The mathematical expectation a \u003d 3 and the standard deviation \u003d 5 of the normally distributed random variable X are set.
Write the density of the distribution of probabilities and schematically build its graph.
Find the probability that x will take a value from the interval (2; 10).
Find the probability that x will be greater than 10.
Find an interval symmetric with respect to the mathematical expectation, in which the values \u200b\u200bof x will be enclosed with a probability \u003d 0.95.
1). Let us compose the distribution density function of a random variable X with parameters a \u003d 3, \u003d 5 using the formula
... Let's build a schematic graph of the function 
... Note that the normal curve is symmetric with respect to the straight line x \u003d 3 and has a 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: 
Function values \u200b\u200bare found from the appendix table.
4) Let's use the formula 
... By condition, the probability of hitting the interval symmetric with respect to the mathematical expectation 
... From the table, we find t, for which Ф (t) \u003d 0.475, t \u003d 2. means 
... Thus, 
... The answer is x (-1; 7).
To problems 31-40.
Find the confidence interval for an estimate with a reliability of 0.95 of the unknown mathematical expectation of a normally distributed feature X of the general population, if the general standard deviation \u003d 5, the sample mean 
and sample size n \u003d 25.
It is required to find the confidence interval 
.
All quantities except t are known. Let us find t from the ratio Ф (t) \u003d 0.95 / 2 \u003d 0.475. From the table in the appendix we find t \u003d 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, the 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 \u003d 0.6. Therefore, the supposed Poisson's law Putting i \u003d 0,1,2,3,4, we find the probabilities P i of the appearance of i non-standard products in 200 lots: Let us find the theoretical frequencies by the formula Let's compare empirical and theoretical frequencies using Pearson's criterion. To do this, we will make a calculation table. Let's combine the few frequencies (4 + 2 \u003d 6) and the corresponding theoretical frequencies (3.96 + 0.6 \u003d 4.56). In practice, most random variables, which are affected by a large number of random factors, obey the normal law of probability distribution. Therefore, in various applications of the theory of probability, this law is of particular importance. A random variable $ X $ obeys the normal probability distribution law if its probability distribution density has the following form $$ f \\ left (x \\ right) \u003d ((1) \\ over (\\ sigma \\ sqrt (2 \\ pi))) e ^ (- (((\\ left (xa \\ right)) ^ 2) \\ over ( 2 (\\ sigma) ^ 2))) $$ The schematic graph of the function $ f \\ left (x \\ right) $ is shown in the figure and has the name "Gaussian curve". To the right of this graph is a 10-mark banknote of the Federal Republic of Germany, which was used even before the advent of the euro. If you look closely, you can see the Gaussian curve and its discoverer, the greatest mathematician Karl Friedrich Gauss, on this banknote. Let's go back to our density function $ f \\ left (x \\ right) $ and give some explanations about the distribution parameters $ a, \\ (\\ sigma) ^ 2 $. The $ a $ parameter characterizes the center of dispersion of the values \u200b\u200bof the random variable, that is, it makes sense of the mathematical expectation. When the $ a $ parameter is changed and the $ (\\ sigma) ^ 2 $ parameter is unchanged, we can observe the displacement of the graph of the function $ f \\ left (x \\ right) $ along the abscissa axis, while the density graph itself does not change its shape. The $ (\\ sigma) ^ 2 $ parameter is the variance and characterizes the shape of the curve of the density graph $ f \\ left (x \\ right) $. When the parameter $ (\\ sigma) ^ 2 $ is changed with the parameter $ a $ unchanged, we can observe how the density graph changes its shape, shrinking or stretching, while not shifting along the abscissa axis. As you know, the probability of a random variable $ X $ falling into the interval $ \\ left (\\ alpha; \\ \\ beta \\ right) $ can be calculated $ P \\ left (\\ alpha< X < \beta \right)=\int^{\beta }_{\alpha }{f\left(x\right)dx}$. Для нормального распределения случайной величины $X$ с параметрами $a,\ \sigma $ справедлива следующая формула: $$ P \\ left (\\ alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right)$$ Here the function $ \\ Phi \\ left (x \\ right) \u003d ((1) \\ over (\\ sqrt (2 \\ pi))) \\ int ^ x_0 (e ^ (- t ^ 2/2) dt) $ is the Laplace function ... The values \u200b\u200bfor this function are taken from. The following properties of the function $ \\ Phi \\ left (x \\ right) $ can be noted. 1
... $ \\ Phi \\ left (-x \\ right) \u003d - \\ Phi \\ left (x \\ right) $, that is, the function $ \\ Phi \\ left (x \\ right) $ is odd. 2
... $ \\ Phi \\ left (x \\ right) $ is a monotonically increasing function. 3
... $ (\\ mathop (lim) _ (x \\ to + \\ infty) \\ Phi \\ left (x \\ right) \\) \u003d 0.5 $, $ (\\ mathop (lim) _ (x \\ to - \\ infty) \\ To calculate the values \u200b\u200bof the $ \\ Phi \\ left (x \\ right) $ function, you can also use the Excel $ f_x $ function wizard: $ \\ Phi \\ left (x \\ right) \u003d NORMDIST \\ left (x; 0; 1; 1 \\ right ) -0.5 $. For example, let's calculate the values \u200b\u200bof the function $ \\ Phi \\ left (x \\ right) $ for $ x \u003d 2 $. The probability of a normally distributed random variable $ X \\ in N \\ left (a; \\ (\\ sigma) ^ 2 \\ right) $ in an interval symmetric with respect to the mathematical expectation $ a $ can be calculated by the formula $$ P \\ left (\\ left | X-a \\ right | The Three Sigma Rule< \delta \right)=2\Phi \left({{\delta }\over {\sigma }}\right).$$ ... It is practically certain that the normally distributed random variable $ X $ will fall into the interval $ \\ left (a-3 \\ sigma; a + 3 \\ sigma \\ right) $.Example 1 ... The random variable $ X $ is subject to the normal probability distribution with the parameters $ a \u003d 2, \\ \\ sigma \u003d 3 $. Find the probability of $ X $ falling into the interval $ \\ left (0,5; 1 \\ right) $ and the probability of the inequality $ \\ left | X-a \\ right |
Using the formula< 0,2$. find $ P \\ left (0,5; 1 \\ right) \u003d \\ Phi \\ left (((1-2) \\ over (3)) \\ right) - \\ Phi \\ left (((0,5-2) \\ left (0.33 \\ right) \u003d 0.191-0.129 \u003d $ 0.062. $$ P \\ left (\\ alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right),$$ example 2 The Three Sigma Rule< 0,2\right)=2\Phi \left({{\delta }\over {\sigma }}\right)=2\Phi \left({{0,2}\over {3}}\right)=2\Phi \left(0,07\right)=2\cdot 0,028=0,056.$$ ... Suppose that during a year the price of a company's shares is a random variable distributed according to the normal law with a mathematical expectation of 50 conventional monetary units and a standard deviation of 10. What is the probability that on a random day of the period under discussion the price for the action will be:
a) more than 70 conventional monetary units? b) below 50 per share? c) between 45 and 58 conventional monetary units per share? let the random variable $ X $ be the stock price of a certain company. By condition, $ X $ is subject to the normal distribution with parameters $ a \u003d 50 $ - mathematical expectation, $ \\ sigma \u003d 10 $ - standard deviation. Probability $ P \\ left (\\ alpha $$ a) \\ P \\ left (X\u003e 70 \\ right) \u003d \\ Phi \\ left (((\\ infty -50) \\ over (10)) \\ right) - \\ Phi \\ left (((70-50) \\ $$ b) \\ P \\ left (X< X < \beta \right)$ попадания $X$ в интервал $\left(\alpha ,\ \beta \right)$ будем находить по формуле: $$ P \\ left (\\ alpha< X < \beta \right)=\Phi \left({{\beta -a}\over {\sigma }}\right)-\Phi \left({{\alpha -a}\over {\sigma }}\right).$$ $$ c) \\ P \\ left (45 The probability that the SV deviation< 50\right)=\Phi \left({{50-50}\over {10}}\right)-\Phi \left({{-\infty -50}\over {10}}\right)=\Phi \left(0\right)+0,5=0+0,5=0,5.$$ X< X < 58\right)=\Phi \left({{58-50}\over {10}}\right)-\Phi \left({{45-50}\over {10}}\right)=\Phi \left(0,8\right)-\Phi \left(-0,5\right)=\Phi \left(0,8\right)+\Phi \left(0,5\right)=$$ from her M.O. a in absolute value will be less than a given positive number, equal to if we put in this equality, then we get s w: space \u003d "720" /\u003e That is, the normally distributed SV {!LANG-1c6b75faf99b58339c14f341249caa64!}"> {!LANG-05de083781b427fcfd0513842f237857!} a deviates from his M.O. if we put in this equality, then we get, 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. The mathematical expectation of a function of one SV. (Tetr) If each possible value of the random variable X
corresponds to one possible value of a random variable Y
then Y
called a random argument function X: Y \u003d φ
(a
). Let us find out how to find the distribution law of a function according to the known distribution law of the argument. 1) Let the argument X
Is a discrete random variable, with different values X
correspond to different values Y
... Then the probabilities of the corresponding values X
and Y
are equal .
2) If different values X
can match the same values Y
, then the probabilities of the argument values \u200b\u200bfor which the function takes the same value are added. 3) If X
- continuous random variable, Y \u003d φ
(a
), φ
(x
) Is a monotone and differentiable function, and ψ
(at
) Is the function inverse to φ
(x
). The mathematical expectation of a function of one random argument. Let be Y \u003d φ
(a
) - random argument function 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 the function Z \u003d X + Y for discrete independent SV X and Y. (tetr) If each pair of possible values \u200b\u200bof random values \u200b\u200bX and Y corresponds to one possible value of a random variable Z, then Z is called a function of two random arguments X and Y and is written Z \u003d φ (X, Y). If X and Y are discrete independent random variables, then, in order to find the distribution of the function Z \u003d X + Y, it is necessary to find all possible values \u200b\u200bof Z, for which it is enough to add each possible value of X with all possible values \u200b\u200bof Y; the probabilities of the found possible values \u200b\u200bof Z are equal to the products of the probabilities of the added values \u200b\u200bof X and Y. If X and Y are continuous independent random variables, then the distribution density g (z) of the sum Z \u003d 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 density of the distribution of arguments; if the possible values \u200b\u200bof the arguments are nonnegative, then the distribution density g (z) of the quantity Z \u003d 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 quantity Z \u003d X + Y, it is advisable to first find the distribution function G (z) and then differentiate it with respect to z : g (z) \u003d 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 hitting a random point (X, Y) in the domain D is equal to the double integral over this domain of the product of the distribution densities: Р [( X, Y) cD] \u003d R 0.3 0.7 R 0.6 0.4 Find the distribution of the random variable Z \u003d X + K. Solution. In order to compose the distribution of the value Z \u003d X + Y, it is necessary to find all possible values \u200b\u200bof Z and their probabilities. Possible values \u200b\u200bof Z are the sums of each possible value of X with all possible values \u200b\u200bof Y: Z 1 \u003d 1 + 2 \u003d 3; z 2 \u003d 1 + 4 \u003d 5; z 3 \u003d 3 + 2 \u003d 5; z4 \u003d 3 + 4 \u003d 7. Find the probabilities of these possible values. For Z \u003d 3, it is sufficient that the value of X takes the value x1 \u003d l and the value of the K-value y1 \u003d 2. The probabilities of these possible values, as follows from these distribution laws, are 0.3 and 0.6, respectively. Since the arguments X and Y are independent, then the events X \u003d 1 and Y \u003d 2 are independent n, therefore, the probability of their joint occurrence (i.e., the probability of the event Z \u003d 3) according to the multiplication theorem Rain 0.3 * 0.6 \u003d 0 ,18. Similarly, we find: I B \u003d! - f4 \u003d 5) \u003d 0.3 0.4 \u003d 0.12; P (Z \u003d 34-2 \u003d 5) \u003d 0.7 0.6 \u003d 0.42; P (Z \u003d 3rd \u003d 7) \u003d 0.7-0.4 \u003d 0.28. Let us write the desired distribution by adding the probabilities of inconsistent events Z \u003d z 2 \u003d 5, Z \u003d z 3 \u003d 5 (0.12 + 0.42 \u003d 0.54): Z 3 5 7; P 0.18 0.54 0.28. Control: 0.18 + 0.54 + 0.28 \u003d 1. As mentioned earlier, examples of probability distributions continuous random variable
X are: Let us give the concept of a normal distribution law, a distribution function of such a law, an order of calculating the probability of a random variable X falling into a certain interval. The length X of some part is a random variable distributed according to the normal distribution law, and has an average value of 20 mm and a standard deviation of 0.2 mm. Decision.
and) We find the probability density of a random variable X, distributed according to the normal law: provided that m x \u003d 20, σ \u003d 0.2. b) For the normal distribution of a random variable, the probability of falling into the interval (19.7; 20.3) is determined: in) We find the probability that the absolute value of the deviation is less than a positive number 0.1: d) 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 have such a deviation, i.e. 38.3%. e) Since the percentage of parts, the deviation of which from the average does not exceed the specified, 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 \u003d 0.74. Hence δ \u003d 0.74 * σ \u003d 0.74 * 0.2 \u003d 0.148 mm. e) Since the sought interval is symmetric about the mean value m x \u003d 20, then it can be defined as the set of values \u200b\u200bX satisfying the inequality 20 - δ< X < 20 + δ или |x − 20| < δ . By hypothesis, the probability of finding X in the desired interval is 0.95, which means 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 \u003d 1.96. Hence δ \u003d 1.96 * σ \u003d 1.96 * 0.2 \u003d 0.392. It is said that SV X has even distribution in the section from a to b, if its density f (x) is constant in this section, that is, For example, a measurement is made using a device with coarse divisions; the nearest integer is taken as an approximate value of the measured value. SV X - the measurement error is evenly distributed over the site, since none of the values \u200b\u200bof the random variable is in any way preferable to others. Exponential (exponential) is called the probability distribution of a continuous random variable, which is described by the density where is a constant positive value. An example of a continuous random variable distributed according to an exponential law is the time between the occurrence of two consecutive events of the simplest flow. Often the duration of the uptime of the elements has an exponential distribution, the distribution function of which is - failure rate (average number of failures per unit of time). Normal law distribution (sometimes called gauss's law) plays an extremely important role in the theory of probability and occupies a special position among other distribution laws. The distribution density of the normal law has the form where m is the mathematical expectation, - standard deviation X. The probability that the normally distributed SV X will take a value belonging to the interval is calculated by the formula: where Ф (X) - laplace function... Its values \u200b\u200bare determined from the table of the appendix of the textbook on probability theory. The probability that the deviation of a normally distributed random variable X from the mathematical expectation in absolute value is less than a given positive number is calculated by the formula EXAMPLE 13.2.41. The value of one division of the ammeter scale is 0.1 A. Readings are rounded to the nearest whole division. Find the probability that an error exceeding 0.02 A. Decision. The round-off error can be thought of as CB X, which is evenly distributed between two adjacent divisions. Density of uniform distribution, where (b-a) is the length of the interval containing the possible values \u200b\u200bof X. In the problem under consideration, this length is 0.1. therefore The counting error will exceed 0.02 if it is enclosed in the interval (0.02; 0.08). According to the formula EXAMPLE 13.2.42. The duration of the uptime of the element has an exponential distribution. Find the probability that over a time duration of hours: a) the element will fail; b) the element will not fail. Decision. a) The function determines the probability of failure of an element for a time duration t, therefore, substituting, we obtain the probability of failure:. b) The events “the element will fail” and “the element will not fail” are opposite, therefore the probability that the element will not fail. EXAMPLE 13.2.43. Random variable X is distributed normally with parameters. Find the probability that SV X deviates from its mathematical expectation m by more than. This probability is very small, that is, such an event can be considered almost impossible (you can make a mistake in about three cases out of 1000). This is the “rule of three sigma”: if a random variable is normally distributed, then the absolute value of its deviation from the mathematical expectation does not exceed three times the standard deviation. EXAMPLE 13.2.44. The mathematical expectation and standard deviation of a normally distributed random variable are, respectively, 10 and 2. Find the probability that, as a result of testing, X will take the value enclosed in the interval (12, 14). Solution: For a normally distributed quantity Substituting, we get We find from the table. Seeking probability. Solve problems using formulas for calculating the probability of continuous random variables and their characteristics 3.2.9.1. Find the mathematical expectation, variance and standard deviation of a random variable X uniformly distributed in the interval (a, b). Resp.: 3.2.9.2. Metro trains run regularly every 2 minutes. The passenger enters the platform at a random time. Find the distribution density of SV T - the time during which he will have to wait for the train; Resp.: 3.2.9.3. The minute hand of the electric clock moves in leaps and bounds at the end of each minute. Find the probability that at a given moment the clock will show a time that differs from the true one by no more than 20 s. Resp.:2/3 3.2.9.4. Random variable X is distributed evenly over the site (a, b). Find the probability that, as a result of the experiment, it deviates from its mathematical expectation by more than. Resp.:0 3.2.9.5. Random variables X and Y are independent and evenly distributed: X - in the interval (a, b), Y - in the interval (c, d). Find the expected value of the XY product. Resp.: 3.2.9.6. Find the mathematical expectation, variance, and standard deviation of an exponentially distributed random variable. Resp.: 3.2.9.7. Write the density and distribution function of the exponential law if parameter. Resp.: 3.2.9.8. The random variable has an exponential distribution with a parameter. To find Resp.:0,233 3.2.9.9. The uptime of the element is distributed according to the exponential law, where t is the time, h. Find the probability that the element will operate reliably for 100 h. Resp.:0,37 3.2.9.10. Three elements are tested that work independently of one another. The duration of the uptime of the elements is distributed according to the exponential law: for the first element Resp.: a) 0.292; b) 0.466; c) 0.19 3.2.9.11. Prove that if a continuous random variable is exponentially distributed, then the probability that X takes a value less than the mathematical expectation M (X) does not depend on the value of the parameter; b) find the probability that X\u003e M (X). Resp.: 3.2.9.12. The mathematical expectation and standard deviation of a normally distributed random variable are, respectively, 20 and 5. Find the probability that, as a result of testing, X will take the value enclosed in the interval (15; 25). Resp.: 0,6826 3.2.9.13. Some substance is weighed without systematic errors. Random weighing errors are subject to the normal law with a standard deviation r. Find the probability that a) weighing will be performed with an error not exceeding 10 r in absolute value; b) of three independent weighings, the error of at least one will not exceed 4d in absolute value. Resp.: 3.2.9.14. Random variable X is distributed normally with mathematical expectation and standard deviation. Find an interval symmetric with respect to the mathematical expectation, in which the value X will fall with a probability of 0.9973 as a result of the test. Resp.:(-5,25) 3.2.9.15. The plant produces balls for bearings, the nominal diameter of which is 10 mm, and the actual diameter is random and distributed according to the normal law with mm and mm. During the inspection, all balls that do not pass through a round hole with a diameter of 10.7 mm and all that pass through a round hole with a diameter of 9.3 mm are rejected. Find the percentage of balls that will be rejected. Resp.:8,02% 3.2.9.16. The machine is stamping parts. The length of the part X is controlled, which is distributed normally with the design length (mathematical expectation) equal to 50 mm. In fact, the length of the manufactured parts is not less than 32 and not more than 68 mm. Find the probability that the length of the part taken at random: a) is greater than 55 mm; b) less than 40 mm. Hint: From equality Resp.: a) 0.0823; b) 0.0027 3.2.9.17. Chocolate boxes are packed automatically; their average weight is 1.06 kg. Find the variance if 5% of the boxes weigh less than 1 kg. It is assumed that the mass of the boxes is distributed according to the normal law. Resp.:0,00133 3.2.9.18. A bomber flying along the bridge, which is 30 meters long and 8 meters wide, dropped bombs. The random variables X and Y (the distance from the vertical and horizontal axes of symmetry of the bridge to the place where the bomb fell) are independent and distributed normally with standard deviations equal to 6 and 4 m, respectively, and mathematical expectations equal to zero. Find: a) the probability of one thrown bomb hitting the bridge; b) the probability of the destruction of the bridge, if two bombs are dropped, and it is known that one hit is enough to destroy the bridge. Resp.: 3.2.9.19. In a normally distributed population, 11% of X values \u200b\u200bare less than 0.5 and 8% of X values \u200b\u200bare greater than 5.8. Find the parameters m and the given distribution. \u003e To problems 41-50.
has the form 
.
,
,
,
,
.
... Substituting the probability values \u200b\u200binto this formula, we obtain 
,
,
,
,
.
Probability of hitting a normally distributed random variable in a given interval
,
... Discrete independent random variables X and Y are given by distributions:№
Index Normal distribution law Note
Definition
Normal is called
the probability distribution of a continuous random variable X, the density of which has the form
where m x is the mathematical expectation of a random variable X, σ x is the standard deviation
2
Distribution function
Probability
hitting the interval (a; b)
- Laplace integral function
Probability
the fact that the absolute value of the deviation is less than the positive number δ
for m x \u003d 0
An example of solving a problem on the topic "Normal distribution law of a continuous random variable"
A task.
It is 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 what percentage are parts whose deviation from the average does not exceed 0.1 mm;
e) find what the deviation should be so that the percentage of parts whose deviation from the average does not exceed the given one increases to 54%;
f) find an interval, symmetric about the mean, in which X will be located with a probability of 0.95.
F ((20.3-20) / 0.2) - F ((19.7-20) / 0.2) \u003d F (0.3 / 0.2) - F (-0.3 / 0, 2) \u003d 2F (0.3 / 0.2) \u003d 2F (1.5) \u003d 2 * 0.4332 \u003d 0.8664.
We found the value Ф (1.5) \u003d 0.4332 in the applications, in the table of values \u200b\u200bof the Laplace integral function Φ (x) ( table 2
)
R (| X-20 |< 0,1) = 2Ф(0,1/0,2) = 2Ф(0,5) = 2*0,1915 = 0,383.
We found the value Ф (0.5) \u003d 0.1915 in applications, in the table of values \u200b\u200bof the integral Laplace function Φ (x) ( table 2
)
The sought interval
: (20 - 0.392; 20 + 0.392) or (19.608; 20.392).
.
determines the probability of an element failure in a time duration t.
,
,
.EXAMPLES OF SOLVING PROBLEMS
... So, 
.
we have 
.Examples and tasks for independent solution
... Find the probability that you will have to wait no more than half a minute.
, 
.
; for the second 
; for the third element 
... Find the probability that in the time interval (0; 5) hours will fail: a) only one element; b) only two elements; c) all three elements.
pre-find.
Examples of problem solving\u003e