Mathematics, 23.06.2019 00:10 burnsmykala23

Suppose you are to throw a dart at a circular dart board with radius 2 inches. let (x, y ) denote the point that you hit on the board (you can assume the board is centered at the origin (0, 0), and that the dart hits somewhere on the board uniformly at random). (a) find the joint pdf of x and y . (b) find the marginal pdfs of x and y . (c) find the conditional pdfs fx|y and fy |x. (d) suppose that the “bulls eye” on the target consists of a small circle centered at the origin with radius 0.25. explain how you would use one of these pdfs to compute the probability of a dart hitting the bulls eye. find this probability an even easier way. (e) what is the probability that x > 1? (f) if you know that x = 1, what is the pdf of y conditioned on this fact? what’s the probability that y > 0.3 conditioned on this fact?

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Answer from: rara54

a. All points on the board are equally likely to be hit with a probability of 1/(area of board), or

$f_{X,Y}(x,y)=\begin{cases}\dfrac1{4\pi}&\text{for }x^2+y^2\le4\\\\0&\text{otherwise}\end{cases}$

b. To find the marginal distribution of , integrate the joint distribution with respect to , and vice versa. We can take advantage of symmetry here to compute the integral:

$\displaystyle\int_y f_{X,Y}(x,y)\,\mathrm dy=2\int_0^{\sqrt{4-x^2}}\frac{\mathrm dy}{4\pi}=\frac{\sqrt{4-x^2}}{2\pi}$

$f_X(x)=\begin{cases}\dfrac{\sqrt{4-x^2}}{2\pi}&\text{for }-2\le x\le2\\\\0&\text{otherwise}\end{cases}$

and by the same computation you would find that

$f_Y(y)=\begin{cases}\dfrac{\sqrt{4-y^2}}{2\pi}&\text{for }-2\le y\le2\\\\0&\text{otherwise}\end{cases}$

c. We get the conditional distributions by dividing the joint distributions by the respective marginal distributions:

$f_{X\mid Y=y}(x)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}$

$f_{X\mid Y=y}(x)=\begin{cases}\dfrac1{2\sqrt{4-y^2}}&\text{for }-2\le y\le2\text{ and }x^2\le4-y^2\\\\0&\text{otherwise}\end{cases}$

and similarly,

$f_{Y\mid X=x}(y)=\begin{cases}\dfrac1{2\sqrt{4-x^2}}&\text{for }-2\le x\le2\text{ and }y^2\le4-x^2\\\\0&\text{otherwise}\end{cases}$

d. You can compute this probability by integrating the joint distribution over a part of the circle (call it "B" for bullseye):

$\displaystyle\iint_Bf_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^{0.25}\frac r{4\pi}\,\mathrm dr\,\mathrm d\theta=\frac1{64}$

(using polar coordinates) The easier method would be to compute the area of a circle with radius 0.25 instead, then divide that by the total area of the dartboard.

$\dfrac{\pi\left(\frac14\right)^2}{\pi\cdot2^2}=\dfrac1{64}$

e. The event that is complementary to the event that $X\le1$ , so

$P(X1)=1-P(X\le1)=1-F_X(1)$

where F_X(x) is the marginal CDF for . We can compute this by integrate the marginal PDF for :

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then

$P(X1)=1-F_X(1)=\dfrac13-\dfrac{\sqrt3}{4\pi}\approx0.1955$

f. We found that either random variable conditioned on the other is a uniform distribution. In particular,

$f_{Y\mid X=1}(y)=\begin{cases}\dfrac1{2\sqrt3}&\text{for }y^2\le3\\\\0&\text{otherwise}\end{cases}$

Then

$P(Y0.3\mid X=1)=1-P(Y\le0.3\mid X=1)=1-F_{Y\mid X=1}(0.3)$

where $F_{Y\mid X=x}(y)$ is the CDF of conditioned on X=x . This is easy to compute:

$F_{Y\mid X=1}(y)=\displaystyle\int_{-\infty}^yf_{Y\mid X=1}(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

and we end up with

$P(Y0.3\mid X=1)=\dfrac{10-\sqrt3}{20}\approx0.4134$

Answer from: Quest

answer: learning objective(s)

· describe the creation and use of systems of equations.

· graph a system of linear equations on the coordinate plane and identify its solution.

introduction

sometimes graphing a single linear equation is all it takes to solve a mathematical problem. other times, one line just doesn’t do it, and a second equation is needed to model the situation. this is often the case when a problem involves two variables. solving these kinds of problems requires working with a system of equations, which is a set of two or more equations containing the same unknowns.

let’s take a look at systems of equations, and see what the graphs of individual equations within a system reveal about the mathematical relationship of the variables.

systems of equations

a system of equations contains two or more linear equations that share two or more unknowns. to find a solution for a system of equations, we must find a value (or range of values) that is true for all equations in the system.

the graphs of equations within a system can tell us how many solutions exist for that system. look at the images below. each show two lines that make up a system of equations (in the graph on the right the two lines are superimposed and look like a single line). how many points in common does each system of lines reveal?

one solution

no solutions

infinite solutions

if the graphs of the equations intersect, then there is one solution that is true for both equations.

if the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.

if the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

remember, the graph of a line represents every point that is a possible solution for the equation of that line. so when the graphs of two equations cross, the point of intersection lies on both lines, meaning that it is a possible solution for both equations. when the graphs of two equations never touch, there are no shared points and there are no possible solutions for the system. when the graphs of two equations lie on top of one another, they share all their points and every one is a possible solution.

graphing as a solution method

graphing equations in order to identify a specific point of intersection is usually not a precise way to solve systems because it is often difficult to see exactly where two lines intersect (unless you are using a computer-based graphing program that allows you to zoom in on a point). however, the graph of a system of equations can still give a good idea of what type of solution, if any, exists. let’s graph a system, and see how it works.

graph the system y = 3x and x + 2y = 4. how many solutions does this system have?

a graph of the two lines y = 3x and x + 2y = 4 shows us that the lines intersect, meaning that there is a single (x, y) value that satisfies both equations. looking at the graph does not tell us exactly where that point is, but we don’t need to know that information here, because the question only asks for the number of solutions.

so a system made of two intersecting lines has one solution. now let’s look at a different situation:

how many solutions exist for the system y -0.5x = 7 and 2y = x − 3?

plotting both equations, it looks like there is no solution—the lines are parallel. to check this finding, we can compare the slopes of the equations. to make that easy, we’ll rearrange the equations into slope-intercept form, or y = mx + b. that gives us the equations y = 0.5x + 7 and y = 0.5x − 1.5. yes, the slope of both lines is 0.5, which means the lines really are parallel. they never intersect, so there is no point that lies on both lines, and no solution to the system.

step-by-step explanation:

Answer from: Quest

alot

step-by-step explanation:

its a wood chuck that can chuck wood

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