Normal law of probability distribution. Probability of normal CV values falling into an interval that is symmetric with respect to the mathematical expectation, the rule of three sigma Find an interval that is symmetric with respect to the mathematical expectation
The mathematical expectation a=3 and the standard deviation =5 of a normally distributed random variable X are given.
Write the probability distribution density and graph it schematically.
Find the probability that x takes a value from the interval (2;10).
Find the probability that x will be greater than 10.
Find an interval that is symmetric with respect to the mathematical expectation, in which, with a probability =0.95, the values of x will be enclosed.
one). Compose the distribution density function of a random variable X with parameters a=3, =5 using the formula
. We construct a schematic graph of the function 
. Let us pay attention to the fact that the normal curve is symmetric about the straight line x=3 and has max at this point equal to 
, i.e. 
and two inflection points 
with ordinate 
Let's build a graph 
2) Let's use the formula: 
The function values are found from the application table.
4) Let's use the formula 
. By condition, the probability of falling into the interval symmetrical with respect to the mathematical expectation 
. According to the table, we find t, at which Ф (t) = 0.475, t = 2. means 
. In this way, 
. The answer is x (-1; 7).
To tasks 31-40.
Find the confidence interval for estimating with a reliability of 0.95 the unknown mathematical expectation a of the normally distributed feature X of the general population, if the general standard deviation =5, the sample mean 
and sample size n=25.
It is required to find the confidence interval 
.
All quantities except t are known. Let's find t from the ratio Ф(t)=0.95/2=0.475. According to the application table, we find t=1.96. Substituting, we finally get the desired confidence interval 12.04 The technical control department checked 200 batches of identical products and obtained the following empirical distribution, frequency n i is the number of batches containing x i non-standard products. It is required at a significance level of 0.05 to test the hypothesis that the number of non-standard products X is distributed according to Poisson's law. Let's find the sample mean: Let us take as an estimate of the parameter of the Poisson distribution the sample mean =0.6. Therefore, the putative Poisson's law Putting i=0,1,2,3,4 we find the probabilities P i of the appearance of i non-standard products in 200 batches: Find the theoretical frequencies by the formula Let's compare the empirical and theoretical frequencies using the Pearson criterion. To do this, we will make a calculation table. Let's combine the few frequencies (4+2=6) and their corresponding theoretical frequencies (3.96+0.6=4.56). The probability that the CB deviation X from her M.O. a in absolute value will be less than a given positive number , is equal to If we put in this equality, then we get w:space="720"/>"> That is, a normally distributed SW X deviates from his M.O. a, as a rule, by less than 3. This is the so-called 3 sigma rule, which is often used in mathematical statistics. Function of one random variable. Mathematical expectation of the function of one SV. (tetr) If each possible value of a random variable X
corresponds to one possible value of the random variable Y
, then Y
called random argument function X: Y=φ
(X
). Let's find out how to find the law of distribution of a function according to the known law of distribution of the argument. 1) Let the argument X
is a discrete random variable, and different values X
correspond to different values Y
. Then the probabilities of the corresponding values X
And Y
equal .
2) If different values X
can match the same values Y
, then the probabilities of the values of the argument for which the function takes the same value are added. 3) If X
is a continuous random variable, Y=φ
(X
), φ
(x
) is a monotone and differentiable function, and ψ
(at
) is the function inverse to φ
(X
). Mathematical expectation of a function of one random argument. Let be Y=φ
(X
) is a function of a random argument X
, and it is required to find its mathematical expectation, knowing the distribution law X
. 1) If X
is a discrete random variable, then 2) If X
is a continuous random variable, then M
(Y
) can be searched in different ways. If the distribution density is known g
(y
), then 21. Function of two random arguments. Distribution of function Z=X+Y for discrete independent SV X and Y.(tetr) If each pair of possible values of random variables X and Y corresponds to one possible value of random variable Z, then Z is called a function of two random arguments X and Y and write Z=φ(X,Y). If X and Y are discrete independent random variables, then in order to find the distribution of the function Z=X+Y, it is necessary to find all possible values of Z, for which it is enough to add each possible value of X to all possible values of Y; the probabilities of the found possible values Z are equal to the products of the probabilities of the added values X and Y. If X and Y are continuous independent random variables, then the distribution density g(z) of the sum Z = X + Y (provided that the distribution density of at least one of the arguments is given in the interval (- oo, oo) by one formula) can be found by the formula , or by an equivalent formula , where f1 and f2 are the distribution densities of the arguments; if the possible values of the arguments are non-negative, then the distribution density g(z) of the value Z=X + Y is found by the formula , or by an equivalent formula . In the case when both densities f1(x) and f2(y) are given on finite intervals, to find the density g(z) of the value Z = X + Y, it is advisable to first find the distribution function G(z) and then differentiate it with respect to z : g(z)=G'(z). If X and Y are independent random variables given by the corresponding distribution densities f1(x) and f2(y), then the probability of a random point (X, Y) falling into the region D is equal to the double integral over this region of the product of the distribution densities: Р [( X, Y)cD] = Р 0.3 0.7 Р 0.6 0.4 Find the distribution of the random variable Z = X + K. Solution. In order to compose the distribution of the value Z=X+Y, it is necessary to find all possible values of Z and their probabilities. The possible Z values are the sums of each possible X value with all possible Y values: Z 1 = 1+2=3; z 2 \u003d 1 + 4 \u003d 5; z 3 \u003d 3 + 2 \u003d 5; z4 = 3+4 = 7. Let's find the probabilities of these possible values. In order for Z=3, it is sufficient that the value X takes the value x1= l and the value K-value y1=2. The probabilities of these possible values, as follows from these distribution laws, are respectively equal to 0.3 and 0.6. Since the arguments X and Y are independent, the events X = 1 and Y = 2 are independent and, therefore, the probability of their joint occurrence (i.e., the probability of the event Z = 3) according to the rain multiplication theorem is 0.3 * 0.6 = 0 ,eighteen. Similarly, we find: I B=!-f4 = 5) = 0.3 0.4 = 0.12; P(Z = 34-2 = 5) = 0.7 0.6 = 0.42; P(Z = 3rd = 7) = 0.7-0.4 = 0.28. Let's write the desired distribution by adding up the probabilities of incompatible events Z = z 2 = 5, Z=z 3 = 5 (0.12+0.42=0.54): Z 3 5 7 ; P 0.18 0.54 0.28. Control: 0.18 + 0.54 + 0.28 = 1. As mentioned earlier, examples of probability distributions continuous random variable
X are: Let's give the concept of a normal distribution law, the distribution function of such a law, the procedure for calculating the probability of hitting a random variable X in a certain interval. The length X of some part is a random variable distributed according to the normal distribution law, and has a mean value of 20 mm and a standard deviation of 0.2 mm. Solution.
but) The probability density of a random variable X, distributed according to the normal law, we find: provided that m x =20, σ =0.2. b) For a normal distribution of a random variable, the probability of falling into the interval (19.7; 20.3) is determined by: in) The probability that the absolute value of the deviation is less than a positive number 0.1 is found: G) Since the probability of a deviation less than 0.1 mm is 0.383, it follows that on average 38.3 parts out of 100 will be with such a deviation, i.e. 38.3%. e) Since the percentage of parts whose deviation from the average does not exceed the specified one has increased to 54%, then P(|X-20|< δ) = 0,54. Отсюда следует, что 2Ф(δ/σ) = 0,54, а значит Ф(δ/σ) = 0,27. Using the application ( table 2
), we find δ/σ = 0.74. Hence δ = 0.74*σ = 0.74*0.2 = 0.148 mm. e) Since the desired interval is symmetric with respect to the mean value m x = 20, it can be defined as a set of X values satisfying the inequality 20 − δ< X < 20 + δ или |x − 20| < δ . By the condition, the probability of finding X in the desired interval is 0.95, which means that P(|x − 20|< δ)= 0,95. С другой стороны P(|x − 20| < δ) = 2Ф(δ/σ), следовательно 2Ф(δ/σ) = 0,95, а значит Ф(δ/σ) = 0,475. Using the application ( table 2
), we find δ/σ = 1.96. Hence δ = 1.96*σ = 1.96*0.2 = 0.392. Example 1 Mathematical expectation of a normally distributed continuous SW XM(X) = 6, and the standard deviation s( X) = 2. Find: 1) the probability of hitting the values of CB X in the interval (2; 9); 3) an interval symmetrical with respect to a X with probability g = 0.9642. Solution. 1) Find the probability of hitting the values of CB X into the interval (2; 9). Values of the Laplace function 2) Define the probability Because a = M(X) = 6 and s = s( X) = 2, then 3) Find an interval symmetrical with respect to a, which includes the values of SW X with probability g = 0.9642. Example 2 Random value T(hours) – the uptime of the device has an exponential distribution. Find the probability that the device will work without repair for at least 600 hours if the mean time of failure-free operation of devices of this type is 400 hours. Solution. M(T) = 400 hours, therefore, according to the formula (1.46) Since for the exponential distribution Example 3 Random value X distributed uniformly on the interval [ a, b]. Find the probability of hitting a random variable X for a segment Solution. Let's use the formula In this way Example 4 Electric trains run strictly according to the schedule with an interval Solution. X– waiting time (min.) for an electric train, can be considered a uniformly distributed random variable with a density: Example 5 The machine produces bushings. The sleeve is considered good if the deviation X its diameter from the design size in absolute value is less than 1 mm. Assuming that the random variable X normally distributed with standard deviation s = 0.5 mm and mathematical expectation a= 0, find how many suitable bushings will be among 100 manufactured ones, as well as the probability that the deviation from the design size will be at least 0.4 mm and not more than 0.8 mm. Solution. Let's use the formula () It follows that approximately 95 bushings out of 100 will be suitable. To find the probability that the deviation from the design size will be at least 0.4 mm and not more than 0.8 mm, we use the formula (1.54) Function values Ф( x) is found from the table. Task options OPTION 1 X (CB X) is given by the distribution series: F(x M(X), dispersion D(XX), fashion M 0 (X); 3) probability P(8≤ X < 30). Построить многоугольник распределения и график F(x). Task 2. Each of the shooters shoots at the target once. The probability that the first, second and third shooters will hit the target with one shot, respectively, are equal to 0.8; 0.6 and 0.9. For Task 3. Probability of occurrence of some event BUT in each experiment is 0.6. It is required: 1) to construct a distribution series of a discrete CB X– number of occurrences of the event BUT in four independent experiments; 2) estimate the probability that in a series of 80 independent experiments this event will occur at least 60 times. Problem 4. Discrete CB X given by a series of distributions: Find distribution series CB Y = –2X 2 + 3, M(Y) And D(Y). Task 5. Continuous CB X Find: a) distribution density f(x); b) M(x); in) Task 6. A function is given A CB X. To find F(x), Task 7. Given M(X) = 14 and s( X SW X. To find: 1) probability 2) probability 3) symmetrical with respect to a CB X with probability g = 0.8385. Task 8. The scale of the stopwatch has a division value of 0.2 s. The countdown is done to the nearest whole division with rounding to the nearest side. The reading error under these conditions can be considered a uniformly distributed random variable. Find the probability to use this stopwatch to count the time with an error of a) less than 0.05 s; b) not less than 0.01 s and not more than 0.05 s. OPTION 2 Problem 1. Discrete random variable X (CB X) is given by the distribution series: Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), dispersion D(X), standard deviation s( X), fashion M 0 (X); 3) probability P(–2≤ X < 5). Построить многоугольник распределения и график F(x). Problem 2. There are 100 tickets in the lottery, 10 of which are winning. Someone buys 4 tickets. For SV X- the number of winning tickets among those that will be purchased, make a distribution series and find F(x), M(X), s( X). Task 3. Reports are compiled independently from one another. The probability of making an error in each report is 0.3. Required: 1) build a distribution series CBX- the number of reports with errors among the four compiled; calculate M(X), D(X) and s( X); 2) estimate the probability that 50 reports will equal 20 reports with errors. Problem 4. It is known that the discrete CB X can only take two values x 1 = -2 and x 2 = 3 and its mathematical expectation M(X) = 1.5. Compile Distribution Series CB X And C.B. Z= Find F(z) and s( Z). Task 5. Continuous CB X given by the distribution function f(x); 2) M(x) And D(X); Task 6. A function is given Define parameter value A, at which this function specifies the probability distribution density of some continuous CB X. To find F(x), Task 7. Given M(X) = 12 and s( X SW X. To find: 1) probability 2) probability 3) symmetrical with respect to a the interval in which the values fall CB X with probability g = 0.4515. Problem 8. Random measurement error of some detail is subject to the normal law with the parameter s = 20 mm. Find the probability that: a) the part was measured with an error not exceeding 22 mm in absolute value; b) in none of the two measurements made, the error will not exceed 22 mm in absolute value. OPTION 3 Problem 1. Discrete random variable X (CB X) is given by the distribution series: Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), dispersion D(X), standard deviation s( X), fashion M 0 (X); 3) probability P(1≤ X < 7). Построить многоугольник распределения и график F(x). Problem 2. Of the three athletes included in the youth team of the country in high jump competitions, one can pass qualified starts with a probability of 0.9, the second with a probability of 0.8 and the third with a probability of 0.6. For CB X- the number of athletes of the national team who will go to the next round of competition, make a distribution series and find M(X), s( X). Task 3. A series of independent shots are fired at the target. The probability of hitting the target with each shot is 0.8. Required: 1) build a distribution series CBX- number of hits with three shots; 2) estimate the probability that with 100 shots there will be at least 90 hits. Problem 4. Discrete random variable X (CB X) is given by the distribution series: Find series and distribution function CB Y = 2X + 1, M(Y) And D(Y). Task 5. Continuous CB X given by the distribution function Find: 1) distribution density f(x); 2) M(x) And D(X); Task 6. A function is given Define parameter value A, at which this function specifies the probability distribution density of some continuous CB X. To find F(x), Task 7. Given M(X) = 13 and s( X SW X. To find: 1) probability 2) probability 3) symmetrical with respect to a the interval in which the values fall CB X with probability g = 0.9973. Problem 8. It is known that the TV repair time is a random variable X, distributed according to the exponential law, while the average TV repair time is two weeks. Find the probability that the repair of the TV brought to the workshop will take: a) less than 10 days; b) 9 to 12 days. OPTION 4 Problem 1. Discrete random variable X (CB X) is given by the distribution series: Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), dispersion D(X), standard deviation s( X), fashion M 0 (X); 3) probability P(–10≤ X < 1). Построить многоугольник распределения и график F(x). Task 2. The attendant has 5 different keys from different rooms. Taking out a key at random, he tries to open the door of one of the rooms. For a discrete CB X- the number of attempts to open the door (the verified key is not used a second time) make a distribution series and find F(x) And M(X). Problem 3. The probability of manufacturing a part with the given accuracy parameters from a standard workpiece for each part is 0.8. Required: 1) build a distribution series CB X- the number of parts with given accuracy characteristics that will be made from five standard blanks; 2) estimate the probability that 70 parts with given accuracy characteristics will be manufactured from 90 workpieces. CB X And Y: Compose a distribution series C.B. Z = Y – X. To find M(Z) And D(Z). Task 5. Continuous CB X given by the distribution function Find: 1) distribution density f(x); 2) M(x); 3) Task 6. A function is given Define parameter value A, at which this function specifies the probability distribution density of some continuous CB X. To find F(x), Task 7. Given M(X) = 16 and s( X) = 2 normally distributed continuous SW X. To find: 1) probability 2) probability 3) symmetrical with respect to a the interval in which the values fall CB X with probability g = 0.9281. Problem 8. The height of an adult male is SV X, distributed according to the normal law with parameters but\u003d 175 cm and s \u003d 10 cm. Find the probability that the height of a randomly selected man will be: a) less than 180 cm; b) not less than 170 cm and not more than 175 cm. OPTION 5 Problem 1. Discrete random variable X (CB X) is given by the distribution series: Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), dispersion D(X), standard deviation s( X), fashion M 0 (X); 3) probability P(40≤ X < 80). Построить многоугольник распределения и график F(x). Problem 2. The target consists of a circle and two concentric rings. Hitting the circle is worth 6 points, hitting ring 2 is worth 4 points, and hitting ring 3 is worth two points. The probabilities of hitting the circle and rings 2 and 3, respectively, are 0.2; 0.3 and 0.5. For a discrete SV X- the sum of points scored as a result of three hits, make a distribution series and find F(x), M(X), s( X). Task 3. The automatic line consists of n independent machines of the same type. The probability that the machine will require adjustment during the shift for each machine is 0.3. Required: 1) build a distribution series CB X- the number of machines that will require adjustment during the shift, if n= 4; 2) estimate the probability that 20 machines will require adjustment per shift, if n = 100. Problem 4. Joint distribution of discrete CB X And Y given by the table: Compose a distribution law C.B. Z = Y + X. To find M(Z) And D(Z). Task 5. Continuous CB X given by the distribution function Find: 1) distribution density f(x); 2) M(x) And D(X); Task 6. A function is given Define parameter value A, at which this function specifies the probability distribution density of some continuous CB X. To find F(x), Task 7. Given M(X) = 10 and s( X) = 4 normally distributed continuous SW X. To find: 1) probability 2) probability 3) symmetrical with respect to a the interval in which the values fall CB X with probability g = 0.5161. Problem 8. The minute hand of an electric clock jumps at the end of each minute. Random value X- the difference between the time shown on the scoreboard and the true time has a uniform distribution. Find the probability that at some point in time the clock will indicate a time that differs from the true one: a) by no less than 10 s and no more than 25 s; b) not less than 25 s. OPTION 6 Problem 1. Discrete random variable X (CB X) is given by the distribution series: Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), dispersion D(X), standard deviation s( X), fashion M 0 (X); 3) probability P(– 3≤ X < 1). Построить многоугольник распределения и график F(x). Task 2. There are 12 students in a group, 5 of which live in a hostel. 4 students are randomly selected from the list. For SV X- the number of students living in the hostel among those who will be selected, make a distribution series and find F(x), M(X) And D(X). Task 3. In the manufacture of parts of the same type on outdated equipment, each part may be defective with a probability of 0.1. Plot distribution series CB X< 3); Task 6. A function is given Define parameter value A, at which this function specifies the probability distribution density of some continuous CB X. To find F(x), Task 7. Given M(X) = 11 and s( X) = 3 normally distributed continuous SW X. To find: 1) probability 2) probability 3) symmetrical with respect to a the interval in which the values fall CB X with probability g = 0.9973. Task 8. The uptime of a TV of a given brand is a random variable distributed according to the normal law with parameters but= 12 years and s = 2 years. Find the probability that the TV will work without repair: a) from 9 to 12 years; OPTION 7 Problem 1. Discrete random variable X (CB X) is given by the distribution series: Find: 1) distribution function F(x); 2) numerical characteristics: mathematical expectation M(X), dispersion D(X), standard deviation s( X), fashion M 0 (X); 3) probability P(2≤ X < 10). Построить многоугольник распределения и график F(x). Task 2. The worker serves 4 independently working machines. The probability that within an hour the machine does not require the attention of the worker for the first machine is 0.7; for the second - 0.75; for the third - 0.8; for the fourth - 0.9. For a discrete SV X- the number of machines that will not require the attention of the worker for an hour, make a distribution series and find F(x), M(X) And D(X). Task 3. Available n independently operating machines. Plot distribution series CB X- the number of machines working at a given time, if n= 6, and the probability that the machine is running at a given time is 0.9; calculate M(X) And D(X). Assess the probability that the company, which has n= 180 and the probability of work for each machine is 0.98, the number of machines currently working will be at least 170. Problem 4. Distribution laws of independent discrete CB X And Y: Compose a distribution series C.B. Z = XY+ 2. Find M(Z) And D(Z). Without exaggeration, it can be called a philosophical law. Observing various objects and processes of the world around us, we often encounter the fact that something is not enough, and that there is a norm: What examples can be given? They are just darkness. This, for example, is the height, weight of people (and not only), their physical strength, mental abilities, etc. There is a "mass" (in one way or another) and there are deviations in both directions. These are different characteristics of inanimate objects (the same dimensions, weight). This is a random duration of processes, for example, the time of a hundred-meter race or the transformation of resin into amber. From physics, air molecules came to mind: among them there are slow ones, there are fast ones, but most of them move at “standard” speeds. Next, we deviate from the center by one more standard deviation and calculate the height: Marking points on the drawing (green color) and we see that this is quite enough. At the final stage, we carefully draw a graph, and especially carefully reflect it convexity / concavity! Well, you probably realized a long time ago that the abscissa axis is horizontal asymptote, and it is absolutely impossible to “climb” for it! With the electronic design of the solution, the graph is easy to build in Excel, and unexpectedly for myself, I even recorded a short video on this topic. But first, let's talk about how the shape of the normal curve changes depending on the values of and . When increasing or decreasing "a" (with unchanged "sigma") the graph retains its shape and moves right / left respectively. So, for example, when the function takes the form In the event of a change in "sigma" (with constant "a"), the graph "remains in place", but changes shape. When enlarged, it becomes lower and elongated, like an octopus stretching its tentacles. And vice versa, when decreasing the graph becomes narrower and taller- it turns out "surprised octopus." Yes, at decrease"sigma" two times: the previous chart narrows and stretches up twice: The normal distribution with unit value "sigma" is called normalized, and if it is also centered(our case), then such a distribution is called standard. It has an even simpler density function, which has already been encountered in local Laplace theorem: Now let's watch a movie:
Yes, quite right - somehow undeservedly we have remained in the shadows probability distribution function. We remember her definition: Inside the integral, a different letter is usually used so that there are no "overlays" with the notation, because here each value is assigned improper integral Almost all values cannot be calculated exactly, but as we have just seen, with modern computing power, this is not difficult. So, for the function =NORMSDIST(z) One, two - and you're done: Now let's recall one of the key tasks of the topic, namely, find out how to find - the probability that a normal random variable will take a value from the interval. Geometrically, this probability is equal to area between the normal curve and the x-axis in the corresponding section: ! also remembers
, what Here you can use Excel again, but there are a couple of significant “buts”: firstly, it is not always at hand, and secondly, “ready-made” values, most likely, will raise questions from the teacher. Why? I have repeatedly talked about this before: at one time (and not very long ago) an ordinary calculator was a luxury, and the “manual” way of solving the problem under consideration is still preserved in the educational literature. Its essence is to standardize the values "alpha" and "beta", that is, reduce the solution to the standard distribution: Note
: the function is easy to obtain from the general case Why is this needed? The fact is that the values were scrupulously calculated by our ancestors and summarized in a special table, which is in many books on terver. But even more common is the table of values, which we have already dealt with in Laplace integral theorem: If we have at our disposal a table of values of the Laplace function I remind you that Answer is required to be given as a percentage, so the calculated probability must be multiplied by 100 and provide the result with a meaningful comment: - with a flight from 5 to 70 m, approximately 15.87% of the shells will fall We train on our own: Example 3 The diameter of bearings manufactured at the factory is a random variable normally distributed with an expectation of 1.5 cm and a standard deviation of 0.04 cm. Find the probability that the size of a randomly taken bearing ranges from 1.4 to 1.6 cm. In the sample solution and below, I will use the Laplace function as the most common option. By the way, note that according to the wording, here you can include the ends of the interval in the consideration. However, this is not critical. And already in this example, we met a special case - when the interval is symmetrical with respect to the mathematical expectation. In such a situation, it can be written in the form and, using the oddness of the Laplace function, simplify the working formula: Well, the solution that fits in one line :) The result of this task turned out to be close to unity, but I would like even more reliability - namely, to find out the boundaries in which the diameter is almost everyone bearings. Is there any criterion for this? Exists! The question is answered by the so-called Its essence is that practically reliable
is the fact that a normally distributed random variable will take a value from the interval Indeed, the probability of deviation from the expectation is less than: In terms of "bearings" - these are 9973 pieces with a diameter of 1.38 to 1.62 cm and only 27 "substandard" copies. In practical research, the “three sigma” rule is usually applied in the opposite direction: if statistically found that almost all values random variable under study fit into an interval of 6 standard deviations, then there are good reasons to believe that this value is distributed according to the normal law. Verification is carried out using the theory statistical hypotheses. We continue to solve the harsh Soviet tasks: Example 4 The random value of the weighing error is distributed according to the normal law with zero mathematical expectation and a standard deviation of 3 grams. Find the probability that the next weighing will be carried out with an error not exceeding 5 grams in absolute value. Solution very simple. By the condition, and we immediately note that at the next weighing (something or someone) we will almost 100% get the result with an accuracy of 9 grams. But in the problem there is a narrower deviation and according to the formula Answer: A solved problem is fundamentally different from a seemingly similar one. Example 3 lesson about uniform distribution. There was an error rounding measurement results, here we are talking about the random error of the measurements themselves. Such errors arise due to the technical characteristics of the device itself. (the range of permissible errors, as a rule, is indicated in his passport), and also through the fault of the experimenter - when, for example, "by eye" we take readings from the arrow of the same scales. Among others, there are also so-called systematic measurement errors. It's already nonrandom errors that occur due to incorrect setup or operation of the device. So, for example, unadjusted floor scales can consistently "add" a kilogram, and the seller systematically underweight buyers. Or not systematically because you can shortchange. However, in any case, such an error will not be random, and its expectation is different from zero. …I am urgently developing a sales training course =) Let's solve the problem on our own: Example 5 The roller diameter is a random normally distributed random variable, its standard deviation is mm. Find the length of the interval, symmetrical with respect to the mathematical expectation, in which the length of the diameter of the bead will fall with probability. Item 5* design layout to help. Please note that the mathematical expectation is not known here, but this does not in the least interfere with solving the problem. And the exam task, which I highly recommend to consolidate the material: Example 6 A normally distributed random variable is given by its parameters (mathematical expectation) and (standard deviation). Required: a) write down the probability density and schematically depict its graph; Such problems are offered everywhere, and over the years of practice I have been able to solve hundreds and hundreds of them. Be sure to practice hand drawing and using paper spreadsheets ;) Well, I will analyze an example of increased complexity: Example 7 The probability distribution density of a random variable has the form Solution: first of all, let's pay attention that the condition does not say anything about the nature of the random variable. By itself, the presence of the exhibitor does not mean anything: it can be, for example, demonstrative or generally arbitrary continuous distribution. And therefore, the “normality” of the distribution still needs to be substantiated: Since the function We present. For this select a full square and organize three-story fraction: In this way: Now let's find the value of the parameter. Since the normal distribution multiplier has the form and , then: Let's plot the density: To tasks 41-50.
has the form 
.
,
,
,
,
.
. Substituting the probability values into this formula, we obtain 
,
,
,
,
.
,
. Discrete independent random variables X and Y are given by distributions:№
Indicator Normal distribution law Note
Definition
Normal is called
the probability distribution of a continuous random variable X, whose density has the form
where m x is the mathematical expectation of the random variable X, σ x is the standard deviation
2
distribution function
Probability
hits in the interval (a; b)
- integral Laplace function
Probability
that the absolute value of the deviation is less than the positive number δ
for m x = 0
An example of solving a problem on the topic "The normal law of distribution of a continuous random variable"
A task.
Necessary:
a) write down the expression for the distribution density;
b) find the probability that the length of the part will be between 19.7 and 20.3 mm;
c) find the probability that the deviation does not exceed 0.1 mm;
d) determine the percentage of parts whose deviation from the average value does not exceed 0.1 mm;
e) find how the deviation should be set so that the percentage of parts whose deviation from the average does not exceed the specified one increases to 54%;
f) find an interval, symmetrical about the mean value, in which X will be located with a probability of 0.95.
F((20.3-20)/0.2) - F((19.7-20)/0.2) = F(0.3/0.2) - F(-0.3/0, 2) \u003d 2Ф (0.3 / 0.2) \u003d 2Ф (1.5) \u003d 2 * 0.4332 \u003d 0.8664.
We found the value Ф(1.5) = 0.4332 in the appendices, in the table of values of the integral Laplace function Φ(x) ( table 2
)
P(|X-20|< 0,1) = 2Ф(0,1/0,2) = 2Ф(0,5) = 2*0,1915 = 0,383.
We found the value Ф(0.5) = 0.1915 in the appendices, in the table of values of the integral Laplace function Φ(x) ( table 2
)
Desired interval
: (20 - 0.392; 20 + 0.392) or (19.608; 20.392).
taken from the table. The property of oddness of the function Ф(– X) = – F( X).
From the table of values of the Laplace function, we find that is, d = 4.2. Then the interval is -4.2< X – 6 < 4,2 и
1,8 < X < 10,2.
then 
0,2233.
, entirely contained inside the segment [ a, b].
where is the probability density 
.
20 minutes. Find the probability that a passenger approaching the platform will wait for the next electric train for more than 10 minutes, as well as the average waiting time.
and this is the average waiting time for an electric train.
at d = 1, s = 0.5 and a = 0.
at a= 0, s = 0.5, a = 0.4, b = 0.8.
x i
pi
0,1
0,2
0,3
0,1
0,3
CB X- the total number of hits on the target under the specified conditions, make a distribution series and find F(x), M(X), s( X) And D(X). x i
–2
–1
pi
0,05
0,10
0,15
?
0,15
0,20
0,10
d) the probability that in three independent trials CB X takes values belonging to the interval exactly twice
M(X) And D(X). Build Graph F(x).
;
; x i
–2
–1
pi
0,2
0,2
0,1
0,3
0,2
3) 
4) the probability that in three independent trials CB X will take on a value belonging to the interval (1; 4) exactly once.
M(X), D(X). Build Graph F(x).
;
; x i
pi
0,3
0,1
0,1
0,4
0,1
x i
–3
–2
–1
pi
0,1
0,2
0,3
0,2
?
3) P(–2,3 < X <1,5);4) вероятность того, что в трех независимых испытаниях CB X exactly twice will take values belonging to the interval (–2.3; 1.5).
And M(X). Build Graph F(x).
;
; x i
–10
–5
pi
0,1
0,1
0,4
0,1
0,3
x i
pi
?
0,5
0,2
y i
pi
0,6
?
CB X takes exactly three times the values belonging to the interval
M(X) And D(X). Build Graph F(x).
;
; x i
pi
0,2
0,1
0,4
0,2
0,1
Y
X
0,20
0,15
0,10
0,30
0,20
0,05
3) P(3 < X < 9); 4) вероятность того, что в четырех независимых испытаниях CB X takes values belonging to the interval (3; 9) exactly twice.
M(X). Build Graph F(x). ;
; x i
–5
–3
–1
pi
0,2
0,2
0,1
0,4
0,1
4) the probability that in four independent trials CB X takes values belonging to the interval (1; 3) exactly twice.
M(X) And D(X). Build Graph F(x).
;
;
b) at least 10 years. x i
pi
0,2
0,1
0,2
0,3
0,2
x i
pi
0,3
?
0,5
y i
–2
–1
pi
?
0,4
Normal law of probability distribution
Here is a basic view density functions normal probability distribution, and I welcome you to this most interesting lesson.
and our graph "moves" 3 units to the left - exactly to the origin: 
A normally distributed quantity with zero mathematical expectation received a completely natural name - centered; its density function 
– even, and the graph is symmetrical about the y-axis.
Everything is in full accordance with geometric transformations of graphs.
. The standard distribution has found wide application in practice, and very soon we will finally understand its purpose.
- the probability that a random variable will take a value LESS than the variable , which "runs" all real values \u200b\u200bto "plus" infinity.
, which is equal to some number from the interval.
of the standard distribution, the corresponding excel function generally contains one argument:
The drawing clearly shows the implementation of all distribution function properties, and from the technical nuances here you should pay attention to horizontal asymptotes and an inflection point.
but each time grind out an approximate value 
is unreasonable, and therefore it is more rational to use "easy" formula:
.
using a linear substitutions. Then and:
and from the replacement just follows the formula 
transition from the values of an arbitrary distribution to the corresponding values of the standard distribution.
, then we solve through it:
Fractional values are traditionally rounded to 4 decimal places, as is done in the standard table. And for control Item 5 layout.
, and to avoid confusion always be in control, table of WHAT function before your eyes.
The delta parameter is called deviation from the mathematical expectation, and the double inequality can be “packed” using module:
is the probability that the value of a random variable deviates from the mathematical expectation by less than .
is the probability that the diameter of a bearing taken at random differs from 1.5 cm by no more than 0.1 cm.three sigma rule
.
or 99.73%
:
- the probability that the next weighing will be carried out with an error not exceeding 5 grams.
b) find the probability that it will take a value from the interval 
;
c) find the probability that the modulo deviates from no more than ;
d) applying the rule of "three sigma", find the values of the random variable .
. Find , mathematical expectation , variance , distribution function , plot density and distribution functions, find .
determined at any real value , and it can be reduced to the form 
, then the random variable is distributed according to the normal law.
Be sure to perform a check, returning the indicator to its original form:
which is what we wanted to see.
- on power rule"pinching off". And here you can immediately write down the obvious numerical characteristics: 
, from which we express and substitute into our function: 
, after which we will once again go over the record with our eyes and make sure that the resulting function has the form 
.
and the plot of the distribution function 
:
If there is no Excel and even a regular calculator at hand, then the last chart is easily built manually! At the point, the distribution function takes on the value 
and here is