Each front tire on a particular type of vehicle is supposed to bo filled to a pr
ID: 3006494 • Letter: E
Question
Each front tire on a particular type of vehicle is supposed to bo filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable- X for the right tire and Y for the left tire, with joint pdf fk(x, y) = {K(x^2 + y^2) 20 lessthanorequalto x lessthanorequalto 30, 20 lessthanorequalto y lessthanorequalto 30 0 otherwise What is the value of K? What is the probabilily lhai boih lires arc underlilled? What is the probabilily that the differene in air pressure between the two tires is at most 2 psi? Determine the (marginal) distribution of air pressure in the right tire alone. Are X and Y independent random variables? Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x, y) = {xe^minusx(1+y) x greaterthanorequalto 0 and y greaterthanorequalto 0 0 otherwise What is the probability that the lifetime X of the first component exceeds 3? What are the marginal pdf's of X and Y? Are the two lifetimes independent? Explain. What is the probability that the lifetime of at least one component exceeds 3? The joint pdf of pressures for right (X) and left (Y) front tires is given by f(x, y) = {K(x^2 + y^2) 20 lessthanorequalto x lessthanorequalto 30, 20 lessthanorequalto y lessthanorequalto 30 0 otherwise Determine the conditional pdf of Y given that X = x and the conditional pdf of X given that Y = y if you are given F_x (x) = 10kx^2 +.05 for 20 lessthanorequalto x lessthanorequalto 30 and F_y (y) = 10ky^2 +.05 for 20 lessthanorequalto y lessthanorequalto 30. If the pressure in the right tire is found to be 22 psi, what is the probability that the left tire has a pressure of at least 25 psi? Compare this to P(Y greaterthanorequalto 25). If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire?Explanation / Answer
2)
Let's find pdf for X (x>0):
........infinity.................................infinity
f(x) = xe-x(1+y) dy = xe-x(-x-1 e-xy) | = e-x
--------0------------------------------------------0
because of some problem in the server inserting equations are not working, for any furthor doubts please comment
Analogously pdf for Y (y>0):
........infinity..............................infinity
g(y) = xe-x(1+y) dx =(z=x(1+y)) = (z*e-z/(1+y)) * (dz/(1+y)) = 1/(1+y)2
----------0---------------------------------------0
Now
a)
...........infinity
p(X>3) = e-x dx = e-3 = 0.049787
-------------3
b)
Obviously X and Y are dependent: joint pdf is not a product of marginal pdf's.
Now
c)
lets find
p(max(X,Y)>3)=1-p(max(x,Y)<=3)
......3......3
=1 - xe-x(e-xydy)dx
......0.......0
.......3
=1 - e-x(1-e-3x)dx = e-3+(1-e-12)/4
.......0
= 0.049787 + 0.249998 = 0.299785
--------------------------------------------------------------------------------------------
1)
(a)
K(x^2 + y^2) dx dy (20 x 30, 20 y 30) = 1
K(x^3/3 + xy^2) dy = 1
K(19000/3 + 10y^2) dy = 1
K[19000y/3 + 10y^3/3] = 1
K[190000/3 + 190000/3] = 1
K = 3/380000
(b)
(3/380000)(x^2 + y^2) dx dy (20 x 26, 20 y 26)
= (3/380000)(x^3/3 + xy^2) dy
= (3/380000)(6150/3 + 6y^2) dy
= (3/2280000)(1025/3 + y^2) dy
= (3/2280000)(1025y/3 + y^3/3)
0.0067
(c)
Three integrals:
[1]
(3/380000)(x^2 + y^2) dy dx (20 y x + 2, 20 x 24) =
(3/380000)(yx^2 + y^3/3) dx =
(3/380000)[(x - 18)x^2 + {(x + 2)^3 - 20^3}/3] dx =
(3/380000)[x^3 - 18x^2 + (1/3){x^3 + 6x^2 + 12x + 8 - 20^3}] dx =
(3/380000)[(4/3)x^3 - 16x^2 + 4x - 2664] dx =
(3/380000)[(1/3)x^4 - 16x^3/3 + 2x^2 - 2664x] =
(1/380000)[74256 - 16(2648) + 6(84) - 7992(2)] =
16408/380000 =
2051/47500 0.043179
[2]
(3/380000)(x^2 + y^2) dy dx (x - 2 y x + 2, 24 x 28) =
(3/380000)(yx^2 + y^3/3) dx =
(3/380000)(4x^2 + [(x + 2)^3 - (x - 2)^3]/3) dx =
(3/380000)(4x^2 + [12x^2 + 16]/3) dx =
(3/380000)(8x^2 + 16/3) dx =
(3/380000)(8x^3/3 + 16x/3) =
(1/380000)[8(11304) + 16(6)] =
2829/11875 0.238232
[3]
(3/380000)(x^2 + y^2) dy dx (x - 2 y 30, 28 x 30) =
(3/380000)(yx^2 + y^3/3) dx =
(3/380000)[(32 - x)x^2 + {30^3 - (x - 2)^3}/3] dx =
(3/380000)[32x^2 - x^3 + {30^3 - (x^3 - 6x^2 + 12x - 8)}/3] dx =
(3/380000)[34x^2 - (4/3)x^3 - 4x + 27008/3] dx =
(3/380000)[(34/3)x^3 - (1/3)x^4 - 2x^2 + 27008x/3] =
(1/380000)[34(5048) - 195344 - 6(116) + 27008(2)] =
3701/47500 0.077916
0.043179 + 0.238232 + 0.077916 0.3593
(d)
(3/380000)(x^2 + y^2) dy (20 y 30) =
(3/380000)(yx^2 + y^3/3) =
(3/380000)(10x^2 + (30^3 - 20^3)/3) =
(3/380000)(10x^2 + 19000/3) =
(3/38000)x^2 + 1/20
e)
No, f(x,y) fX(x) · fY(y), so X and Y are not independent.
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