Mathematics Probability and Statistics, Mathematics Probability and Statistics

Probability theory and mathematical statistics are important branches of modern mathematics. In the past 20 years, with the development of computers and various statistical software, probability and statistical methods have been widely used in the fields of finance, insurance, biology, medicine, economics, operations research management and engineering technology. These include:

Limit Theory, Stochastic Process Theory, Mathematical Statistics, Application of Probability Theory Methods, Applied Statistics, etc. Limit theory includes strong limit theory and weak limit theory. Stochastic process theory includes Mahalanobis process theory, martingale theory, stochastic calculus, stationary process and other related theories. The application of probability theory method is a very wide field, including stochastic mechanics, statistical physics, insurance, stochastic networks, queuing theory, reliability theory, stochastic signal processing and other related aspects.

The application of statistical methods is mainly produced in the research activities of substantive disciplines, for example, the least squares method and normal distribution theory are derived from astronomical observation error analysis, correlation and regression analysis are derived from biological research, principal component analysis and factor analysis are derived from the research of education and psychology, and the sampling survey method is derived from the collection of statistical survey data. On the basis of learning the basic theories of probability theory, statistics, and stochastic process theory, this research direction is committed to the research of the intersection of probability and statistical theory and methods with other disciplines, as well as the research of data mining generated by the combination of statistics and computer science. In addition, mathematical finance is also a major research direction of the major.

It mainly uses mathematical modeling, theoretical analysis, derivation, numerical calculation and computer simulation and other theoretical analysis, statistical analysis and simulation analysis to study and analyze the theoretical and practical problems involved.

1 The vertical axis brackets are 2, 4, 6, and 8 from the bottom to the top

2 Add a white bar after the most rolling royal shout, and the excellent black bar is high on one side (height is 5)3 As you can see from the picture, the class size is: 5+4+6+10+5+5 = Ohno 35 (people).

The number of outstanding people is: 5 + 4 = 9 people.

The excellent rate is: 9 35 100% =

Take 4 steps and cross 6 steps, which means that the result is two positive and two negative, in this case there are c(4,2)=6 cases, a total of 2 4=16 cases, so the probability is 6 16 = 3 8

3 8 is equivalent to the appearance of two heads; The probability of two opposites.

(1) x can take 1,2,3, where x=1 only takes one time to obtain qualified products, so only take once, choose one out of 10, the denominator is 10, obtain qualified products, choose one out of 8, the numerator is 8, then p(x=1)=8 10=4 5

x=2, then the first time to take the defective product and the second time to take **, the first denominator is 10, 10 is 1, the numerator is 2, 2 is 1; The second time, because it is not put back, the denominator is 9, 9 chooses 1, and the numerator is 8, 8 chooses 1, so p(x=2)=2 10·8 9=8 45

x=3, then the first and second defective products are taken and the third time is **, the first denominator is 10, 10 is 1, the numerator is 2, 2 is 1; The second time, because it is not put back, the denominator is 9, 9 is 1, the numerator is 1, 1 is 1, and the third time, because it is not put back, the denominator is 8, 8 is 1, and the numerator is 8, 8 is 1, so p(x=3)=2 10·1 9·8 8=1 45

2）p=p(x=2)+p(x=3)=8/45+1/45=1/5（3）e(x)=1·4/5+2·8/45+3·1/45=11/9

5 of the 11 eggs were bad and 6 were good.

First of all, you have to understand that these 5 bad eggs are not the same, and the 6 good eggs are also different. And the selection is not about the order of priority.

Secondly, choose 8 out of 11, whether good or bad, a total of c (11 8) (you should understand this statement).

And 5 of the bad ones are selected, which means that 3 of the 6 good ones are selected, and there is a total of c(6, 3).

So, the final answer is: p=c(6 3)hu sleepy c(11 8)don't do the pose, you know? Hehe.

Hope it helps.

8 of these 11 eggs were randomly selected, and all the possibilities were: c(11,8) species;

These 5 bad ones are all selected for the following situations, all of which are:

That is, select all of these 5, and then choose 3 of the 6 from the remaining 6) Liang or.

c(5,5)c(6,3)

So the probability: p=c(5,5)c(6,3) c(11,8)= 4 33

Wrong. If you set x to 9 or more shots.

When p(x=1), you only consider one situation, in fact, there are three cases: the first shot, the second shot, or the third shot.

So p(x=1)=(, the rest of the same analysis, may be the first and second, the first and third, the second and third.

p(x=2)=3*(, and p(x=3) has only one case, all in p(x=3)=

Each time: the probability of more than 9 rings.

Each time: the probability of less than 9 rings.

Shoot three times, all with a probability of less than 9 rings.

p=c(3,3)*(

If you shoot three times, the probability of hitting more than 9 rings (including 9 rings) is.

(1) Solution: (x's pull-u) ( (n 1 2)) n(0,1) (standard normal distribution) (n 1 2: represents n square).

So: p=p=p=2* (1 )-1= (In this case, you need to check the table of the standard normal function (1)=.)

2) Solution: (x's pull-u) (s (n (1 2)) t(n-1) (t-distribution with degrees of freedom n-1) where s is the sample variance squared.

Therefore: p=1-p=1-(2*The value here is taken as an approximation) In this case, you need to check the table of the distribution of the statistic t)=

A event is ** B is defective.

There are four scenarios.

bab=(5 20)*(15 19)*(4 18)bbb=(5 20)*(4 19)*(3 18)abb=(15 20)*(5 19)*(4 18)aab=(15 20)*(14 19)*(5 18) Add them up to equal.