A mathematical model of college probability of about 800 words, urgent !! 200

7. Evaluation of the model.

In addition to calculating the optimal insurance in the scheme given by the problem, this model takes into account the impact of the amount of insurance funds on the insurance profit, and introduces quantification. But this model is based on our assumptions, such as that the interest rate of a bank cannot be fixed for many years, and does not take into account the probability of death per year. In this way, the impact of these aspects on the model can be considered in terms of model improvement.

This model is very meaningful for the actual insurance problem, which can be used as a reference tool for the insurance company and provide certain information for the policyholder. In this paper, the sensitivity analysis of the model caused by the change of life expectancy is also analyzed. However, there are also shortcomings: the model does not have graphics, ** and other parts, which cannot make the problem clearer and intuitively expressed.

Upstairs drunk

Question (1) f(x) is wrong to determine that the integral is not equal to 1 on the field of real numbers, and there is nothing wrong with the interval to be extended to 0.

In addition, the explanation is not clear enough, and the solution is not complete enough to make mistakes.

Insurance solution: first find the distribution function for x:

p(x<=x) (use an uppercase x instead of e, I'm used to it) =fx(x)=|0,x<0

x^2,01

The first question y=2x, then when 0 is paired.

The third question says y=x 2, then fy(y)=p(y<=y)=p(x 2<=y)=p(-root y<=x<=root y)=fx(root y)-fx(-root y)=(root y) 2-0=y, (0 then there is the result upstairs.)

Keep in mind that fx(x) cannot be directly deduced to fy(y) when solving this kind of problem, and it is possible to make mistakes in defining domains if you are experienced like the one upstairs, and you can only rely on the equivalence relationship of p(y<=y)=p(2x<=y)=p(x<=y 2) to derive it reliably.

(1) When an Aix. So the answer to this question is.

f(x)= |1 2)x, 0=1-x)=1-p(e<=1-x)=1-fe(1-x), let.

The probability density of e+1 is f(x), so at this time, the probability density of -e+1 f(x)=-p(1-x)*(1)=p(1-x)=2-2x, so the answer to the second question is.

f(x)= | 2-2x , 0|0, rest of the place.

3) When 0e 2 probability density is f(x), then e 2 probability density f(x) = p(x (1 2))*1 2) * (x (-1 2)) = 1 So the answer to three questions.

f(x)= | 1, 0|0, rest of the place.

Just drank two beers for your own reference.

Method for finding the derivative (1) Steps to find the derivative of the function y=f(x) at x0: Find the increment of the function δy=f(x0+δx)-f(x0) Find the average rate of change Take the limit and get the derivative. (2) Derivative formulas for several common functions:

c'=0 (c is constant); x^n)'=nx^(n-1) (n∈q)； sinx)'=cosx； ④cosx)'=-sinx； ⑤e^x)'=e^x； ⑥a^x)'=a xlna (3) The four rules of operation for derivatives: u v).'=u'±v' ②(uv)'=u'v+uv' ③(u/v)'=(u'v-uv'v 2 (4) The derivative of a composite function The derivative of a composite function to an independent variable is equal to the derivative of a known function to an intermediate variable, multiplied by the derivative of an intermediate variable to an independent variable—known as the chain rule. Derivatives are an important pillar of calculus!

Derivative Formulas and Proof Here are a few basic derivatives of functions and how they are derived: is a constant) y'=0 y'=nx^(n-1) y'=a^xlna y=e^x y'=e^x y'=logae/x y=lnx y'=1/x y'=cosx y'=-sinx y'=1/cos^2x y'=-1/sin^2x y'=1/√1-x^2 y'=-1/√1-x^2 y'=1/1+x^2 y'=-1/1+x^2

The details are as follows:

p (damaged. x<=200) =

p (damaged. 200=240) =

1)x~n(220，25)

z=(x-220)/25

p(x<=200)= p(z<

p(200=240) = p(z>

p(damaged) = p(damaged|x<=200)*p(x<=200)+p(damaged|.) 200<=x<=240)*p(200=240)*p(x>=240) =

2)p(x>=240|Corrupt) = p (Corrupt |x>=240)*p(x>=240) p(damaged)=

If you don't understand, you can ask.

Good luck with your studies! o(∩_o~

|Damaged|"x between 200-400 volts.

bai"=>"x between 200-240 volts"?

p (damaged. du|zhi x<=200) =

p (damaged. 200=240) =

1)x~daon(220，25)

z=(x-220)/25

p(x<=200)= p(z<

p(200=240) = p(z>

p(damaged) = p(damaged|x<=200)*p(x<=200)+p(damaged|.) 200<=x<=240)*p(200=240)*p(x>=240) =

2)p(x>=240|Corrupt) = p (Corrupt |x>=240)*p(x>=240) p(damaged)=