ssrs 2008 r2 barcode font Random Variables and Probability Distributions in Software

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CHAPTER 2 Random Variables and Probability Distributions
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Applications to geometric probability 2.31. A person playing darts finds that the probability of the dart striking between r and r
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P(r R r dr) c B1 r 2 a R dr
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Here, R is the distance of the hit from the center of the target, c is a constant, and a is the radius of the target (see Fig. 2-18). Find the probability of hitting the bull s-eye, which is assumed to have radius b. Assume that the target is always hit.
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The density function is given by f (r) Since the target is always hit, we have c B1 r 2 a R 1
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r 2 a R dr
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Fig. 2-18
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3 > 2a. Then the probability of hitting the bull s-eye is
30 f (r) dr
2.32. Two points are selected at random in the interval 0 squares is less than 1.
3 b B1 2a 30
r 2 a R dr
b (3a2 b2) 2a3
1. Determine the probability that the sum of their
Let X and Y denote the random variables associated with the given points. Since equal intervals are assumed to have equal probabilities, the density functions of X and Y are given, respectively, by (1) f1(x) e 1 0 0 x 1 otherwise e f2 ( y) e 1 0 0 y 1 otherwise
Then since X and Y are independent, the joint density function is given by (2) f (x, y) f1(x) f2(y) 1 0 0 x 1, 0 otherwise y 1
It follows that the required probability is given by (3) P(X2 Y2 1) 6 dx dy
where r is the region defined by x2 y2 1, x 0, y 0, which is a quarter of a circle of radius 1 (Fig. 2-19). Now since (3) represents the area of r, we see that the required probability is p > 4.
CHAPTER 2 Random Variables and Probability Distributions
Fig. 2-19
Miscellaneous problems 2.33. Suppose that the random variables X and Y have a joint density function given by
f (x, y) e c (2x 0 y) 2 x 6, 0 otherwise y 5
Find (a) the constant c, (b) the marginal distribution functions for X and Y, (c) the marginal density functions for X and Y, (d) P(3 X 4, Y 2), (e) P(X 3), (f) P(X Y 4), (g) the joint distribution function, (h) whether X and Y are independent.
(a) The total probability is given by
6 5 2 y 6
c(2x
y) dx dy
3x 53
c 2xy
c 10x
For this to equal 1, we must have c
1 > 210.
25 dx 2
y2 5 2 dx 2 0
210c
(b) The marginal distribution function for X is
F1(x)
3 ` v
f (u, v) du dv
x u 6
0 du dv
x 2x 2
2 5x 84 x 6 18 2 x 6
3 2 v
2u v du dv 0 210 2u v du dv 210
3 2 v
The marginal distribution function for Y is
F2( y)
3 ` v
f (u, v) du dv
6 u 6
3 ` v
0 du dv
0 y2
0 0 5 y 5
3 0 v
2u v du dv 0 210 2u v du dv 210
16y 105 y
3 2 v
CHAPTER 2 Random Variables and Probability Distributions
(c) The marginal density function for X is, from part (b), f1(x) d F (x) dx 1 e (4x 0 5)>84 2 x 6 otherwise
The marginal density function for Y is, from part (b), f2( y) d F (y) dy 2 e (2y 0 1 4 210 3x
16)>105
0 y 5 otherwise y) dx dy 23 28 3 20
(d) (e) (f )
X P(X
4, Y 3) P(X
3 (2x 3 y 2
1 6 210 3x Y 4)
3 (2x 3 y 0
y) dx dy
6 f (x, y) dx dy
where r is the shaded region of Fig. 2-20. Although this can be found, it is easier to use the fact that P(X Y 4) 1 P(X Y 4) 1 6 f (x, y) dx dy
where rr is the cross-hatched region of Fig. 2-20. We have P(X Thus P(X Y 4) 33 > 35. Y 4) 1 4 210 3x
3 2 y
y) dx dy
2 35
Fig. 2-20
Fig. 2-21
(g) The joint distribution function is
F(x, y)
x, Y
3 ` v
f (u, v) du dv
In the uv plane (Fig. 2-21) the region of integration is the intersection of the quarter plane u x, v y and the rectangle 2 u 6, 0 v 5 [over which f (u, v) is nonzero]. For (x, y) located as in the figure, we have
F(x, y)
3 2 v
2u v du dv 0 210
16y y 2 105
CHAPTER 2 Random Variables and Probability Distributions
When (x, y) lies inside the rectangle, we obtain another expression, etc. The complete results are shown in Fig. 2-22. (h) The random variables are dependent since f (x, y) 2 f1(x) f2 ( y) or equivalently, F(x, y) 2 F1(x)F2(y).
2.34. Let X have the density function f (x) Find a function Y e 6x (1 0 x) 0 x 1 otherwise
h(X) which has the density function g(y) e 12y 3(1 0 y 2) 0 y 1 otherwise
Fig. 2-22
We assume that the unknown function h is such that the intervals X x and Y y h(x) correspond in a one-one, continuous fashion. Then P(X x) P(Y y), i.e., the distribution functions of X and Y must be equal. Thus, for 0 x, y 1,
306u(1 or By inspection, x Y !X. y2 or y h(x) 3x2
u) du 2x3
3 3012v (1
v2) dv
!x is a solution, and this solution has the desired properties. Thus
2.35. Find the density function of U
Method 1 Let U XY and V
XY if the joint density function of X and Y is f(x, y).
xy, v x or x v, y u > v. Then the Jacobian is given by
X, corresponding to which u 'x 'u 4 'y 'u 'x 'v 4 'y 'v
0 v 1
1 2 uv 2
CHAPTER 2 Random Variables and Probability Distributions
Thus the joint density function of U and V is g(u, v)
from which the marginal density function of U is obtained as g(u) 3 `g(u, v) dv
1 u f v, v u vu
` u vu
Method 2 The distribution function of U is
G(u) For u B3
u f v, v dv
6 f (x, y) dx dy
xy u
0, the region of integration is shown shaded in Fig. 2-23. We see that
0 ` ` u>x
G(u)
f (x, y) dyR dx
30 B 3 `f (x, y) dyR dx
Fig. 2-23
Fig. 2-24
Differentiating with respect to u, we obtain
g(u)
The same result is obtained for u Fig. 2-24.
1 u x f x, x dx `
0, when the region of integration is bounded by the dashed hyperbola in
` 1 u 30 x f x, x dx
1 u f x, x dx u xu
2.36. A floor has parallel lines on it at equal distances l from each other. A needle of length a l is dropped at random onto the floor. Find the probability that the needle will intersect a line. (This problem is known as Buffon s needle problem.)
Let X be a random variable that gives the distance of the midpoint of the needle to the nearest line (Fig. 2-24). Let be a random variable that gives the acute angle between the needle (or its extension) and the line. We denote by x and u any particular values of X and . It is seen that X can take on any value between 0 and l > 2, so that 0 x l > 2. Also can take on any value between 0 and p > 2. It follows that P(x i.e., the density functions of X and X x dx) 2 dx l P(u 2 > l, f2(u) 2 p du 1 du) 2 p du
are given by f1(x)
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