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) = K(x2 + y2) 20 ≤ x ≤ 30, 20 ≤ y ≤ 30 0 otherwise (a) What is the value of K? (Enter your answer as a fraction.) K = 3/380000 Correct: Your answer is correct. (b) What is the probability that both tires are underfilled? (Round your answer to four decimal places.) .430426 Correct: Your answer is correct. (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.) .3593 Correct: Your answer is correct. (d) Determine the (marginal) distribution of air pressure in the right tire alone.

[tex]f\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}{K\left( {{x^2} + {y^2}} \right)}&{20 \le x \le 30}\\0&{{\rm{Otherwise}}}\end{array}} \right\\[/tex]

Now by the definition of the pdf

[tex]\int\limits_x {\int\limits_y {f\left( {x,y} \right)dydx} } = 1\\\int\limits_x {\left( {\int\limits_y {f\left( {x,y} \right)dy} } \right)dx} = 1\\K\int\limits_x {\left( {\int\limits_y {\left( {{x^2} + {y^2}} \right)dy} } \right)dx} = 1\\K\int\limits_x {\left( {\left[ {{x^2}y + \dfrac{{{y^3}}}{3}} \right]_{20}^{30}} \right)dx} = 1[/tex]

[tex]K\int\limits_x {\left( {10{x^2} + \dfrac{{19000}}{3}} \right)dx} = 1\\K\left[ {10\dfrac{{{x^3}}}{3} + \dfrac{{19000}}{3}x} \right]_{20}^{30} = 1\\K\left( {10\dfrac{{19000}}{3} + \dfrac{{19000}}{3}\left( {10} \right)} \right) = 1\\\dfrac{{380000}}{3}K = 1\\K = \dfrac{3}{{380000}}[/tex]

So the value of K is found as 3/380000.

Part b

The probability of both tyres being underfilled is given such that the X≤27 psi thus

[tex]P\left( {X \le 27,Y \le 27} \right) \\= \int\limits_{20}^{27} {\int\limits_{20}^{27} {K\left( {{x^2} + {y^2}} \right)dydx} } \\=\int\limits_{20}^{27} {\int\limits_{20}^{27} {\left( {{x^2} + {y^2}} \right)dydx} } \\=\int\limits_{20}^{27} {\left(\frac{21}{380000}x^2+\frac{11683}{380000}\right)dx} } \\=\frac{81781}{190000}=0.4304[/tex]

So the probability that both the tires will be underfilled is 0.4304.

Part c

The probability that the difference between the two tyres is more than 2 psi. For this the cases are as

[tex]\begin{array}{l}20 \le x \le 22::\,\,\,\,20 \le y \le x + 2\\22 \le x \le 28::\,\,\,\,x - 2 \le y \le x + 2\\28 \le x \le 30::\,\,\,\,x - 2 \le y \le 30\end{array}[/tex]

For these cases the probability is given as [tex]P\left( {\left| {X - Y} \right| \le 2} \right) = \left\{ {\int\limits_{28}^{30} {\int\limits_{x - 2}^{30} {K\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{20}^{22} {\int\limits_{20}^{x + 2} {K\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{22}^{28} {\int\limits_{x - 2}^{x + 2} {K\left( {{x^2} + {y^2}} \right)dydx} } } \right\}[/tex][tex]\\ = K\left[ {\left\{ {\int\limits_{28}^{30} {\int\limits_{x - 2}^{30} {\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{20}^{22} {\int\limits_{20}^{x + 2} {\left( {{x^2} + {y^2}} \right)dydx} } } \right\} + \left\{ {\int\limits_{22}^{28} {\int\limits_{x - 2}^{x + 2} {\left( {{x^2} + {y^2}} \right)dydx} } } \right\}} \right]\\ = K\left[ {\left\{ {\dfrac{{14804}}{{15}}} \right\} + \left\{ {\dfrac{{8204}}{{15}}} \right\} + \left\{ {\dfrac{{46864}}{{15}}} \right\}} \right][/tex][tex]= \dfrac{3}{{38000}}\left[ {\left\{ {\dfrac{{14804}}{{15}}} \right\} + \left\{ {\dfrac{{8204}}{{15}}} \right\} + \left\{ {\dfrac{{46864}}{{15}}} \right\}} \right]\\ = \dfrac{{4267}}{{11875}}\\ = 0.3593[/tex]

So the probability that the difference of the air pressure between the two tyres is 2psi is 0.3593.

Part d

The marginal distribution of air pressure in right tire

[tex]\begin{array}{l}{f_x}\left( x \right) = \int\limits_y {f\left( {x,y} \right)dy} \\ = K\int\limits_y {\left( {{x^2} + {y^2}} \right)dy} \\ = K\left( {{x^2}y + \dfrac{{{y^3}}}{3}} \right)_{20}^{30}\\ = K\left( {{x^2}\left( {30 - 20} \right) + \dfrac{{{{30}^3} - {{20}^3}}}{3}} \right)\\ = K\left( {10{x^2} + \dfrac{{19000}}{3}} \right)\\ = \dfrac{3}{{38000}}\left( {10{x^2} + \dfrac{{19000}}{3}} \right)\\ = \dfrac{1}{{20}} + \dfrac{{3{x^2}}}{{38000}}\end{array}[/tex]

So the marginal distribution of air pressure in the right tyre is [tex]\dfrac{1}{{20}} + \dfrac{{3{x^2}}}{{38000}}[/tex]