Each front tire on a particular type of vehicle is supposed to be tilled to a pr

Question

Each front tire on a particular type of vehicle is supposed to be tilled to a pressure of 26 psi. Suppose the actual pressure in each tire i* 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 a. What is the value of K? b. What is the probability that both tires are underfilled? c. What is the probability that the difference in air pressure between the two tires is at most 2 psi? d Determine the (marginal) distribution of air pressure in the right tire alone. c. Are X and Y independent random variables?

Explanation / Answer

(a) P(20 x 30, 20 y 30) =1
K(x^2 + y^2) dx dy = 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)
P(20 x 24, 20 y 24) = (3/380000)(x^2 + y^2) dx dy
(3/380000)(x^3/3 + xy^2) dy =
(3/380000)(5824/3 + 4y^2) dy =
(3/95000)(1456/3 + y^2) dy =
(3/95000)(1456y/3 + y^3/3) =
11648/95000 =
1456/11875 0.1226

(c)

[1]
P(20 y x + 2, 20 x 22) =
(3/380000)(x^2 + y^2) dy dx =
(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]
P(x - 2 y x + 2, 22 x 28) =
(3/380000)(x^2 + y^2) dy dx =
(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

required probabilility = 0.043179 + 0.238232 + 0.077916 0.3593

(d) mpdf of X

f(x) = f(x,y)dy=
(3/380000)(x^2 + y^2) dy =
(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

Similarly f(y) = (3/38000)y^2 + 1/20

e) Here f(x,y) fX(x) · fY(y), so X and Y are not independent.

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