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Test 7
Normal distribution law. The probability of a normally distributed random variable (NDSV) falling into 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 CB X will take a value belonging to the interval 
: 
, Where 
- Laplace function. Function 
determined from tables.
At 
=0 curve symmetrical relative to the op-amp axis 
is the standard (or standardized) normal distribution.
Since the probability density function of the NRSV is symmetrical with respect to the mathematical expectation, it is possible to construct the so-called dispersion scale:
It can be seen that with a probability of 0.9973 it can be stated that the NRSV will take values within the interval 
. This statement is called the “Three Sigma Rule” in probability theory.
1. Compare the values
for two NRSV curves. 
1) 
2) 
2. Continuous random variable X is specified by the probability distribution density
. Then the mathematical expectation of this normally distributed random variable is equal to: 1) 3 2) 18 3) 4 4) 
3. NRSV X is given by the distribution density: 
.
Expected value 
and the dispersion of this SV are equal to:
1) =1 2) =5 3) =5
=25 =1 =25
4. The three sigma rule means that:
1) Probability of SV hitting the interval 
, that is, close to unity;
2) NRSV cannot go beyond 
;
3) The NRSV density graph is symmetrical with respect to the mathematical expectation
5. SV X is distributed normally with a mathematical expectation equal to 5 and standard deviation equal to 2 units. The expression for the distribution density of this NRSV has the form:
1) 
2) 
3) 
6. The mathematical expectation and standard deviation of NRSV X are equal to 10 and 2. The probability that, as a result of the test, SV X will take the value contained in the interval is:
1) 0,1915 2) 0,3830 3) 0,6211
7. The part is considered suitable if the deviation X of the actual size from the size in the drawing according to absolute value less than 0.7 mm. Deviations X from the size in the drawing are NRSV with the value
=0.4 mm. 100 parts produced; Of these, the following will be suitable: 1) 92 2) 64 3) 71
8. The mathematical expectation and standard deviation of NRSV X are equal to 10 and 2. The probability that, as a result of the test, SV X will take the value contained in the interval is:
1) 0,1359 2) 0,8641 3) 0,432
9. The error X of manufacturing a part is NRSV with the value a=10 and =0.1. Then, with a probability of 0.9973, the interval of part sizes that is symmetrical with respect to a=10 will be:
1) 9,7; 10,3 2) 9,8; 10,2 3) 9,9; 10,1
10. Weigh all products 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 carried out with an error not exceeding 15 g in absolute value is:
1) 0,8664 2) 0,1336 3) 0,4332
11. NRSV X has a mathematical expectation a=10 and standard deviation =5. With a probability of 0.9973, the value of X will fall into the interval:
1) (5; 15) 2) (0; 20) 3) (-5; 25)
12. NRSV X has a mathematical expectation a=10. It is known that the probability of X falling into the interval is 0.3. Then the probability of CB X falling into the interval will be equal to:
1) 0,1 2) 0,2 3) 0,3
13. NRSV X has a mathematical expectation a=25. The probability of X falling into the interval is 0.2. Then the probability of X falling into the interval will be equal to:
1) 0,1 2) 0,2 3) 0,3
14. The room temperature is maintained by a 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 a 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 that are 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 NRSV, the mathematical expectation a=4. |
(3) |
|
3 |
In accordance with the equation for the density of NRSV we have: =1; =5, that is  .
|
(1) |
|
4 |
Answer (1) is correct. |
(1) |
|
5 |
The expression for the NRSV distribution density has the form:  . By condition: =2; a
=5, that is, answer (1) is correct. |
(1) |
|
6 |
By condition =10; =2. The interval is . Then:  ;  .
According to the Laplace function tables: |
(2) |
|
7 |
By condition: =0;  ;=0.4. This means the interval will be [-0.7; 0.7].
That is, out of 100 parts, 92 pieces are most likely to be suitable. |
(1) |
; . Then the desired probability: 
; 
.
;
|
8 |
By condition: =10 and =2. The interval is . Then:  ;  . According to the Laplace function tables:  ;  ;
|
(1) |
|
9 |
In an interval symmetrical with respect to the mathematical expectation a =10 with probability 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: |
; 
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Any fractal is constructed 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 a 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 cubes adjacent to it along the faces 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 consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.
Probability of falling into a given interval of a normal random variableIt is already known that if a random variable X is given by the distribution density f (x), then the probability that X will take a value belonging to the interval (a, b) is as follows:
Let the random variable X be distributed according to the normal law. Then the probability that X will take a value belonging to the interval (a,b) is equal to
Let's transform this formula so that you can use ready-made tables. Let's introduce a new variable z = (x--а)/--s. Hence x = sz+a, dx = sdz. Let's find new limits of integration. If x= a, then z=(a-a)/--s; if x = b, then z = (b-a)/--s.
Thus we have
Using the Laplace function
we'll finally get it
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