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

Question

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 27 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

f(x,y) =

Please show all steps. Thanks!

5. 3/5 points I Previous Answers Devorestat8 5.E.009 My Notes Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 27 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 f(x, y SK(x2 y2) 20 s x s 30, 20 s y s 30 otherwise (a) What is the value of K? (Enter your answer as a fraction.) K 3380000 (b) What is the probability that both tires are underfilled? (Round your answer to four decimal places.) 0.4304 (c) What is the probability that the difference in air pressure between the two tires is at most 2 psi? (Round your answer to four decimal places (d) Determine the (marginal) distribution of air pressure in the right tire alone. for 20 s x s 30 (e) Are X and Y independent rv's? O Yes, f(x,y) fx(x) fYy), so X and Y are independent. O Yes, f(x,y) fx(x) fry), so X and Yare independent. O No, f(x,y) fx(x) fry), so X and Y are not dependent. No, f(x,y) fx(x). fYy), so X and Y are not independent.

Explanation / Answer

(c)
Three integrals:
[1]
(3/380000)(x^2 + y^2) dy dx (20 y x + 2, 20 x 22) =
(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, 22 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

Now [1] + [2] + [3] is 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) Correct Answre: Option (D)

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