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 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, wiht joint pdf

1) What is the probability that the difference in air pressure between the two tires is at most 2 psi?

2) Determine the marginal distribution of air pressure in the right tire alone.

Each front tire on a particular type of vehicle is supposed to be 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 What is the probability that the difference in air pressure between the two tires is at most 2 psi? Determine the marginal distribution of air pressure in the right tire alone. f(X,y) = { K (x2 + y2) 20 le x le 30 and 20 le y le 30 0 otherwise

Explanation / Answer

(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 ? 24, 20 ? y ? 24) =

? (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)

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

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