### Application question on probability theory!

Let y be the number of incoming shipments, x be a uniform distribution of [10,30], and the probability density be 1/20

When x ≥ y, the profit is 500y+300(x-y)=300x+200y

When x<y, the profit is 500x-100(y-x)=600x-100y

Profit’s Expectation = 1/20∫600x-100ydx + 1/20∫300x+200ydx

The first integral has a lower bound of 10 and an upper bound of y; the second has a lower bound of y and an upper bound of 30

The result obtained = 5250+350y-7.5y^2, which is derived so that it is 0; 15y=350 and y=23.33333< /p>

Calculate the results for y=23 and 24, and compare them, knowing that the profit is greater for x=23 = 12482.5

### A probability density to find the distribution law of the problem: let the customer in a bank window waiting for service time X (unit: minutes) has a probability density f (x)

Photograph of a solving process, I do not know to see clearly not. (1) Distribution law can be calculated according to the general formula of Y = 0, 1, 2, 3, 4, 5 when the probability of the distribution law table.

### Let the probability density of random variable x be f(x)=e^-x,x>0,f(x)=0,other,find the probability density of y=x^2

F(y)=P(Y<y)=P(x^2<y)=P(-y^0.5<x<y^0.5)=Fx(y^0.5)-Fx(-y^0.5). where Fx(x)=1-e^-x carries over

Differentiating gives f(y)=(0.5y^-0.5)(e^(y^0.5)+e^(-y^0.5)).

Or do it with Jacobian.

x=(+or-y^0.5),|Jacobian|=|dx/dy|=1/2y^-0.5

f(y)=(0.5y^-0.5)(fx(y^0.5)+fx(-y^0.5))=(0.5y^-0.5)(e^(y^0.5)+e^(-y^ 0.5))

In fact the distribution of any random variable x, y=x^2 is (0.5y^-0.5)(fx(y^0.5)+fx(-y^0.5)) Next time just apply this formula directly, the above proof works for all random variables x

### Application questions about probability trouble everyone

Solution:

(1) This is a uniform distribution in the random variable, F(X)=x-a/b-a. a ≤ x < b0, others.

The method is [in the probability density non-zero interval] on the probability density function to find the definite integral, such as this question, the probability density non-zero interval is a ≤ x < b. To find the definite integral if you do not know how to do it, please look up the relevant books.

(2) [In the probability density non-0 interval], E(x) = ∫xf(x)dx. (The upper and lower limits will not be hit.) Variance D(x) = E(x²) – [E(x)]² where E(x²) = ∫x²f(x)dx To find the definite integral [in the probability density non-zero interval]. (Upper and lower bounds will not be hit.)

### A reintegration application problem Let the plane sheet D be a closed region bounded by the line x+y=1 and the two coordinate axes, and its density function be u(x,y)=xy, then the mass is

∫∫∫(D)u(x,y)dS

=∫∫∫(D)xydxdy

=∫∫(0,1)xdx∫(0,1-x)ydy

=(1/2)∫(0,1)x[y²](0,1-x)dx

=(1/2)∫(0,1)x(1-x)²dx

=(1/2)∫(0,1)∫(0,1)(x one 2x² ten x³)dx

=(1/2)[x²/2 one 2x³/3 ten x^4/4](0,1)

= (1/2)(1/2-2/3 ten 1/4)

=(1/2)(3/4 one 2/3)

=(1/2)(1/12)

=124

Extension

Because the values of the random variable X depend only on the integral of the probability density function, the values of the probability density function at individual points do not not affect how the random variable behaves. More precisely, a function can also be a probability density function of X if there are only a finite, countably infinite number of points at which the function and the probability density function of X take different values, or if the measure is zero with respect to the entire axis of real numbers.

The probability that a continuous-type random variable takes values at any point is 0. As a corollary, the probability that a continuous-type random variable takes values on an interval is independent of whether the interval is open or closed. Note that the probability P{x=a} = 0, but {X=a} is not an impossible event.