2. Determination of unknown distribution parameters.
Using a histogram, we can approximately plot the distribution density 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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(66) |
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(67) |
From the two obtained equations () find unknown parameters A And B. So, for example, if a random variable obeys normal law probability distribution, then its probability distribution density
depends on two parameters a And . These parameters, as we know, are respectively mathematical expectation and average square deviation 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, are 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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(71) |
Equations with parameters are rightfully considered one of the most complex tasks I know school mathematics. It is these tasks that end up year after year on the list of tasks of type B and C at the unified state Unified State Examination. However, among large number equations with parameters are those that can be easily solved graphically. Let's consider this method using the example of solving several problems.
Find the sum of integer values of the number a for which the equation |x 2 – 2x – 3| = a has four roots.
Solution.
To answer the question of the problem, let’s build on one coordinate plane function graphs
y = |x 2 – 2x – 3| and y = a.
Graph of the first function y = |x 2 – 2x – 3| will be obtained from the graph of the parabola y = x 2 – 2x – 3 by symmetrically displaying with respect to the x-axis that part of the graph that is below the Ox axis. The part of the graph located above the x-axis will remain unchanged.
Let's do this step by step. The graph of the function y = x 2 – 2x – 3 is a parabola, the branches of which are directed upward. To build its graph, we find the coordinates of the vertex. This can be done using the formula x 0 = -b/2a. Thus, x 0 = 2/2 = 1. To find the coordinate of the vertex of the parabola along the ordinate axis, we substitute the resulting value for x 0 into the equation of the function in question. We get that y 0 = 1 – 2 – 3 = -4. This means that the vertex of the parabola has coordinates (1; -4).
Next, you need to find the intersection points of the parabola branches with the coordinate axes. At the points of intersection of the branches of the parabola with the abscissa axis, the value of the function is zero. Therefore, we solve the quadratic equation x 2 – 2x – 3 = 0. Its roots will be the required points. By Vieta’s theorem we have x 1 = -1, x 2 = 3.
At the points of intersection of the parabola branches with the ordinate axis, the value of the argument is zero. Thus, the point y = -3 is the point of intersection of the branches of the parabola with the y-axis. The resulting graph is shown in Figure 1.
To obtain a graph of the function y = |x 2 – 2x – 3|, let us display the part of the graph located below the x-axis symmetrically relative to the x-axis. The resulting graph is shown in Figure 2.
The graph of the function y = a is a straight line parallel to the abscissa axis. It is depicted in Figure 3. Using the figure, we find that the graphs have four common points (and the equation has four roots) if a belongs to the interval (0; 4).
Integer values of number a from the resulting interval: 1; 2; 3. To answer the question of the problem, let’s find the sum of these numbers: 1 + 2 + 3 = 6.
Answer: 6.
Find the arithmetic mean of integer values of the number a for which the equation |x 2 – 4|x| – 1| = a has six roots.
Let's start by plotting the function y = |x 2 – 4|x| – 1|. To do this, we use the equality a 2 = |a| 2 and select perfect square in a submodular expression written on the right side of the function:
x 2 – 4|x| – 1 = |x| 2 – 4|x| - 1 = (|x| 2 – 4|x| + 4) – 1 – 4 = (|x |– 2) 2 – 5.
Then the original function will have the form y = |(|x| – 2) 2 – 5|.
To construct a graph of this function, we construct sequential graphs of functions:
1) y = (x – 2) 2 – 5 – parabola with vertex at point with coordinates (2; -5); (Fig. 1).
2) y = (|x| – 2) 2 – 5 – part of the parabola constructed in step 1, which is located to the right of the ordinate axis, is symmetrically displayed to the left of the Oy axis; (Fig. 2).
3) y = |(|x| – 2) 2 – 5| – the part of the graph constructed in point 2, which is located below the x-axis, is displayed symmetrically relative to the x-axis upward. (Fig. 3).
Let's look at the resulting drawings:
The graph of the function y = a is a straight line parallel to the abscissa axis.
Using the figure, we conclude that the graphs of functions have six common points (the equation has six roots) if a belongs to the interval (1; 5).
This can be seen in the following figure:
Let's find the arithmetic mean of the integer values of parameter a:
(2 + 3 + 4)/3 = 3.
Answer: 3.
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Equations with parameters are rightfully considered one of the most difficult problems in school mathematics. It is precisely such tasks that end up year after year in the list of tasks of type B and C on a single state exam Unified State Exam. However, among the large number of equations with parameters, there are those that can easily be solved graphically. Let's consider this method using the example of solving several problems.
Find the sum of integer values of the number a for which the equation |x 2 – 2x – 3| = a has four roots.
Solution.
To answer the question of the problem, let’s construct graphs of functions on one coordinate plane
y = |x 2 – 2x – 3| and y = a.
Graph of the first function y = |x 2 – 2x – 3| will be obtained from the graph of the parabola y = x 2 – 2x – 3 by symmetrically displaying with respect to the x-axis that part of the graph that is below the Ox axis. The part of the graph located above the x-axis will remain unchanged.
Let's do this step by step. The graph of the function y = x 2 – 2x – 3 is a parabola, the branches of which are directed upward. To build its graph, we find the coordinates of the vertex. This can be done using the formula x 0 = -b/2a. Thus, x 0 = 2/2 = 1. To find the coordinate of the vertex of the parabola along the ordinate axis, we substitute the resulting value for x 0 into the equation of the function in question. We get that y 0 = 1 – 2 – 3 = -4. This means that the vertex of the parabola has coordinates (1; -4).
Next, you need to find the intersection points of the parabola branches with the coordinate axes. At the points of intersection of the branches of the parabola with the abscissa axis, the value of the function is zero. Therefore, we solve the quadratic equation x 2 – 2x – 3 = 0. Its roots will be the required points. By Vieta’s theorem we have x 1 = -1, x 2 = 3.
At the points of intersection of the parabola branches with the ordinate axis, the value of the argument is zero. Thus, the point y = -3 is the point of intersection of the branches of the parabola with the y-axis. The resulting graph is shown in Figure 1.
To obtain a graph of the function y = |x 2 – 2x – 3|, let us display the part of the graph located below the x-axis symmetrically relative to the x-axis. The resulting graph is shown in Figure 2.
The graph of the function y = a is a straight line parallel to the abscissa axis. It is depicted in Figure 3. Using the figure, we find that the graphs have four common points (and the equation has four roots) if a belongs to the interval (0; 4).
Integer values of number a from the resulting interval: 1; 2; 3. To answer the question of the problem, let’s find the sum of these numbers: 1 + 2 + 3 = 6.
Answer: 6.
Find the arithmetic mean of integer values of the number a for which the equation |x 2 – 4|x| – 1| = a has six roots.
Let's start by plotting the function y = |x 2 – 4|x| – 1|. To do this, we use the equality a 2 = |a| 2 and select the complete square in the submodular expression written on the right side of the function:
x 2 – 4|x| – 1 = |x| 2 – 4|x| - 1 = (|x| 2 – 4|x| + 4) – 1 – 4 = (|x |– 2) 2 – 5.
Then the original function will have the form y = |(|x| – 2) 2 – 5|.
To construct a graph of this function, we construct sequential graphs of functions:
1) y = (x – 2) 2 – 5 – parabola with vertex at point with coordinates (2; -5); (Fig. 1).
2) y = (|x| – 2) 2 – 5 – part of the parabola constructed in step 1, which is located to the right of the ordinate axis, is symmetrically displayed to the left of the Oy axis; (Fig. 2).
3) y = |(|x| – 2) 2 – 5| – the part of the graph constructed in point 2, which is located below the x-axis, is displayed symmetrically relative to the x-axis upward. (Fig. 3).
Let's look at the resulting drawings:
The graph of the function y = a is a straight line parallel to the abscissa axis.
Using the figure, we conclude that the graphs of functions have six common points (the equation has six roots) if a belongs to the interval (1; 5).
This can be seen in the following figure:
Let's find the arithmetic mean of the integer values of parameter a:
(2 + 3 + 4)/3 = 3.
Answer: 3.
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TO tasks with parameter can include, for example, the search for solutions to linear and quadratic equations V general view, study of the equation for the number of roots available depending on the value of the parameter.
Without bringing detailed definitions, as examples, consider the following equations:
y = kx, where x, y are variables, k is a parameter;
y = kx + b, where x, y are variables, k and b are parameters;
ax 2 + bx + c = 0, where x are variables, a, b and c are a parameter.
Solving an equation (inequality, system) with a parameter means, as a rule, solving infinite set equations (inequalities, systems).
Tasks with a parameter can be divided into two types:
A) the condition says: solve the equation (inequality, system) - this means, for all values of the parameter, find all solutions. If at least one case remains uninvestigated, such a solution cannot be considered satisfactory.
b) required to specify possible values parameters for which the equation (inequality, system) has certain properties. For example, has one solution, has no solutions, has solutions, belonging to the interval etc. In such tasks, it is necessary to clearly indicate at what parameter value the required condition is met.
The parameter, being an unknown fixed number, has a kind of special duality. First of all, it is necessary to take into account that the assumed popularity indicates that the parameter must be perceived as a number. Secondly, the freedom to manipulate the parameter is limited by its obscurity. So, for example, the operations of dividing by an expression that contains a parameter or extracting the root of an even degree from similar expression require preliminary studies. Therefore, care is required when handling the parameter.
For example, to compare two numbers -6a and 3a, you need to consider three cases:
1) -6a will be greater than 3a if a is a negative number;
2) -6a = 3a in the case when a = 0;
3) -6a will be less than 3a if a is a positive number 0.
The solution will be the answer.
Let the equation kx = b be given. This equation is short note an infinite number of equations with one variable.
When solving such equations there may be cases:
1. Let k be any real number not equal to zero and b is any number from R, then x = b/k.
2. Let k = 0 and b ≠ 0, original equation will take the form 0 x = b. Obviously, this equation has no solutions.
3. Let k and b be numbers, equal to zero, then we have the equality 0 x = 0. Its solution is any real number.
An algorithm for solving this type of equation:
1. Determine the “control” values of the parameter.
2. Solve the original equation for x for the parameter values that were determined in the first paragraph.
3. Solve the original equation for x for parameter values different from those chosen in the first paragraph.
4. You can write the answer in the following form:
1) for ... (parameter values), the equation has roots ...;
2) for ... (parameter values), there are no roots in the equation.
Example 1.
Solve the equation with the parameter |6 – x| = a.
Solution.
It is easy to see that a ≥ 0 here.
According to the rule of module 6 – x = ±a, we express x:
Answer: x = 6 ± a, where a ≥ 0.
Example 2.
Solve the equation a(x – 1) + 2(x – 1) = 0 with respect to the variable x.
Solution.
Let's open the brackets: aх – а + 2х – 2 = 0
Let's write the equation in standard form: x(a + 2) = a + 2.
If the expression a + 2 is not zero, that is, if a ≠ -2, we have the solution x = (a + 2) / (a + 2), i.e. x = 1.
If a + 2 is equal to zero, i.e. a = -2, then we have the correct equality 0 x = 0, so x is any real number.
Answer: x = 1 for a ≠ -2 and x € R for a = -2.
Example 3.
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a = 0, then we transform the equation to the form a + x = a 2 + ax or (a – 1)x = -a(a – 1). The last equation for a = 1 has the form 0 x = 0, therefore x is any number.
If a ≠ 1, then the last equation will take the form x = -a.
This solution can be illustrated on the coordinate line (Fig. 1)
Answer: there are no solutions for a = 0; x – any number with a = 1; x = -a for a ≠ 0 and a ≠ 1.
Graphical method
Let's consider another way to solve equations with a parameter - graphically. This method is used quite often.
Example 4.
Depending on the parameter a, how many roots does the equation ||x| – 2| = a?
Solution.
To solve using the graphical method, we construct graphs of the functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows possible cases location of the straight line y = a and the number of roots in each of them.
Answer: the equation will not have roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case of a = 2; four roots – at 0< a < 2.
Example 5.
At what a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let us depict the graphs of the functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x – 1|, expanding the modules using the interval method, we obtain:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x – 1, for x > 1.
On Figure 3 It is clearly seen that the equation will have a single root only when a = 1.
Answer: a = 1.
Example 6.
Determine the number of solutions to the equation |x + 1| + |x + 2| = a depending on parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at points (-2; 1) and (-1; 1) (Figure 4).
Answer: if the parameter a is less than one, then the equation will not have roots; if a = 1, then the solution to the equation is an infinite set of numbers from the segment [-2; -1]; if the values of parameter a are greater than one, then the equation will have two roots.
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