2. Determination of unknown distribution parameters.

Using a histogram, we can approximately plot the distribution density of a random variable. The appearance of this graph often allows us to make an assumption about the probability density distribution of a random variable. The expression of this distribution density usually includes some parameters that need to be determined from experimental data.
Let us dwell on the particular case when the distribution density depends on two parameters.
So let x 1 , x 2 , ..., x n- observed values ​​of a continuous random variable, and let its probability distribution density depend on two unknown parameters A And B, i.e. looks like . One of the methods for finding unknown parameters A And B consists in the fact that they are chosen in such a way that the mathematical expectation and variance of the theoretical distribution coincide with the sample means and variance:

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From the two obtained equations () the unknown parameters are found A And B. So, for example, if a random variable obeys the normal probability distribution law, then its probability distribution density

depends on two parameters a And . These parameters, as we know, are, respectively, the mathematical expectation and the standard deviation of a random variable; therefore equalities () will be written like this:

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Therefore, the probability distribution density has the form

Note 1. We have already solved this problem in . The measurement result is a random variable that obeys the normal distribution law with parameters a And . For approximate value a we chose the value , and for the approximate value - the value .

Note 2. At large quantities experiments, finding quantities and using formulas () is associated with cumbersome calculations. Therefore, they do this: each of the observed values ​​of the quantity , falling into i th interval ] X i-1 , X i [ statistical series, is considered approximately equal to the middle c i this interval, i.e. c i =(X i-1 +X i)/2. Consider the first interval ] X 0 , X 1 [. It hit him m 1 observed values ​​of the random variable, each of which we replace with a number from 1. Therefore, the sum of these values ​​is approximately equal to m 1 s 1. Similarly, the sum of values ​​falling into the second interval is approximately equal to m 2 with 2 etc. That's why

In a similar way we obtain the approximate equality

So, let's show that

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Equations with parameters: graphical solution method

8-9 grades

The article discusses a graphical method for solving some equations with parameters, which is very effective when you need to establish how many roots an equation has depending on the parameter a.

Problem 1. How many roots does the equation have? | | x | – 2 | = a depending on parameter a?

Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | | x | – 2 | and y = a. Graph of the function y = | | x | – 2 | shown in the figure.

The graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (if a = 0).

From the drawing it can be seen that:

If a= 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | | x | – 2 | two common points; This means that the original equation has two roots (in in this case roots can be found: x 1.2 = d 2).
If 0< a < 2, то прямая y = a имеет с графиком функции y = | | x | – 2 | четыре общие точки и, следовательно, исходное уравнение имеет четыре корня.
If a= 2, then the line y = 2 has three common points with the graph of the function. Then the original equation has three roots.
If a> 2, then straight line y = a will have two points with the graph of the original function, that is, this equation will have two roots.

If a < 0, то корней нет;
If a = 0, a> 2, then there are two roots;
If a= 2, then three roots;
if 0< a < 2, то четыре корня.

Problem 2. How many roots does the equation have? | x 2 – 2| x | – 3 | = a depending on parameter a?

Solution. In the coordinate system (x; y) we will construct graphs of the functions y = | x 2 – 2| x | – 3 | and y = a.

Graph of the function y = | x 2 – 2| x | – 3 | shown in the figure. The graph of the function y = a is a straight line parallel to Ox or coinciding with it (when a = 0).

From the drawing you can see:

If a= 0, then straight line y = a coincides with the Ox axis and has the graph of the function y = | x2 – 2| x | – 3 | two common points, as well as the straight line y = a will have with the graph of the function y = | x 2 – 2| x | – 3 | two common points at a> 4. So, when a= 0 and a> 4 the original equation has two roots.
If 0< a < 3, то прямая y = a has with the graph of the function y = | x 2 – 2| x | – 3 | four common points, as well as the straight line y= a will have four common points with the graph of the constructed function at a= 4. So, at 0< a < 3, a= 4 the original equation has four roots.
If a= 3, then straight line y = a intersects the graph of a function at five points; therefore, the equation has five roots.
If 3< a < 4, прямая y = a пересекает график построенной функции в шести точках; значит, при этих значениях параметра исходное уравнение имеет шесть корней.
If a < 0, уравнение корней не имеет, так как прямая y = a не пересекает график функции y = | x 2 – 2| x | – 3 |.

If a < 0, то корней нет;
If a = 0, a> 4, then there are two roots;
if 0< a < 3, a= 4, then four roots;
If a= 3, then five roots;
if 3< a < 4, то шесть корней.

Problem 3. How many roots does the equation have?

depending on parameter a?

Solution. Let us construct a graph of the function in the coordinate system (x; y) but first let's present it in the form:

The lines x = 1, y = 1 are asymptotes of the graph of the function. Graph of the function y = | x | + a obtained from the graph of the function y = | x | displacement by a units along the Oy axis.

Function graphs intersect at one point at a> – 1; This means that equation (1) for these parameter values ​​has one solution.

At a = – 1, a= – 2 graphs intersect at two points; This means that for these parameter values, equation (1) has two roots.
At – 2< a < – 1, a < – 2 графики пересекаются в трех точках; значит, уравнение (1) при этих значениях параметра имеет три решения.

If a> – 1, then one solution;
If a = – 1, a= – 2, then there are two solutions;
if – 2< a < – 1, a < – 1, то три решения.

Comment. When solving equation (1) of problem 3, special attention should be paid to the case when a= – 2, since the point (– 1; – 1) does not belong to the graph of the function but belongs to the graph of the function y = | x | + a.

Let's move on to solving another problem.

Problem 4. How many roots does the equation have?

x + 2 = a| x – 1 | (2)

depending on parameter a?

Solution. Note that x = 1 is not a root of this equation, since the equality 3 = a· 0 cannot be true for any parameter value a. Let's divide both sides of the equation by | x – 1 |(| x – 1 | No. 0), then equation (2) will take the form In the coordinate system xOy we will plot the function

The graph of this function is shown in the figure. Graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (if a = 0).

If aЈ – 1, then there are no roots;
if – 1< aЈ 1, then one root;
If a> 1, then there are two roots.

Let's consider the most complex equation.

Problem 5. At what values ​​of the parameter a equation

a x 2 + | x – 1 | = 0 (3)

has three solutions?

Solution. 1. The control value of the parameter for this equation will be the number a= 0, at which equation (3) takes the form 0 + | x – 1 | = 0, whence x = 1. Therefore, when a= 0, equation (3) has one root, which does not satisfy the conditions of the problem.

2. Consider the case when a № 0.

Let us rewrite equation (3) in the following form: a x 2 = – | x – 1 |. Note that the equation will have solutions only when a < 0.

In the coordinate system xOy we will construct graphs of the functions y = | x – 1 | and y = a x 2 . Graph of the function y = | x – 1 | shown in the figure. Graph of the function y = a x 2 is a parabola whose branches are directed downward, since a < 0. Вершина параболы - точка (0; 0).

Equation (3) will have three solutions only when the straight line y = – x + 1 is tangent to the graph of the function y= a x 2 .

Let x 0 be the abscissa of the point of tangency of the straight line y = – x + 1 with the parabola y = a x 2 . The tangent equation has the form

y = y(x 0) + y "(x 0)(x – x 0).

Let's write down the tangency conditions:

This equation can be solved without using the concept of derivative.

Let's consider another method. Let us use the fact that if the straight line y = kx + b has a single common point with the parabola y = a x 2 + px + q, then the equation a x 2 + px + q = kx + b must have a unique solution, that is, its discriminant is zero. In our case we have the equation a x 2 = – x + 1 ( a No. 0). Discriminant equation

Problems to solve independently

6. How many roots does the equation have depending on the parameter a?

1)| | x | – 3 | = a;
2)| x + 1 | + | x + 2 | = a;
3)| x 2 – 4| x | + 3 | = a;
4)| x 2 – 6| x | + 5 | = a.

1) if a<0, то корней нет; если a=0, a>3, then two roots; If a=3, then three roots; if 0<a<3, то четыре корня;
2) if a<1, то корней нет; если a=1, then there is an infinite set of solutions from the interval [– 2; – 1]; If a> 1, then there are two solutions;
3) if a<0, то корней нет; если a=0, a<3, то четыре корня; если 0<a<1, то восемь корней; если a=1, then six roots; If a=3, then there are three solutions; If a>3, then there are two solutions;
4) if a<0, то корней нет; если a=0, 4<a<5, то четыре корня; если 0<a< 4, то восемь корней; если a=4, then six roots; If a=5, then three roots; If a>5, then there are two roots.

7. How many roots does the equation have | x + 1 | = a(x – 1) depending on parameter a?

Note. Since x = 1 is not a root of the equation, this equation can be reduced to the form .

Answer: if a J –1, a > 1, a=0, then one root; if – 1<a<0, то два корня; если 0<aЈ 1, then there are no roots.

8. How many roots does the equation x + 1 = have? a| x – 1 |depending on parameter a?

Draw a graph (see figure).

Answer: if aЈ –1, then there are no roots; if – 1<aЈ 1, then one root; If a>1, then there are two roots.

9. How many roots does the equation have?

2| x | – 1 = a(x – 1)

depending on parameter a?

Note. Reduce the equation to form

Answer: if a J –2, a>2, a=1, then one root; if –2<a<1, то два корня; если 1<aЈ 2, then there are no roots.

10. How many roots does the equation have?

depending on parameter a?

Answer: if aЈ 0, a i 2, then one root; if 0<a<2, то два корня.

11. At what values ​​of the parameter a equation

x 2 + a| x – 2 | = 0

has three solutions?

Note. Reduce the equation to the form x 2 = – a| x – 2 |.

Answer: when a J –8.

12. At what values ​​of the parameter a equation

a x 2 + | x + 1 | = 0

has three solutions?

Note. Use problem 5. This equation has three solutions only if the equation a x 2 + x + 1 = 0 has one solution, and the case a= 0 does not satisfy the conditions of the problem, that is, the case remains when

13. How many roots does the equation have?

x | x – 2 | = 1 – a

depending on parameter a?

Note. Reduce the equation to the form –x |x – 2| + 1 = a

depending on parameter a?

Note. Construct graphs of the left and right sides of this equation.

Answer: if a<0, a>2, then there are two roots; if 0Ј aЈ 2, then one root.

16. How many roots does the equation have?

depending on parameter a?

Note. Construct graphs of the left and right sides of this equation. To graph a function Let's find the intervals of constant sign of the expressions x + 2 and x:

Answer: if a>– 1, then one solution; If a= – 1, then there are two solutions; if – 3<a<–1, то четыре решения; если aЈ –3, then there are three solutions.