contracts (out of four) for which the insured amounts were paid:
.
The IC distribution series (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. Draw up a law of distribution of a random variable expressing the number of white roses among two simultaneously taken.
Solution: In a selection 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. Probabilities that X takes these values, we find it using the formula:
Where -- number of roses;
-- number of white roses;
– number of roses taken at the same time;
-- the number of white roses among those taken.
.
.
.
Then the distribution law of the random variable will be as follows:
Example 2.13. Among the 15 assembled units, 6 require additional lubrication. Draw up a distribution law for the number of units that need additional lubrication among five randomly selected from the total number.
Solution: Random value X– the number of units that require additional lubrication among the five selected – can take the following values: 0, 1, 2, 3, 4, 5 and has a hypergeometric distribution. Probabilities that X takes these values, we find it using the formula:
Where -- number of assembled units;
-- the number of units that require additional lubrication;
– number of selected units;
-- the number of units that require additional lubrication among those selected.
.
.
.
.
.
Then the distribution law of the random variable will be as follows:
Example 2.14. Of the 10 watches received for repair, 7 require general cleaning of the mechanism. The watches are not sorted by type of repair. The master, wanting to find watches that need cleaning, examines them one by one and, having found such watches, 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. Probabilities that X takes these values, we find it using 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 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 times he dials a phone number before reaching the desired number, if he dials the last digit at random and does not subsequently dial the dialed digit.
Solution: The random variable can take the following values: . Since the subscriber does not dial the dialed digit in the future, the probabilities of these values are equal.
Let's compile 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 reliability tests for each device in the series is equal to p. Determine the mathematical expectation of the number of devices that failed if they were tested N devices.
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 p, distributed according to the binomial law. Expected value 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 distribution law of a discrete random variable:
We find: .
Example 2.18. The technical control department checks products for standardness. The probability that the product is standard is 0.9. Each batch contains 5 products. 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 inspection.
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 products.
We find the probability using Bernoulli's 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 values of SV X: 0, 1, 2. Using the Bernoulli formula, we determine the probabilities of these events:
, , .
Then the distribution law X has the form:
From the definition of mathematical expectation, we determine the probability:
Let's find the dispersion of SV X:
.
2) You can use the formula:
.
Answer: 
.
Example 2.20. Expectation and standard deviation of a normally distributed random variable X respectively equal to 20 and 5. 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 through the Laplace function:
Example 2.21. Given function: 
At what parameter value 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:
Let's calculate the mathematical expectation using the formula:
.
Let's calculate the variance using the formula:
T is equal 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 event occurring is equal to , 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 with three shots.
Solution: Since three independent trials are performed, and the probability of the occurrence of event A (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 probability of the 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 arrive in the next 5 minutes.
Average number of clients 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 (random) request will wait on the processor for more than 35 seconds.
Solution: In this example, the mathematical expectation 
, and the failure rate is equal to .
Then the desired probability:
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 place in the hall randomly. What is the probability that no more than three people will be in the seventh place of the row?
Solution:
Example 2.31.
Then, according to the classical definition of probability:
Where -- number of parts in the batch;
-- number of non-standard parts in the batch;
– number of selected parts;
-- number of non-standard parts among those selected.
Then the distribution law of the random variable will be as follows.
Distribution density probabilities X call the function f(x)– the first derivative of the distribution function F(x):
The concept of probability distribution density of a random variable X For discrete value not applicable.
Probability distribution density f(x)– called the differential distribution function:
Property 1. Distribution density is a non-negative quantity:
Property 2. Improper integral from the distribution density in the range from to is equal to unity:
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), That:
f(x)– distribution density of a random variable.
Dispersion continuous random variable X, the possible values of which belong to the entire axis, is determined by the equality:
A special case. If the values of a random variable belong to the interval ( a,b), That:
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 determine the distribution density of a continuous random variable X:
a) Mathematical expectation 
:
b) Variance 
V) 
Tasks for independent work:
1. Random variable X given by the distribution function:
M(x);
b) variance D(x);
X into the interval (2,3).
2. Random variable X
Find: a) mathematical expectation M(x);
b) variance D(x);
c) determine the probability of a random variable hitting X into the interval (1;1.5).
3. Random variable X is given by the cumulative distribution function:
Find: a) mathematical expectation M(x);
b) variance D(x);
c) determine the probability of a random variable hitting 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 segment [ a,b], if on this segment the probability distribution density of the random variable is constant, and outside it it is equal to zero, i.e.:
Rice. 4.
; 
; 
.
Example 1.27. A bus on a certain route moves uniformly at intervals 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 appear at the stop within 2 to 5 minutes after the previous bus leaves, i.e. random value X must fall into the interval (2;5). That. required 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 the random variable X, distributed uniformly in the interval (2;8).
2. The minute hand of an electric clock moves abruptly at the end of every minute. Find the probability that at a given moment the clock will show a time that differs from the true time by no more than 20 seconds.
1.4.2. Exponential distribution
Continuous random variable X is distributed according to the exponential law if its probability density has the form:
where is the parameter of the exponential distribution.
Thus 
Rice. 5.
Numerical characteristics:
Example 1.28. Random value X– operating time of a light bulb - has an exponential distribution. Determine the probability that the operating time of the light bulb will be at least 600 hours if the average operating time is 400 hours.
Solution: According to the conditions of the problem, the mathematical expectation of a random variable X equals 400 hours, therefore:
; 
The required 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 variable X is 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
– Laplace function.
A distribution for which ; , i.e. with probability density 
called standard.
Rice. 6.
The probability that the absolute value is rejected is less positive number :
.
In particular, when a= 0 the equality is true:
Example 1.29. Random value X normally distributed. 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 normal distribution random variable X, knowing that M(x)= 3, D(x)= 16.
2. Expectation and standard deviation of a normally distributed random variable X respectively equal to 20 and 5. 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 out of 3 independent measurements the error of at least one will not exceed 4 mm in absolute value.
4. A certain substance is weighed without systematic errors. Random weighing errors are subject to the normal law with a standard deviation r. Find the probability that weighing will be carried out with an error not exceeding 10 g in absolute value.
Chapter 1. Discrete random variable
§ 1. Concepts of a random variable.
Distribution law of a discrete random variable.
Definition : Random is a quantity that, as a result of testing, takes only one value out of a possible set of its values, unknown in advance and depending on random reasons.
There are two types of random variables: discrete and continuous.
Definition : The random variable X is called discrete (discontinuous) if the set of its values is finite or infinite but countable.
In other words, the possible values of a discrete random variable can be renumbered.
A random variable can be described using its distribution law.
Definition : Distribution law of a discrete random variable call the correspondence between possible values of a random variable and their probabilities.
The distribution law of a discrete random variable X can be specified in the form of a table, in the first row of which all possible values of the random variable are indicated in ascending order, and in the second row the corresponding probabilities of these values, i.e.
where р1+ р2+…+ рn=1
Such a table is called a distribution series of a discrete random variable.
If the set of possible values of a random variable is infinite, then the series p1+ p2+…+ pn+… converges and its sum is equal to 1.
The distribution law of a discrete random variable X can be depicted graphically, for which a broken line is constructed in a rectangular coordinate system, connecting sequentially points with coordinates (xi; pi), i=1,2,…n. The resulting line is called distribution polygon (Fig. 1).
Organic chemistry" href="/text/category/organicheskaya_hiimya/" rel="bookmark">organic chemistry are 0.7 and 0.8, respectively. Draw up a distribution law for the random variable X - the number of exams that the student will pass.
Solution. The considered random variable X as a result of the exam can take one of the following values: x1=0, x2=1, x3=2.
Let's find the probability of these values. Let's denote the events:
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 |
So, the distribution law of the random variable X is given by the table:
Control: 0.6+0.38+0.56=1.
§ 2. Distribution function
A complete description of a random variable is also given by the distribution function.
Definition: Distribution function of a discrete random variable X is called a function F(x), which determines for each value x the probability that the random variable X will take a value less than x:
F(x)=P(X<х)
Geometrically, the distribution function is interpreted as the probability that the random variable X will take the value that is represented on the number line by a point lying to the left of point x.
1)0≤ F(x) ≤1;
2) F(x) is a non-decreasing function on (-∞;+∞);
3) F(x) - continuous on the left at points x= xi (i=1,2,...n) and continuous at all other points;
4) F(-∞)=P (X<-∞)=0 как вероятность невозможного события Х<-∞,
F(+∞)=P(X<+∞)=1 как вероятность достоверного события Х<-∞.
If the distribution law of a discrete random variable X is given in the form of a table:
then the distribution function F(x) is determined by the formula:
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0 for x≤ x1,
р1 at x1< х≤ x2,
F(x)= р1 + р2 at x2< х≤ х3
1 for x> xn.
Its graph is shown in Fig. 2:
§ 3. Numerical characteristics of a discrete random variable.
One of the important numerical characteristics is the mathematical expectation.
Definition: Mathematical expectation M(X) discrete random variable X is the sum of the products of all its values and their corresponding probabilities:
M(X) = ∑ xiрi= x1р1 + x2р2+…+ xnрn
The mathematical expectation serves as a characteristic of the average value of a random variable.
Properties of mathematical expectation:
1)M(C)=C, where C is a constant value;
2)M(C X)=C M(X),
3)M(X±Y)=M(X) ±M(Y);
4)M(X Y)=M(X) M(Y), where X, Y are independent random variables;
5)M(X±C)=M(X)±C, where C is a constant value;
To characterize the degree of dispersion of possible values of a discrete random variable around its mean value, dispersion is used.
Definition: Variance D ( X ) random variable X is the mathematical expectation of the squared deviation of the random variable from its mathematical expectation:
Dispersion properties:
1)D(C)=0, where C is a constant value;
2)D(X)>0, where X is a random variable;
3)D(C X)=C2 D(X), where C is a constant value;
4)D(X+Y)=D(X)+D(Y), where X, Y are independent random variables;
To calculate variance it is often convenient to use the formula:
D(X)=M(X2)-(M(X))2,
where M(X)=∑ xi2рi= x12р1 + x22р2+…+ xn2рn
The variance D(X) has the dimension of a squared random variable, which is not always convenient. Therefore, the value √D(X) is also used as an indicator of the dispersion of possible values of a random variable.
Definition: Standard deviation σ(X) random variable X is called the square root of the variance:
Task No. 2. The discrete random variable X is specified by the distribution law:
Find P2, the distribution function F(x) and plot its graph, as well as M(X), D(X), σ(X).
Solution: Since the sum of the probabilities of possible values of the random variable X is equal to 1, then
Р2=1- (0.1+0.3+0.2+0.3)=0.1
Let's find the distribution function F(x)=P(X Geometrically, this equality can be interpreted as follows: F(x) is the probability that the random variable will take the value that is represented on the number axis by the point lying to the left of the point x. If x≤-1, then F(x)=0, since there is not a single value of this random variable on (-∞;x); If -1<х≤0, то F(х)=Р(Х=-1)=0,1, т. к. в промежуток (-∞;х) попадает только одно значение x1=-1; If 0<х≤1, то F(х)=Р(Х=-1)+ Р(Х=0)=0,1+0,1=0,2, т. к. в промежуток (-∞;x) there are two values x1=-1 and x2=0; If 1<х≤2, то F(х)=Р(Х=-1) + Р(Х=0)+ Р(Х=1)= 0,1+0,1+0,3=0,5, т. к. в промежуток (-∞;х) попадают три значения x1=-1, x2=0 и x3=1; If 2<х≤3, то F(х)=Р(Х=-1) + Р(Х=0)+ Р(Х=1)+ Р(Х=2)= 0,1+0,1+0,3+0,2=0,7, т. к. в промежуток (-∞;х) попадают четыре значения x1=-1, x2=0,x3=1 и х4=2; If x>3, then F(x)=P(X=-1) + P(X=0)+ P(X=1)+ P(X=2)+P(X=3)= 0.1 +0.1+0.3+0.2+0.3=1, because four values x1=-1, x2=0, x3=1, x4=2 fall into the interval (-∞;x) and x5=3. https://pandia.ru/text/78/455/images/image006_89.gif" width="14 height=2" height="2"> 0 at x≤-1, 0.1 at -1<х≤0, 0.2 at 0<х≤1, F(x)= 0.5 at 1<х≤2, 0.7 at 2<х≤3, 1 at x>3 Let's represent the function F(x) graphically (Fig. 3): https://pandia.ru/text/78/455/images/image014_24.jpg" width="158 height=29" height="29">≈1.2845. §
4. Binomial distribution law discrete random variable, Poisson's law. Definition: Binomial
is called the law of distribution of a discrete random variable X - the number of occurrences of event A in n independent repeated trials, in each of which event A may occur with probability p or not occur with probability q = 1-p. Then P(X=m) - the probability of occurrence of event A exactly m times in n trials is calculated using the Bernoulli formula: Р(Х=m)=Сmnpmqn-m The mathematical expectation, dispersion and standard deviation of a random variable X distributed according to a binary law are found, respectively, using the formulas: https://pandia.ru/text/78/455/images/image016_31.gif" width="26"> The probability of event A - “rolling out a five” in each trial is the same and equal to 1/6, i.e. . P(A)=p=1/6, then P(A)=1-p=q=5/6, where - “failure to get an A.” The random variable X can take the following values: 0;1;2;3. We find the probability of each of the possible values of X using Bernoulli’s formula: Р(Х=0)=Р3(0)=С03р0q3=1 (1/6)0 (5/6)3=125/216; Р(Х=1)=Р3(1)=С13р1q2=3 (1/6)1 (5/6)2=75/216; Р(Х=2)=Р3(2)=С23р2q =3 (1/6)2 (5/6)1=15/216; Р(Х=3)=Р3(3)=С33р3q0=1 (1/6)3 (5/6)0=1/216. That. the distribution law of the random variable X has the form: Control: 125/216+75/216+15/216+1/216=1. Let's find the numerical characteristics of the random variable X: M(X)=np=3 (1/6)=1/2, D(X)=npq=3 (1/6) (5/6)=5/12, Task No. 4. An automatic machine stamps parts. The probability that a manufactured part will be defective is 0.002. Find the probability that among 1000 selected parts there will be: a) 5 defective; b) at least one is defective. Solution:
The number n=1000 is large, the probability of producing a defective part p=0.002 is small, and the events under consideration (the part turns out to be defective) are independent, therefore the Poisson formula holds: Рn(m)= e-
λ
λm Let's find λ=np=1000 0.002=2. a) Find the probability that there will be 5 defective parts (m=5): Р1000(5)= e-2
25
= 32 0,13534
= 0,0361 b) Find the probability that there will be at least one defective part. Event A - “at least one of the selected parts is defective” is the opposite of the event - “all selected parts are not defective.” Therefore, P(A) = 1-P(). Hence the desired probability is equal to: P(A)=1-P1000(0)=1- e-2
20
= 1- e-2=1-0.13534≈0.865. Tasks for independent work.
1.1
1.2.
The dispersed random variable X is specified by the distribution law: Find p4, the distribution function F(X) and plot its graph, as well as M(X), D(X), σ(X). 1.3.
There are 9 markers in the box, 2 of which no longer write. Take 3 markers at random. Random variable X is the number of writing markers among those taken. Draw up a law of distribution of a random variable. 1.4.
There are 6 textbooks randomly arranged on a library shelf, 4 of which are bound. The librarian takes 4 textbooks at random. Random variable X is the number of bound textbooks among those taken. Draw up a law of distribution of a random variable. 1.5.
There are two tasks on the ticket. The probability of correctly solving the first problem is 0.9, the second is 0.7. Random variable X is the number of correctly solved problems in the ticket. Draw up a distribution law, calculate the mathematical expectation and variance of this random variable, and also find the distribution function F(x) and build its graph. 1.6.
Three shooters are shooting at a target. The probability of hitting the target with one shot is 0.5 for the first shooter, 0.8 for the second, and 0.7 for the third. Random variable X is the number of hits on the target if the shooters fire one shot at a time. Find the distribution law, M(X),D(X). 1.7.
A basketball player throws the ball into the basket with a probability of hitting each shot of 0.8. For each hit, he receives 10 points, and if he misses, no points are awarded to him. Draw up a distribution law for the random variable X - the number of points received by a basketball player in 3 shots. Find M(X),D(X), as well as the probability that he gets more than 10 points. 1.8.
Letters are written on the cards, a total of 5 vowels and 3 consonants. 3 cards are chosen at random, and each time the taken card is returned back. Random variable X is the number of vowels among those taken. Draw up a distribution law and find M(X),D(X),σ(X). 1.9.
On average, under 60% of contracts, the insurance company pays insurance amounts in connection with the occurrence of an insured event. Draw up a distribution law for the random variable X - the number of contracts for which the insurance amount was paid among four contracts selected at random. Find the numerical characteristics of this quantity. 1.10.
The radio station sends call signs (no more than four) at certain intervals until two-way communication is established. The probability of receiving a response to a call sign is 0.3. Random variable X is the number of call signs sent. Draw up a distribution law and find F(x). 1.11.
There are 3 keys, of which only one fits the lock. Draw up a law for the distribution of the random variable X-number of attempts to open the lock, if the tried key does not participate in subsequent attempts. Find M(X),D(X). 1.12.
Consecutive independent tests of three devices are carried out for reliability. Each subsequent device is tested only if the previous one turned out to be reliable. The probability of passing the test for each device is 0.9. Draw up a distribution law for the random variable X-number of tested devices. 1.13
.Discrete random variable X has three possible values: x1=1, x2, x3, and x1<х2<х3. Вероятность того, что Х примет значения х1 и х2, соответственно равны 0,3 и 0,2. Известно, что М(Х)=2,2, D(X)=0,76. Составить закон распределения случайной величины. 1.14.
The electronic device block contains 100 identical elements. The probability of failure of each element during time T is 0.002. The elements work independently. Find the probability that no more than two elements will fail during time T. 1.15.
The textbook was published in a circulation of 50,000 copies. The probability that the textbook is bound incorrectly is 0.0002. Find the probability that the circulation contains: a) four defective books, b) less than two defective books. 1
.16.
The number of calls arriving at the PBX every minute is distributed according to Poisson's law with the parameter λ=1.5. Find the probability that in a minute the following will arrive: a) two calls; b) at least one call. 1.17.
Find M(Z),D(Z) if Z=3X+Y. 1.18.
The laws of distribution of two independent random variables are given: Find M(Z),D(Z) if Z=X+2Y. Answers:
https://pandia.ru/text/78/455/images/image007_76.gif" height="110"> 1.1.
p3=0.4; 0 at x≤-2, 0.3 at -2<х≤0, F(x)= 0.5 at 0<х≤2, 0.9 at 2<х≤5, 1 at x>5 0.3 at -1<х≤0, 0.4 at 0<х≤1, F(x)= 0.6 at 1<х≤2, 0.7 at 2<х≤3, 1 at x>3 M(X)=1; D(X)=2.6; σ(X) ≈1.612. https://pandia.ru/text/78/455/images/image025_24.gif" width="2 height=98" height="98"> 0 at x≤0, 0.03 at 0<х≤1, F(x)= 0.37 at 1<х≤2, 1 for x>2 M(X)=2; D(X)=0.62 M(X)=2.4; D(X)=0.48, P(X>10)=0.896 1.
8
.
M(X)=15/8; D(X)=45/64; σ(X) ≈ M(X)=2.4; D(X)=0.96 https://pandia.ru/text/78/455/images/image008_71.gif" width="14"> 1.11.
M(X)=2; D(X)=2/3 1.14.
1.22 e-0.2≈0.999 1.15.
a)0.0189; b) 0.00049 1.16.
a)0.0702; b)0.77687 1.17.
3,8; 14,2 1.18.
11,2; 4. Chapter 2. Continuous random variable
Definition: Continuous
is a quantity whose all possible values completely fill a finite or infinite span of the number line. Obviously, the number of possible values of a continuous random variable is infinite. A continuous random variable can be specified using a distribution function. Definition: F distribution function
a continuous random variable X is called a function F(x), which determines for each value xhttps://pandia.ru/text/78/455/images/image028_11.jpg" width="14" height="13">R The distribution function is sometimes called the cumulative distribution function. Properties of the distribution function:
1)1≤ F(x) ≤1 2) For a continuous random variable, the distribution function is continuous at any point and differentiable everywhere, except, perhaps, at individual points. 3) The probability of a random variable X falling into one of the intervals (a;b), [a;b], [a;b], is equal to the difference between the values of the function F(x) at points a and b, i.e. R(a)<Х 4) The probability that a continuous random variable X will take one separate value is 0. 5) F(-∞)=0, F(+∞)=1 Specifying a continuous random variable using a distribution function is not the only way. Let us introduce the concept of probability distribution density (distribution density). Definition
:
Probability distribution density
f
(
x
)
of a continuous random variable X is the derivative of its distribution function, i.e.: The probability density function is sometimes called the differential distribution function or differential distribution law. The graph of the probability density distribution f(x) is called probability distribution curve
.
Properties of probability density distribution:
1) f(x) ≥0, at xhttps://pandia.ru/text/78/455/images/image029_10.jpg" width="285" height="141">DIV_ADBLOCK92"> https://pandia.ru/text/78/455/images/image032_23.gif" height="38 src="> +∞ 2 6 +∞ 6 6 ∫ f(x)dx=∫ 0dx+ ∫ c(x-2)dx +∫ 0dx= c∫ (x-2)dx=c(x2/2-2x) =c(36/2-12-(4/ 2-4))=8s; b) It is known that F(x)= ∫ f(x)dx Therefore, x if x≤2, then F(x)= ∫ 0dx=0; https://pandia.ru/text/78/455/images/image032_23.gif" height="38 src="> 2 6 x 6 6 if x>6, then F(x)= ∫ 0dx+∫ 1/8(x-2)dx+∫ 0dx=1/8∫(x-2)dx=1/8(x2/2-2x) = 1/8(36/2-12-(4/2+4))=1/8 8=1. Thus, 0 at x≤2, F(x)= (x-2)2/16 at 2<х≤6, 1 for x>6. The graph of the function F(x) is shown in Fig. 3 https://pandia.ru/text/78/455/images/image034_23.gif" width="14" height="62 src="> 0 at x≤0, F(x)= (3 arctan x)/π at 0<х≤√3, 1 for x>√3. Find the differential distribution function f(x) Solution:
Since f(x)= F’(x), then DIV_ADBLOCK93"> · Mathematical expectation M (X)
continuous random variable X are determined by the equality: M(X)= ∫ x f(x)dx, provided that this integral converges absolutely. · Dispersion
D
(
X
)
continuous random variable X is determined by the equality: D(X)= ∫ (x-M(x)2) f(x)dx, or D(X)= ∫ x2 f(x)dx - (M(x))2 · Standard deviation σ(X)
continuous random variable is determined by the equality: All properties of mathematical expectation and dispersion, discussed earlier for dispersed random variables, are also valid for continuous ones. Task No. 3. The random variable X is specified by the differential function f(x): https://pandia.ru/text/78/455/images/image036_19.gif" height="38"> -∞ 2 X3/9 + x2/6 = 8/9-0+9/6-4/6=31/18, https://pandia.ru/text/78/455/images/image032_23.gif" height="38"> +∞ D(X)= ∫ x2 f(x)dx-(M(x))2=∫ x2 x/3 dx+∫1/3x2 dx=(31/18)2=x4/12 + x3/9 - - (31/18)2=16/12-0+27/9-8/9-(31/18)2=31/9- (31/18)2==31/9(1-31/36)=155/324, https://pandia.ru/text/78/455/images/image032_23.gif" height="38"> P(1<х<5)= ∫ f(x)dx=∫ х/3 dx+∫ 1/3 dx+∫ 0 dx= х2/6 +1/3х = 4/6-1/6+1-2/3=5/6. Problems for independent solution.
2.1.
A continuous random variable X is specified by the distribution function: 0 at x≤0, F(x)= https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤ π/6, F(x)= - cos 3x at π/6<х≤ π/3, 1 for x> π/3. Find the differential distribution function f(x), and also Р(2π /9<Х< π /2). 2.3.
0 at x≤2, f(x)= c x at 2<х≤4, 0 for x>4. 2.4.
A continuous random variable X is specified by the distribution density: 0 at x≤0, f(x)= c √x at 0<х≤1, 0 for x>1. Find: a) number c; b) M(X), D(X). 2.5.
https://pandia.ru/text/78/455/images/image041_3.jpg" width="36" height="39"> at x, 0 at x. Find: a) F(x) and construct its graph; b) M(X),D(X), σ(X); c) the probability that in four independent trials the value of X will take exactly 2 times the value belonging to the interval (1;4). 2.6.
The probability distribution density of a continuous random variable X is given: f(x)= 2(x-2) at x, 0 at x. Find: a) F(x) and construct its graph; b) M(X),D(X), σ (X); c) the probability that in three independent trials the value of X will take exactly 2 times the value belonging to the segment . 2.7.
The function f(x) is given as: https://pandia.ru/text/78/455/images/image045_4.jpg" width="43" height="38 src=">.jpg" width="16" height="15">[-√ 3/2; √3/2]. 2.8.
The function f(x) is given as: https://pandia.ru/text/78/455/images/image046_5.jpg" width="45" height="36 src="> .jpg" width="16" height="15">[- π /4 ; π /4]. Find: a) the value of the constant c at which the function will be the probability density of some random variable X; b) distribution function F(x). 2.9.
The random variable X, concentrated on the interval (3;7), is specified by the distribution function F(x)= . Find the probability that random variable X will take the value: a) less than 5, b) not less than 7. 2.10.
Random variable X, concentrated on the interval (-1;4), is given by the distribution function F(x)= . Find the probability that random variable X will take the value: a) less than 2, b) not less than 4. 2.11.
https://pandia.ru/text/78/455/images/image049_6.jpg" width="43" height="44 src="> .jpg" width="16" height="15">. Find: a) number c; b) M(X); c) probability P(X> M(X)). 2.12.
The random variable is specified by the differential distribution function: https://pandia.ru/text/78/455/images/image050_3.jpg" width="60" height="38 src=">.jpg" width="16 height=15" height="15"> . Find: a) M(X); b) probability P(X≤M(X)) 2.13.
The Rem distribution is given by the probability density: https://pandia.ru/text/78/455/images/image052_5.jpg" width="46" height="37"> for x ≥0. Prove that f(x) is indeed a probability density function. 2.14.
The probability distribution density of a continuous random variable X is given: DIV_ADBLOCK96"> https://pandia.ru/text/78/455/images/image055_3.jpg" width="187 height=136" height="136">(Fig. 5) 2.16.
The random variable X is distributed according to the law “ right triangle"in the interval (0;4) (Fig. 5). Find an analytical expression for the probability density f(x) on the entire number line. Answers
0 at x≤0, f(x)= https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤ π/6, F(x)= 3sin 3x at π/6<х≤ π/3,
Непрерывная случайная величина Х имеет равномерный закон распределения на некотором интервале (а;b), которому принадлежат все возможные значения Х, если плотность распределения вероятностей f(x) постоянная на этом интервале и равна 0 вне его, т. е. 0 for x≤a, f(x)= for a<х 0 for x≥b. The graph of the function f(x) is shown in Fig. 1 F(x)= https://pandia.ru/text/78/455/images/image077_3.jpg" width="30" height="37">, D(X)=, σ(X)=. Task No. 1. The random variable X is uniformly distributed on the segment. Find: a) probability distribution density f(x) and plot it; b) the distribution function F(x) and plot it; c) M(X),D(X), σ(X). Solution:
Using the formulas discussed above, with a=3, b=7, we find: https://pandia.ru/text/78/455/images/image081_2.jpg" width="22" height="39"> at 3≤х≤7, 0 for x>7 Let's build its graph (Fig. 3): https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86 src="> 0 at x≤3, F(x)= https://pandia.ru/text/78/455/images/image084_3.jpg" width="203" height="119 src=">Fig. 4 D(X) = ==https://pandia.ru/text/78/455/images/image089_1.jpg" width="37" height="43">==https://pandia.ru/text/ 78/455/images/image092_10.gif" width="14" height="49 src="> 0 at x<0, f(x)= λе-λх for x≥0. The distribution function of a random variable X, distributed according to the exponential law, is given by the formula: DIV_ADBLOCK98"> https://pandia.ru/text/78/455/images/image095_4.jpg" width="161" height="119 src="> Fig. 6 The mathematical expectation, variance and standard deviation of the exponential distribution are respectively equal to: M(X)= , D(X)=, σ (Х)= Thus, the mathematical expectation and the standard deviation of the exponential distribution are equal to each other. The probability of X falling into the interval (a;b) is calculated by the formula: P(a<Х Task No. 2. The average failure-free operation time of the device is 100 hours. Assuming that the failure-free operation time of the device has an exponential distribution law, find: a) probability distribution density; b) distribution function; c) the probability that the device’s failure-free operation time will exceed 120 hours. Solution:
According to the condition, the mathematical distribution M(X)=https://pandia.ru/text/78/455/images/image098_10.gif" height="43 src="> 0 at x<0, a) f(x)= 0.01e -0.01x for x≥0. b) F(x)= 0 at x<0, 1-e -0.01x at x≥0. c) We find the desired probability using the distribution function: P(X>120)=1-F(120)=1-(1- e -1.2)= e -1.2≈0.3. §
3.Normal distribution law Definition:
A continuous random variable X has normal distribution law (Gauss's law),
if its distribution density has the form: where m=M(X), σ2=D(X), σ>0. The normal distribution curve is called normal or Gaussian curve
(Fig.7) The distribution function of a random variable X, distributed according to the normal law, is expressed through the Laplace function Ф (x) according to the formula: where is the Laplace function. Comment:
The function Ф(x) is odd (Ф(-х)=-Ф(х)), in addition, for x>5 we can assume Ф(х) ≈1/2. The graph of the distribution function F(x) is shown in Fig. 8 https://pandia.ru/text/78/455/images/image106_4.jpg" width="218" height="33"> The probability that the absolute value of the deviation is less than a positive number δ is calculated by the formula: In particular, for m=0 the following equality holds: "Three Sigma Rule"
If a random variable X has a normal distribution law with parameters m and σ, then it is almost certain that its value lies in the interval (a-3σ; a+3σ), because https://pandia.ru/text/78/455/images/image110_2.jpg" width="157" height="57 src=">a) b) Let's use the formula: https://pandia.ru/text/78/455/images/image112_2.jpg" width="369" height="38 src="> From the table of function values Ф(х) we find Ф(1.5)=0.4332, Ф(1)=0.3413. So, the desired probability: P(28 Tasks for independent work
3.1.
The random variable X is uniformly distributed in the interval (-3;5). Find: b) distribution function F(x); c) numerical characteristics; d) probability P(4<х<6). 3.2.
The random variable X is uniformly distributed on the segment. Find: a) distribution density f(x); b) distribution function F(x); c) numerical characteristics; d) probability P(3≤х≤6). 3.3.
There is an automatic traffic light on the highway, in which the green light is on for 2 minutes, yellow for 3 seconds, red for 30 seconds, etc. A car drives along the highway at a random moment in time. Find the probability that a car will pass a traffic light without stopping. 3.4.
Subway trains run regularly at intervals of 2 minutes. A passenger enters the platform at a random time. What is the probability that a passenger will have to wait more than 50 seconds for a train? Find the mathematical expectation of the random variable X - the waiting time for the train. 3.5.
Find the variance and standard deviation of the exponential distribution given by the distribution function: F(x)= 0 at x<0, 1st-8x for x≥0. 3.6.
A continuous random variable X is specified by the probability distribution density: f(x)= 0 at x<0, 0.7 e-0.7x at x≥0. a) Name the distribution law of the random variable under consideration. b) Find the distribution function F(X) and the numerical characteristics of the random variable X. 3.7.
The random variable X is distributed according to the exponential law specified by the probability distribution density: f(x)= 0 at x<0, 0.4 e-0.4 x at x≥0. Find the probability that as a result of the test X will take a value from the interval (2.5;5). 3.8.
A continuous random variable X is distributed according to the exponential law specified by the distribution function: F(x)= 0 at x<0, 1st-0.6x at x≥0 Find the probability that, as a result of the test, X will take a value from the segment. 3.9.
The expected value and standard deviation of a normally distributed random variable are 8 and 2, respectively. Find: a) distribution density f(x); b) the probability that as a result of the test X will take a value from the interval (10;14). 3.10.
The random variable X is normally distributed with a mathematical expectation of 3.5 and a variance of 0.04. Find: a) distribution density f(x); b) the probability that as a result of the test X will take a value from the segment . 3.11.
The random variable X is normally distributed with M(X)=0 and D(X)=1. Which of the events: |X|≤0.6 or |X|≥0.6 is more likely? 3.12.
The random variable X is distributed normally with M(X)=0 and D(X)=1. From which interval (-0.5;-0.1) or (1;2) is it more likely to take a value during one test? 3.13.
The current price per share can be modeled using the normal distribution law with M(X)=10 den. units and σ (X)=0.3 den. units Find: a) the probability that the current share price will be from 9.8 den. units up to 10.4 days units; b) using the “three sigma rule”, find the boundaries within which the current stock price will be located. 3.14.
The substance is weighed without systematic errors. Random weighing errors are subject to the normal law with the mean square ratio σ=5g. Find the probability that in four independent experiments an error in three weighings will not occur in absolute value 3r. 3.15.
The random variable X is normally distributed with M(X)=12.6. The probability of a random variable falling into the interval (11.4;13.8) is 0.6826. Find the standard deviation σ. 3.16.
The random variable X is distributed normally with M(X)=12 and D(X)=36. Find the interval into which the random variable X will fall as a result of the test with a probability of 0.9973. 3.17.
A part manufactured by an automatic machine is considered defective if the deviation X of its controlled parameter from the nominal value exceeds modulo 2 units of measurement. It is assumed that the random variable X is normally distributed with M(X)=0 and σ(X)=0.7. What percentage of defective parts does the machine produce? 3.18.
The X parameter of the part is distributed normally with a mathematical expectation of 2 equal to the nominal value and a standard deviation of 0.014. Find the probability that the deviation of X from the nominal value will not exceed 1% of the nominal value. Answers
https://pandia.ru/text/78/455/images/image116_9.gif" width="14" height="110 src="> b) 0 for x≤-3, F(x)= left"> 3.10.
a)f(x)= , b) Р(3.1≤Х≤3.7) ≈0.8185. 3.11.
|x|≥0.6. 3.12.
(-0,5;-0,1). 3.13.
a) P(9.8≤Х≤10.4) ≈0.6562. 3.14.
0,111. 3.15.
σ=1.2. 3.16.
(-6;30). 3.17.
0,4%. A continuous random variable can be specified using the distribution function F(x)
. This method of assignment is not the only one. A continuous random variable can also be specified using another function called the distribution density or probability density (sometimes called a differential function). Definition4.1:
Distribution density of a continuous random variable X call the function f
(x)
- the first derivative of the distribution function F(x)
: f
(
x
)
=
F
"(
x
)
.
From this definition it follows that the distribution function is an antiderivative of the distribution density. Note that the distribution density is not applicable to describe the probability distribution of a discrete random variable. Probability of a continuous random variable falling into a given interval
Knowing the distribution density, you can calculate the probability that a continuous random variable will take a value belonging to a given interval. Theorem:
The probability that a continuous random variable X will take values belonging to the interval (a,
b), is equal to a certain integral of the distribution density, taken in the range fromabeforeb :
Proof: We use the ratio P(a
≤
Xb)
=
F(b)
–
F(a).
According to the Newton-Leibniz formula, Thus, Because P(a
≤
X b)=
P(a
X b)
, then we finally get
Geometrically, the obtained result can be interpreted as follows: the probability that a continuous random variable will take a value belonging to the interval (a,
b), equal to the area of a curvilinear trapezoid bounded by the axisOx, distribution curvef(x) and straightx =
aAndx =
b.
Comment: In particular, if f(x)
– the function is even and the ends of the interval are symmetrical relative to the origin, then Example. The probability density of a random variable is given X Find the probability that as a result of the test X will take values belonging to the interval (0.5, 1). Solution: Required probability Finding the distribution function from a known distribution density
Knowing the distribution density f(x)
, we can find the distribution function F(x)
according to the formula Really, F(x)
=
P(X
x) = P(-∞
X x)
.
Hence, Thus, Knowing the distribution density, you can find the distribution function. Of course, from a known distribution function one can find the distribution density, namely: f(x)
=
F"(x).
Example. Find the distribution function for the given distribution density: Solution: Let's use the formula
If x
≤
a, That f(x)
= 0
, hence, F(x)
= 0
. If a , then f(x) = 1/(b-a),
hence, If x
>
b, That So, the required distribution function Comment: We obtained the distribution function of a uniformly distributed random variable (see uniform distribution). Properties of distribution density
Property 1: Distribution density is a non-negative function: f
(
x
)
≥ 0
. Property 2: The improper integral of the distribution density in the range from -∞ to ∞ is equal to unity: Comment: The distribution density graph is called distribution curve. Comment: The distribution density of a continuous random variable is also called the distribution law. Example. The distribution density of the random variable has the following form: Find a constant parameter a. Solution: The distribution density must satisfy the condition , so we will require that the equality be satisfied From here
Let's calculate the improper integral: Thus, the required parameter Probable meaning of distribution density
Let F(x)
– distribution function of a continuous random variable X. By definition of distribution density, f(x)
=
F"(x)
, or Difference F(x+∆x) -F(x)
determines the probability that X will take a value belonging to the interval (x,
x+∆x). Thus, the limit of the probability ratio that a continuous random variable will take a value belonging to the interval (x,
x+∆x), to the length of this interval (at ∆х→0) is equal to the value of the distribution density at the point X. So the function f(x)
determines the probability distribution density for each point X. From differential calculus it is known that the increment of a function is approximately equal to the differential of the function, i.e. Because F"(x)
=
f(x)
And dx = ∆
x, That F(x+∆
x)
-
F(x)
≈
f(x)∆
x. The probabilistic meaning of this equality is: the probability that a random variable will take a value belonging to the interval (x,
x+∆
x) is approximately equal to the product of the probability density at point x and the length of the interval ∆x. Geometrically, this result can be interpreted as follows:
the probability that a random variable will take a value belonging to the interval (x,
x+∆
x) is approximately equal to the area of a rectangle with base ∆х and heightf(x).
Definition5.1:
Random value X, taking two values 1
And 0
with probabilities (“success”) p and (“failure”) q, called Bernoullievskaya: ,
Where k=0,1.
Let it be produced n
independent trials, in each of which the event A may or may not appear. The probability of an event occurring in all trials is constant and equal p(hence the probability of non-occurrence q = 1 -
p). Consider the random variable X– number of occurrences of the event A in these tests. Random value X takes values 0,1,2,…
n with probabilities calculated using the Bernoulli formula:
, Where k = 0,1,2,…
n. Definition5.2:
Binomial is called the probability distribution determined by Bernoulli's formula. Example. Three shots are fired at the target, and the probability of hitting each shot is 0.8. Consider a random variable X– number of hits on the target. Find its distribution series. Solution: Random value X takes values 0,1,2,3
with probabilities calculated using the Bernoulli formula, where n = 3,
p
= 0,8
(probability of hit), q
= 1 - 0,8 = = 0,2
(probability of missing). Thus, the distribution series has the following form: Use Bernoulli's formula for large values n quite difficult, therefore, to calculate the corresponding probabilities, use the local Laplace theorem, which allows you to approximately find the probability of the occurrence of an event exactly k once every n tests, if the number of tests is large enough. Local Laplace theorem: If the probability p occurrence of an event A
Note1: Tables containing function values
Example: Find the probability that the event A
will come exactly 80
once every 400
trials if the probability of occurrence of this event in each trial is equal to 0,2.
Solution: By condition n = 400,
k = 80,
p
= 0,2
, q = 0,8
. Let's calculate the value determined by the task data x:
If you need to calculate the probability that an event A will appear in n tests no less k 1
once and no more k 2
times, then you need to use Laplace’s integral theorem: Laplace's integral theorem: If the probability p occurrence of an event A in each trial is constant and different from zero and one, then the probability
In other words, the probability that an event A
will appear in n tests from k 1
before k 2
times, approximately equal Where
Note2: Function
Example: Find the probability that among
400
randomly selected parts will turn out to be untested from 70 to 100 parts, if the probability that the part did not pass the quality control inspection is equal to 0,2.
Solution: By condition n = 400,
p = 0,2
, q
= 0,8,
k 1
=
70,
k 2
=
100
. Let's calculate the lower and upper limits of integration: Thus we have: From the table in Appendix 2 we find that
Note3: In a series of independent trials (when n is large, p is small), the Poisson formula is used to calculate the probability of an event occurring exactly k times (see Poisson distribution). Definition5.3:
A discrete random variable is called Poisson, if its distribution law has the following form: Examples of Poisson random variables: Number of calls to an automatic station over a period of time T.
The number of decay particles of some radioactive substance over a period of time T.
Number of TVs that arrive at the workshop over a period of time T in the big city .
Number of cars that will arrive at the stop line of an intersection in a large city .
Note1: Special tables for calculating these probabilities are given in Appendix 3. Note2: In a series of independent tests (when n great, p is not enough) to calculate the probability of an event occurring exactly k times using Poisson's formula:
,
Where
,
that is, the average number of occurrences of events remains constant. Note3: If there is a random variable that is distributed according to the Poisson law, then there is necessarily a random variable that is distributed according to the exponential law and, vice versa (see Exponential distribution). Example. The plant sent to the base 5000
good quality products. The probability that the product will be damaged in transit is equal to 0,0002
. Find the probability that exactly three unusable products will arrive at the base. Solution: By condition n = 5000,
p
= 0,0002,
k = 3.
We'll find λ: λ =
n.p.= 5000·0.0002 = 1. According to the Poisson formula, the desired probability is equal to: ,
where is the random variable X– number of unusable products. Let independent tests be carried out, in each of which the probability of the event occurring is A equal to p(0 p q
= 1 -
p. Challenges end as soon as the event appears A. Thus, if an event A appeared in k-th test, then in the previous k
– 1
it did not appear in tests. Let us denote by X discrete random variable - the number of trials that need to be carried out before the first occurrence of the event A. Obviously, the possible values X are integers x 1 = 1, x 2 = 2, ... Let first k-1
testing event A did not come, but in k-th test appeared. The probability of this “complex event”, according to the theorem of multiplication of probabilities of independent events,
P
(X =
k) =
q
k -1
p.
Definition5.4:
A discrete random variable has geometric distribution, if its distribution law has the following form: P
(
X
=
k
) =
q
k
-1
p
,
Where
.
Note1: Believing k = 1,2,…
, we get a geometric progression with the first term p and denominator q (0q. For this reason, the distribution is called geometric.
Note2: Row
converges and its sum is equal to one. Indeed, the sum of the series is equal to
. Example. The gun is fired at the target until the first hit is made. Probability of hitting target p
= 0,6
. Find the probability that a hit will occur on the third shot. Solution: By condition p = 0,6,
q
= 1 – 0,6 = 0,4,
k = 3.
The required probability is: P
(X = 3) = 0,4
2
·0.6 = 0.096. Let's consider the following problem. Let the party out N products available M standard (MN).
Randomly taken from the batch n products (each product can be extracted with the same probability), and the selected product is not returned to the batch before selecting the next one (therefore, the Bernoulli formula is not applicable here). Let us denote by X random variable - number m standard products among n selected. Then the possible values X will be 0, 1, 2,…, min; Let's label them and... By values of the independent variable (Fonds) use the button ( chapter ... ... methodological instructions By performing practical work 5.1 Methodical recommendations By implementation of educational projects 5.2 Methodical recommendations By... sensitivity), one-dimensional and multidimensional... random component in size... With section"Performance... ... sections in textbooks. Problem solving By each topic. Elaboration methodological instructions for laboratory work By ... random and instrumental measurement error 1.8 Topics tests And methodological instructions By...Particle in one-dimensional potential hole. ... ... Methodical instructions for LABORATORY WORK By ... size, and the largest amount quantities... array random numbers... 3.0 4.0 3.0 -2.5 14.3 16.2 18.0 1.0 a) one-dimensional array b) two-dimensional array Fig. 2– Files... are described in section implementation after...
1.2.
p4=0.1; 0 at x≤-1,
https://pandia.ru/text/78/455/images/image038_17.gif" width="14" height="86"> 0 for x≤a,
,
The normal curve is symmetrical with respect to the straight line x=m, has a maximum at x=a, equal to .
,
4. Probability density of a continuous random variable
.
.
.
.
.
.
.
. Let's find the indefinite integral:
.
.5. Typical distributions of discrete random variables
5.1. Bernoulli distribution
5.2. Binomial distribution
that the event A
will appear in n tests exactly k times, approximately equal (the more accurate, the more n) function value
,
Where
,
.
,
are given in Appendix 1, and
.
Function
is the density of the standard normal distribution (see normal distribution).
.
From the table in Appendix 1 we find
.
Then the required probability will be:
that the event A
will appear in n tests from k 1
before k 2
times, approximately equal to a certain integral
,
Where
And
.
,
And
.
called the Laplace function (see normal distribution). Tables containing function values
,
are given in Appendix 2, and
.
;
.
And
.
Then the required probability is:5.3. Poisson distribution
,
Where
And
(constant value).5.4. Geometric distribution
5.5. Hypergeometric distribution
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