Continuous probability distribution random variable X, which takes all values from the interval , is called uniform, if its probability density on this segment is constant, and outside it is equal to zero. Thus, the probability density of a continuous random variable X, distributed evenly on the segment , looks like:
Let's define expected value, dispersion and for a random variable with a uniform distribution.
, 
, 
.
Example. All values of a uniformly distributed random variable lie on the segment . Find the probability of a random variable falling into the interval (3;5) .
a=2, b=8, 
.
Binomial distribution
Let it be produced n tests, and the probability of occurrence of an event A in each test is p and does not depend on the outcome of other trials (independent trials). Since the probability of an event occurring A in one test is p, then the probability of its non-occurrence is equal to q=1-p.
Let the event A came in n trials m once. This complex event can be written as a product:
.
Then the probability that n test event A will come m times is calculated by the formula:
or 
(1)
Formula (1) is called Bernoulli formula.
Let X is a random variable equal to the number of occurrences of the event A V n tests, which takes values with probabilities:
The resulting law of distribution of a random variable is called binomial distribution law.
| X | m | n | ||||
| P |
Expected value, dispersion And standard deviation random variables distributed according to the binomial law are determined by the formulas:
, , .
Example. Three shots are fired at the target, and the probability of hitting each shot is 0.8. We consider a random variable X- the number of hits on the target. Find its distribution law, mathematical expectation, variance and standard deviation.
p=0.8, q=0.2, n=3, 
, , 
.
- probability of 0 hits;
Probability of one hit;
Probability of two hits;
is the probability of three hits.
We get the distribution law:
| X | ||||
| P | 0,008 | 0,096 | 0,384 | 0,512 |
Tasks
1. A coin is tossed 7 times. Find the probability that it will fall upside down 4 times.
2. A coin is tossed 8 times. Find the probability that the coat of arms will appear no more than three times.
3. The probability of hitting the target when firing from a gun p=0.6. Find the mathematical expectation of the total number of hits if 10 shots are fired.
4. Find the mathematical expectation of the number of lottery tickets that will win if 20 tickets are purchased, and the probability of winning for one ticket is 0.3.
A uniform distribution is such a distribution of a random variable, when it can take any value within the given limits with the same probability.
The uniform distribution of a random variable is shown in fig. 5.9.
Rice. 5.9.
The uniform distribution probability density has the form:
where a and b are the parameters of the law that determine the limits of variation of the random variable X.
The law of uniform distribution is subject, in particular, to errors due to friction in instrument supports, non-excluded residuals of systematic errors, discretization errors in digital instruments, dimensional errors within the same sorting group during selective assembly, errors in the parameters of products selected within narrower limits compared to technological tolerance, the total processing error caused by
Integral
is called the normalized Laplace function, and its values for x - X different / \u003d - are tabulated. The value of the normalized Laplace function Ф(/) with an error less than 10"5 can be determined by the formula
If />0, Ф(/) = 7", and if /< 0, то Ф(/) = 1-7". Функция Лапласа нечетная, т. е.
For negative values / tabular data are taken with a minus sign.
The probability that a random variable obeying the law normal distribution, during measurements will take a value within (x, x,), can be written through Ф (/) as follows:
In a theoretical normal distribution curve, its branches asymptotically approach the abscissa axis, i.e., the dispersion zone of the random variable x lies within ± oo. In practice, the dispersion zone of the random variable x is limited by finite limits.
For example, the probability that a random variable will be within
linear change in time of the dominant factor (wear of the cutting tool, thermal deformation, etc.), errors arising due to rounding of values obtained when measuring on instruments, etc.
The distribution function F(x) of the uniform distribution (cumulative distribution function) is expressed by the following equation for (a< х < Ь):
The form of the distribution function is shown in fig. 5.10.
The mathematical expectation A / (x), the variance 0 (x) and the standard deviation (a) of a random variable obeying a uniform distribution, respectively, are equal to:
In practice, the limiting stray field co with a uniform distribution is equal to b - a or, taking into account (5.48), i.e.
co = b - a = 2m/Ze.
Rice. 5.10.
Rice. 5.11.
Simpson's law
The view of the triangular distribution curve is shown in fig. 5.11. The probability density has the form:
According to this law, for example, the errors of the sum (difference) of two uniformly distributed quantities are distributed. If, for example, the deviations of the dimensions of the hole and the shaft are evenly distributed within the tolerance fields, and the tolerances of the shaft and the hole are approximately the same, then the gaps within the clearance tolerance will be distributed according to the triangle law. In this case, the gap probability density will have the following form:
where 5m(n, 5^ - respectively, the minimum and maximum values of the gap in the joint; .$m = ^"^^"la _ the average value of the gap in the joint; /G5 = 5m1p - clearance tolerance; l - current value of the gap.
The distribution function of Simpson's law has the form:
A graphical representation of the integral distribution function is shown in fig. 5.12.
The mathematical expectation, variance and standard deviation of a random variable obeying Simpson's law are respectively equal to:
In practice, the limiting stray field with the distribution of a random variable according to Simpson's law is 2/, i.e.
With the help of which many real processes are modeled. And the most common example is the schedule of public transport. Suppose a bus (trolleybus / tram) walks at intervals of 10 minutes, and at a random time you come to a stop. What is the probability that the bus will arrive within 1 minute? Obviously 1/10th. And the probability that you have to wait 4-5 minutes? Same . What is the probability that the bus will have to wait more than 9 minutes? One tenth!
Consider some finite interval, let for definiteness it will be a segment . If random value has constant probability density on a given segment and zero density outside it, then we say that it is distributed evenly. In this case, the density function will be strictly defined: 
Indeed, if the length of the segment (see drawing) is , then the value is inevitably equal - in order to get the unit area of \u200b\u200bthe rectangle, and it was observed known property:
Let's check it formally:
, h.t.p. From a probabilistic point of view, this means that the random variable reliably will take one of the values of the segment ..., eh, I'm slowly becoming a boring old man =)
The essence of uniformity is that no matter what internal gap fixed length we have not considered (remember the "bus" minutes)- the probability that a random variable will take a value from this interval will be the same. In the drawing, I have shaded three such probabilities - once again I draw attention to the fact that they are determined by the areas, not function values !
Consider a typical task:
Example 1
A continuous random variable is given by its distribution density: 
Find the constant , calculate and compose the distribution function. Build charts. Find
In other words, everything you could dream of :)
Solution: since on the interval (terminal interval) 
, then the random variable has a uniform distribution, and the value of "ce" can be found by the direct formula 
. But it’s better in a general way - using a property:
…why is it better? No more questions ;)
So the density function is: 
Let's do the trick. Values impossible
, and therefore bold dots are placed at the bottom: 
As a quick check, let's calculate the area of the rectangle: 
, h.t.p.
Let's find expected value, and, probably, you already guess what it is equal to. Recall the "10-minute" bus: if randomly come to a stop for many, many days, save me, then average you have to wait 5 minutes.
Yes, that's right - the expectation should be exactly in the middle of the "event" interval:
, as expected.
We calculate the dispersion by formula 
. And here you need an eye and an eye when calculating the integral:
Thus, dispersion: 
Let's compose distribution function 
. Nothing new here:
1) if , then and 
;
2) if , then and:
3) and, finally, at 
, That's why:
As a result: 
Let's execute the drawing: 
On the "live" interval, the distribution function grows linearly, and this is another sign that we have a uniformly distributed random variable. Well, still, after all derivative linear function- is a constant.
The required probability can be calculated in two ways, using the found distribution function:
either with the help definite integral from density: 
Whoever likes it.
And here you can also write answer: ,
, graphs are built along the solution.
... "it is possible", because they usually do not punish for its absence. Usually;)
There are special formulas for calculating and uniform random variable, which I suggest you derive yourself:
Example 2
Continuous random variable defined by density 
.
Calculate the mathematical expectation and variance. Simplify the results (abbreviated multiplication formulas to help).
It is convenient to use the obtained formulas for verification, in particular, check the problem you just solved by substituting the specific values \u200b\u200bof “a” and “b” into them. Brief solution at the bottom of the page.
And at the end of the lesson, we will analyze a couple of “text” tasks:
Example 3
The division value of the scale of the measuring instrument is 0.2. Instrument readings are rounded to the nearest whole division. Assuming that the rounding errors are evenly distributed, find the probability that during the next measurement it will not exceed 0.04.
For better understanding solutions imagine that this is some kind of mechanical device with an arrow, for example, scales with a division value of 0.2 kg, and we have to weigh a pig in a poke. But not in order to find out his fatness - now it will be important WHERE the arrow will stop between two adjacent divisions.
Consider a random variable - distance arrows off nearest left division. Or from the nearest right, it doesn't matter.
Let's compose the probability density function:
1) Since the distance cannot be negative, then on the interval . Logically.
2) It follows from the condition that the arrow of the scales with equally likely can stop anywhere between divisions *
, including the divisions themselves, and therefore on the interval : 
* This is an essential condition. So, for example, when weighing pieces of cotton wool or kilogram packs of salt, uniformity will be observed at much narrower intervals.
3) And since the distance from the CLOSEST left division cannot be more than 0.2, then for is also zero.
Thus: 
It should be noted that no one asked us about the density function, and it full build I cited exclusively in cognitive circuits. When finishing the task, it is enough to write down only the 2nd paragraph.
Now let's answer the question of the problem. When does the rounding error to the nearest division not exceed 0.04? This will happen when the arrow stops no further than 0.04 from the left division on right or no further than 0.04 from the right division left. In the drawing, I shaded the corresponding areas: 
It remains to find these areas with the help of integrals. In principle, they can also be calculated “in a school way” (like the areas of rectangles), but simplicity does not always find understanding;)
By addition theorem for the probabilities of incompatible events:
- the probability that the rounding error will not exceed 0.04 (40 grams for our example)
It is easy to understand that the maximum possible rounding error is 0.1 (100 grams) and therefore the probability that the rounding error will not exceed 0.1 is equal to one. And from this, by the way, follows another, more easy way solution in which you need to consider a random variable – rounding error to the nearest division. But the first one came to my mind first :)
Answer: 0,4
And one more point on the task. The condition may contain errors. Not rounding, but o random errors measurements themselves, which are usually (but not always), are distributed according to the normal law. Thus, Just one word can change your mind! Be alert and delve into the meaning of tasks!
And as soon as everything goes in a circle, then our feet bring us to the same stop:
Example 4
Buses of a certain route go strictly according to the schedule and with an interval of 7 minutes. Compose a function of the density of a random variable - the waiting time for the next bus by a passenger who randomly approached the bus stop. Find the probability that he will wait for the bus no more than three minutes. Find the distribution function and explain its meaningful meaning.
As an example of a continuous random variable, consider a random variable X uniformly distributed over the interval (a; b). We say that the random variable X evenly distributed on the interval (a; b), if its distribution density is not constant on this interval:
From the normalization condition, we determine the value of the constant c . The area under the distribution density curve should be equal to one, but in our case it is the area of a rectangle with a base (b - α) and a height c (Fig. 1). 
Rice. 1 Uniform distribution density
From here we find the value of the constant c: 
So, the density of a uniformly distributed random variable is equal to 
Let us now find the distribution function by the formula:
1) for 
2) for
3) for 0+1+0=1.
Thus, 
The distribution function is continuous and does not decrease (Fig. 2). 
Rice. 2 Distribution function of a uniformly distributed random variable
Let's find mathematical expectation of a uniformly distributed random variable according to the formula:
Uniform distribution variance is calculated by the formula and is equal to
Example #1. The scale division value of the measuring instrument is 0.2 . Instrument readings are rounded to the nearest whole division. Find the probability that an error will be made during the reading: a) less than 0.04; b) big 0.02
Solution. The rounding error is a random variable uniformly distributed over the interval between adjacent integer divisions. Consider the interval (0; 0.2) as such a division (Fig. a). Rounding can be carried out both towards the left border - 0, and towards the right - 0.2, which means that an error less than or equal to 0.04 can be made twice, which must be taken into account when calculating the probability: 
P = 0.2 + 0.2 = 0.4
For the second case, the error value can also exceed 0.02 on both division boundaries, that is, it can be either greater than 0.02 or less than 0.18. 
Then the probability of an error like this:
Example #2. It was assumed that the stability of the economic situation in the country (the absence of wars, natural disasters, etc.) over the past 50 years can be judged by the nature of the distribution of the population by age: in a calm situation, it should be uniform. As a result of the study, the following data were obtained for one of the countries.
Is there any reason to believe that there was an unstable situation in the country?We carry out the decision using the calculator Hypothesis testing. Table for calculating indicators.
| Groups | Interval middle, x i | Quantity, fi | x i * f i | Cumulative frequency, S | |x - x cf |*f | (x - x sr) 2 *f | Frequency, f i /n |
| 0 - 10 | 5 | 0.14 | 0.7 | 0.14 | 5.32 | 202.16 | 0.14 |
| 10 - 20 | 15 | 0.09 | 1.35 | 0.23 | 2.52 | 70.56 | 0.09 |
| 20 - 30 | 25 | 0.1 | 2.5 | 0.33 | 1.8 | 32.4 | 0.1 |
| 30 - 40 | 35 | 0.08 | 2.8 | 0.41 | 0.64 | 5.12 | 0.08 |
| 40 - 50 | 45 | 0.16 | 7.2 | 0.57 | 0.32 | 0.64 | 0.16 |
| 50 - 60 | 55 | 0.13 | 7.15 | 0.7 | 1.56 | 18.72 | 0.13 |
| 60 - 70 | 65 | 0.12 | 7.8 | 0.82 | 2.64 | 58.08 | 0.12 |
| 70 - 80 | 75 | 0.18 | 13.5 | 1 | 5.76 | 184.32 | 0.18 |
| 1 | 43 | 20.56 | 572 | 1 |
weighted average
Variation indicators.
Absolute Variation Rates.
The range of variation is the difference between the maximum and minimum values of the attribute of the primary series.
R = X max - X min
R=70 - 0=70
Dispersion- characterizes the measure of spread around its mean value (measure of dispersion, i.e. deviation from the mean).
Standard deviation.
Each value of the series differs from the average value of 43 by no more than 23.92
Testing hypotheses about the type of distribution.
4. Testing the hypothesis about uniform distribution the general population.
In order to test the hypothesis about the uniform distribution of X, i.e. according to the law: f(x) = 1/(b-a) in the interval (a,b)
necessary:
1. Estimate the parameters a and b - the ends of the interval in which the possible values of X were observed, according to the formulas (the * sign denotes the estimates of the parameters):
2. Find the probability density of the estimated distribution f(x) = 1/(b * - a *)
3. Find theoretical frequencies:
n 1 \u003d nP 1 \u003d n \u003d n * 1 / (b * - a *) * (x 1 - a *)
n 2 \u003d n 3 \u003d ... \u003d n s-1 \u003d n * 1 / (b * - a *) * (x i - x i-1)
n s = n*1/(b * - a *)*(b * - x s-1)
4. Compare the empirical and theoretical frequencies using the Pearson test, assuming the number of degrees of freedom k = s-3, where s is the number of initial sampling intervals; if, however, a combination of small frequencies, and therefore the intervals themselves, was made, then s is the number of intervals remaining after the combination.
Solution:
1. Find the estimates of the parameters a * and b * of the uniform distribution using the formulas:
2. Find the density of the assumed uniform distribution:
f(x) = 1/(b * - a *) = 1/(84.42 - 1.58) = 0.0121
3. Find the theoretical frequencies:
n 1 \u003d n * f (x) (x 1 - a *) \u003d 1 * 0.0121 (10-1.58) \u003d 0.1
n 8 \u003d n * f (x) (b * - x 7) \u003d 1 * 0.0121 (84.42-70) \u003d 0.17
The remaining n s will be equal:
n s = n*f(x)(x i - x i-1)
| i | n i | n*i | n i - n * i | (n i - n* i) 2 | (n i - n * i) 2 /n * i |
| 1 | 0.14 | 0.1 | 0.0383 | 0.00147 | 0.0144 |
| 2 | 0.09 | 0.12 | -0.0307 | 0.000943 | 0.00781 |
| 3 | 0.1 | 0.12 | -0.0207 | 0.000429 | 0.00355 |
| 4 | 0.08 | 0.12 | -0.0407 | 0.00166 | 0.0137 |
| 5 | 0.16 | 0.12 | 0.0393 | 0.00154 | 0.0128 |
| 6 | 0.13 | 0.12 | 0.0093 | 8.6E-5 | 0.000716 |
| 7 | 0.12 | 0.12 | -0.000701 | 0 | 4.0E-6 |
| 8 | 0.18 | 0.17 | 0.00589 | 3.5E-5 | 0.000199 |
| Total | 1 | 0.0532 |
Therefore, the critical region for this statistic is always right-handed :)