The probability of a random variable hitting a given interval. Normal distribution law. the probability of a normally distributed random variable (NRSV) falling into a given interval. Calculation of the probability of a given deviation of the normal random
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Test 7
Normal distribution law. The probability of hitting a normally distributed random variable (NRSV) in a given interval.
Basic information from the theory.
The probability distribution of a random variable (RV) is called normal X, if the distribution density is determined by the equation:
Where a- mathematical expectation of SV X; 
- standard deviation.
Schedule 
symmetrical about a vertical line 
... The more, the greater the range of the curve ... Function values are available in the tables.
The probability that SV X will take a value belonging to the interval 
: 
, where 
is the Laplace function. Function 
determined by tables.
At 
= 0 curve symmetric about the op-amp axis 
is the standard (or normalized) normal distribution.
Since the probability density function of the LRSV is symmetric with respect to the mathematical expectation, it is possible to construct the so-called scattering scale:
It can be seen that with a probability of 0.9973 it can be argued that the NRSV will take values within the interval 
... This statement is called the Three Sigma Rule in probability theory.
1. Compare the quantities
for the two LRSV curves. 
1) 
2) 
2. Continuous random variable X is given by the probability distribution density
... Then the mathematical expectation of this normally distributed random variable is: 1) 3 2) 18 3) 4 4) 
3. LRSV X is given by the distribution density: 
.
Expected value 
and the variance of this RV are equal:
1) =1 2) =5 3) =5
=25 =1 =25
4. The rule of three sigma means that:
1) The probability of hitting the SW in the interval 
, that is, close to one;
2) LDCV cannot go beyond 
;
3) The density graph of the LRSV is symmetric with respect to the mathematical expectation
5. SV X is distributed normally with mathematical expectation equal to 5 and standard deviation equal to 2 units. The expression for the distribution density of this LDCV has the form:
1) 
2) 
3) 
6. The mathematical expectation and standard deviation of LRSV X are equal to 10 and 2. The probability that, as a result of the test, SV X will take the value enclosed in the interval is:
1) 0,1915 2) 0,3830 3) 0,6211
7. A part is considered suitable if the deviation X of the actual size from the size in the drawing in absolute value is less than 0.7 mm. The deviations of X from the size in the drawing are NRSV with the value
= 0.4 mm. Manufactured 100 parts; of which will be suitable: 1) 92 2) 64 3) 71
8. The mathematical expectation and standard deviation of the LRSV X are equal to 10 and 2. The probability that as a result of the test the SV X will take the value enclosed in the interval is:
1) 0,1359 2) 0,8641 3) 0,432
9. Uncertainty X of manufacturing a part is NRSV with the value a= 10 and = 0.1. Then, with a probability of 0.9973, the range of sizes of parts, symmetric with respect to a= 10 would be:
1) 9,7; 10,3 2) 9,8; 10,2 3) 9,9; 10,1
10. Weigh all items without systematic errors. Random errors of X measurements are subject to the normal law with the value = 10 g. The probability that weighing will be performed with an error not exceeding 15 g in absolute value is:
1) 0,8664 2) 0,1336 3) 0,4332
11. LRSV X has a mathematical expectation a= 10 and RMS = 5. With a probability of 0.9973, X will fall into the interval:
1) (5; 15) 2) (0; 20) 3) (-5; 25)
12. LDCV X has a mathematical expectation a= 10. It is known that the probability of X hitting the interval is 0.3. Then the probability of hitting SV X in the interval will be equal to:
1) 0,1 2) 0,2 3) 0,3
13. LRSV X has a mathematical expectation a= 25. The probability of X hitting the interval is 0.2. Then the probability of getting X into the interval will be equal to:
1) 0,1 2) 0,2 3) 0,3
14. The room temperature is maintained by the heater and has a normal distribution with
and 
... The probability that the temperature in this room will be between 
before 
is: 1) 0,95 2) 0,83 3) 0,67
15. For the standardized normal distribution, the value is:
1) 1 2) 2 3) 
16. An empirical normal distribution is formed when:
1) there are a large number of independent random causes that have approximately the same statistical weight;
2) there are a large number of random variables strongly dependent on each other;
3) the sample size is small.
|
1 |
Meaning determines the range of the distribution density curve relative to the mathematical expectation. For curve 2, the range is larger, that is |
(2) |
|
2 |
In accordance with the equation for the density of the LRSV, the mathematical expectation a=4. |
(3) |
|
3 |
In accordance with the equation for the LRSV density, we have: =1; = 5, that is  .
|
(1) |
|
4 |
Answer (1) is correct. |
(1) |
|
5 |
The expression for the density distribution of the LRSV is:  ... By condition: = 2; a
= 5, that is, answer (1) is correct. |
(1) |
|
6 |
By condition =10; = 2. The interval is equal. Then:  ;  .
According to the tables of the Laplace function: |
(2) |
|
7 |
By condition: =0;  ;= 0.4. So the interval will be [-0.7; 0.7].
That is, out of 100 parts, 92 are most likely to be usable. |
(1) |
; ... Then the required probability is: 
; 
.
;
|
8 |
By condition: = 10 and = 2. The interval is equal. Then:  ;  ... According to the tables of the Laplace function:  ;  ;
|
(1) |
|
9 |
Into an interval symmetric with respect to the mathematical expectation a = 10 with a probability of 0.9973, all parts with dimensions equal to  , that is ; ... Thus: |
(1) |
|
10 |
By condition  ,that is = 0, and the interval will be [-15; 15] Then: |
; 
. Let us find the distribution function of the random variable NS subject to the normal distribution law:
we make a replacement in the integral and bring it to the form:
.
Integral is not expressed in terms of elementary functions, but it can be calculated through a special function expressing a definite integral of the expression or. Let us express the function through the Laplace function Ф (х):
.
The probability of hitting a random variable X on the section (α, β) is expressed by the formula:
.
Using the last formula, we can estimate the probability of deviation of a normal random variable from its mathematical expectation by a predetermined arbitrarily small positive value ε:
.
Let, then and. At t= 3 we get, i.e. the event that the deviation of a normally distributed random variable from the mathematical expectation will be less is practically reliable.
This is the three sigma rule: if the random variable is normally distributed, then the absolute value of the deviation of its values from the mathematical expectation does not exceed three times the standard deviation.
Task. Let the diameter of the part produced by the shop be a random variable distributed normally, m = 4.5 cm, cm. Find the probability that the size of the diameter of a randomly taken part differs from its mathematical expectation by no more than 1 mm.
Solution... This problem is characterized by the following values of the parameters that determine the desired probability:, 
, F (0.2) = 0.0793,
Control questions
1. What probability distribution is called uniform?
2. What is the form of the distribution function of a random variable uniformly distributed on the segment [ a; b]?
3. How to calculate the probability of the values of a uniformly distributed random variable falling within a given interval?
4. How is the exponential distribution of a random variable determined?
5. What is the form of the distribution function of a random variable distributed according to the exponential law?
6. What probability distribution is called normal?
7. What properties does the normal distribution density have? How do the normal distribution parameters affect the appearance of the normal distribution density graph?
8. How to calculate the probability of falling values of a normally distributed random variable in a given interval?
9. How to calculate the probability of deviation of the values of a normally distributed random variable from its mathematical expectation?
10. What is the Three Sigma Rule?
11. What are the mathematical expectation, variance and standard deviation of a random variable distributed according to a uniform law on the segment [ a; b]?
12. What are the mathematical expectation, variance and standard deviation of a random variable distributed according to the exponential law with the parameter λ?
13. What are the mathematical expectation, variance and standard deviation of a random variable distributed according to the normal law with parameters m and ?
Control tasks
1. Random variable NS distributed uniformly on the segment [−3, 5]. Find the distribution density and distribution function NS... Plot both functions. Find the probabilities and. Calculate expectation, variance and standard deviation NS.
2. Buses of route number 21 run regularly at intervals of 10 minutes. The passenger goes to the stop at a random time. Consider a random variable NS- waiting time for the bus passenger (in minutes). Find the distribution density and distribution function NS... Plot both functions. Find the probability that the passenger will have to wait no more than five minutes for the bus. Find the average waiting time for the bus and the variance of the waiting time for the bus.
3. It has been established that the repair time of the video recorder (in days) is a random variable NS, distributed according to the exponential law. The average time to repair a VCR is 10 days. Find the distribution density and distribution function NS... Plot both functions. Find the probability that it will take at least 11 days to repair the VCR.
4. Draw the graphs of the density and distribution function of a random variable NS distributed according to the normal law with parameters m= = - 2 and = 0.2.
FORMS OF DEFINING THE DISTRIBUTION LAW FOR CONTINUOUS RANDOM VALUES
FORMS OF DETERMINING THE LAW OF DISCRETE RANDOM VALUES DISTRIBUTION
1). Distribution table (row) - the simplest form of setting the law of distribution of discrete random variables.
Since the table lists all possible values of the random variable.
2). Distribution polygon ... When graphically displaying a distribution series in a rectangular coordinate system, all possible values of a random variable are plotted along the abscissa axis, and the probabilities corresponding to them are plotted along the ordinate axis. Then points are applied and connected with straight line segments. The resulting figure - the distribution polygon - is also a form of specifying the distribution law of a discrete random variable.
3). Distribution function - the probability that a random variable X will take on a value less than some given x, i.e.
| . |
From a geometric point of view, it can be considered as the probability of hitting a random point NS to the segment of the numerical axis located to the left of the fixed point NS.
2) ; ;
Task 2.1. Random value NS- the number of hits on the target with 3 shots (see problem 1.5). Construct a distribution series, distribution polygon, calculate the values of the distribution function and plot its graph.
Solution:
1) A series of distribution of a random variable NS presented in the table
| At | , |
| At | , |
| At | , |
| At | |
| at | . |
Plotting along the abscissa the values NS, and along the ordinate - values and choosing a certain scale, we get a graph of the distribution function (Fig. 2.2). The distribution function of a discrete random variable has jumps (discontinuities) at those points at which the random variable NS takes the specific values indicated in the distribution table. The sum of all jumps in the distribution function is equal to one.
Rice. 2.2 - Distribution function of a discrete quantity
1). Distribution function .
For a continuous random variable, the graph of the distribution function (Fig. 2.3) has the shape of a smooth curve.
Distribution function properties:
c) if.
Rice. 2.3 - Distribution function of continuous quantity
2). Distribution density defined as derivative of the distribution function, i.e.
| . |
Curve depicting the distribution density of a random variable is called distribution curve (fig. 2.4).
Density properties:
and those. density is a non-negative function;
b), i.e. limited area distribution curve and the abscissa is always equal to 1.
If all possible values of the random variable NS are between a before b, then the second density property will take the form:
Rice. 2.4 - Distribution curve
In practice, it is often necessary to know the probability that a random variable NS will take a value within certain limits, for example, from a to b. Seeking probability for discrete random variable NS is determined by the formula
since the probability of any single value of a continuous random variable is zero:.
The probability of hitting a continuous random variable NS for the interval (a, b) is also determined by the expression:
Task 2.3. Random value NS given by the distribution function
Find the density, as well as the probability that, as a result of the test, a random variable NS will take the value enclosed in the interval.
Solution:
2. The probability of hitting a random variable NS in the interval is determined by the formula. Taking and, we find
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Any fractal is built according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.
The iterative algorithm for constructing the Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 adjacent cubes are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set, already consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.
Rice. 4. Density of normal distribution.
Example 6. Determination of the numerical characteristics of a random variable by its density is considered on an example. A continuous random variable is given by the density
Determine the type of distribution, find the mathematical expectation M (X) and the variance D (X).
Solution. Comparing the given distribution density with (1.16), we can conclude that the normal distribution law is given with m = 4. Therefore, the mathematical expectation
M (X) = 4, variance D (X) = 9.
Standard deviation σ = 3.
The normal distribution function (1.17) is related to the Laplace function, which has the form:
relation: Φ (- x) = −Φ (x). (The Laplace function is odd). The values of the functions f (x) and Ф (х) can be calculated using the table.
The normal distribution of a continuous random variable plays an important role in probability theory and in describing reality; it is very widespread in random natural phenomena. In practice, random variables are very often encountered, which are formed precisely as a result of the summation of many random terms. In particular, the analysis of measurement errors shows that they are the sum of various kinds of errors. Practice shows that the probability distribution of measurement errors is close to the normal law.
Using the Laplace function, one can solve the problem of calculating the probability of falling into a given interval and a given deviation of a normal random variable.
3.4. Probability of hitting a given interval of a normal random variable
If a random variable X is given by the distribution density f (x), then the probability that X will take a value belonging to a given interval is calculated by formula (1.9a). Substituting into formula (1.9a) the value of the distribution density from (1.16) for the normal distribution N (a, σ) and making a number of transformations, the probability that X takes a value belonging to a given interval will be equal to:
P (x 1 ≤ X ≤ x 2) = Φ (x 2 σ - a)
where: a - mathematical expectation.
−Φ( |
x1 - a |
||
Example 7. A random variable X is distributed according to the normal law. The mathematical expectation is a = 60, the standard deviation is σ = 20. Find the probability of hitting a random variable X in a given interval (30; 90).
Solution. The required probability is calculated by the formula (1.18).
We get: P (30< X < 90) = Ф((90 – 60) / 20) –Ф((30 – 60)/20) = 2Ф(1,5).
According to the table in Appendix 1: Ф (1.5) = 0.4332 .. P (30< X < 90)=2 Ф(1,5) = 2 0,4332 = 0,8664.
The probability of hitting a random variable X in a given interval (30; 90) is: P (30< X < 90) = 0,8664.
3.5. Calculation of the probability of a given deviation of a normal random variable
The tasks of calculating the probability of deviation of a normal random variable from a given value are associated with various kinds of errors (measurements, weighing). Errors of various kinds are indicated by the variable ε.
Let ε be the deviation of a normally distributed random variable X modulo. It is required to find the probability that the deviation of the random variable X from the mathematical expectation does not exceed a given value ε. This probability is written as: P (| X – a | ≤ ε). It is assumed that in the formula (1.18) the segment [x1; x2] is symmetric with respect to the mathematical expectation a. Thus: a – х1 = ε; х2 –a = ε. If these expressions are added, we can write: x2 - x1 = 2ε. The boundaries of the interval [x1; x2] will look like:
x1 = a –ε; x2 = a + ε. |
The values x1, x2 from (1.19) are substituted into the right side of (1.18), and the expression in curly braces is rewritten as two inequalities:
1) х 1 ≤ X and is replaced in it х1 according to (1.19), we get: а – ε ≤ X or а – X ≤ ε.
2) X ≤ х 2, similarly replaced by х2, it turns out: X ≤ a + ε or X – a ≤ ε.
Example 8. Measurement of the diameter of the part. Random measurement errors are taken as a random value X and are subject to the normal law with mathematical expectation a = 0, with standard deviation σ = 1 mm. Find the probability that the measurement will be made with an error not exceeding 2 mm in absolute value.
Solution. Given: ε = 2, σ = 1mm, a = 0.
According to the formula (5.20): P (| X – 0 | ≤ 2) = 2Ф (ε / σ) = 2Ф (2/1) = 2Ф (2,0).
The probability that the measurement will be made with an error not exceeding 1mm in absolute value is equal to:
P (| X | ≤ ε) = 2 0.4772 = 0.9544.
Example 9. A random variable distributed according to the normal law with the parameters: a = 50 and σ = 15. Find the probability that the deviation random variable from its mathematical expectation - and will be less than 5, i.e. P (| X – a |<5).
Solution. Taking into account (1.18), we have: P (| X– a |< ε )=2Ф(ε /σ );
