Question

explain as much as possible, thanks

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 24 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 y2) 20 SX 30, 20 y 30 0 otherwise Alxv) = {««x2 + x3) (a) What is the value of K? (Enter your answer as a fraction.) K = (b) What is the probability that both tires are underfilled? (Round your answer to four decimal places.) (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 SX 30 (e) Are X and Y independent rv's? O Yes, f(x,y) = f(x) .fyy), so X and Y are independent. Yes, f(x,y) = fx(x) .frly), so X and Y are independent. O No, f(x,y) = fx(x) fyy), so X and Y are not independent. No, f(x,y) + fx(x) • fyly), so X and Y are not independent.

Answer 1

f(2,y) dody = 1 1. "S51. y)dady SA S Ⓡ K(x2 + y)drdy = 1 *LT (x2 + y2)dxdy 30 1 K* 380000 3 = 1 3 K 7.89 * 10-6 380000

(b)

$\\ P(both\ tyre\ are\ under-filled)=\int_{20}^{24}\int_{20}^{24}\frac{3}{380000}(x^2+y^2)dxdy\\ \\ \\ =\frac{1456}{11875}\\ \\ \\ =0.1226$

$\\ (c)\\ \\ Required\ probability=\\ \\ \\ =1-\int_{22}^{30}\int_{20}^{x-2}\frac{3}{380000}(x^2+y^2)dydx+\int_{22}^{30}\int_{20}^{y-2}\frac{3}{380000}(x^2+y^2)dxdy\\ \\ \\ =1-0.320337-0.320337\\ \\ \\ =0.359326\\ \\ \\ =0.3593$

$\\ (d)\\ \\ f(x)=\int_{20}^{30}\frac{3}{380000}(x^2+y^2)dy\\ \\ \\ =\frac{3x^2+1900}{38000}\ \ \ \ \ \ \ \ 20\leq x\leq 30$

$\\ (e)\\ \\ Similarly;\\ \\ f(y)=\int_{20}^{30}\frac{3}{380000}(x^2+y^2)dx\\ \\ \\ =\frac{3y^2+1900}{38000}\ \ \ \ \ \ \ \ 20\leq y\leq 30\\ \\ \\ \\ So,\ \ \ \ f(x,y)\neq f(x)f(y)\\ \\ \\ Answer:\ \ \ No,\ f(x,y)\neq f(x)f(y)\ ,\ so\ X\ and\ Y\ are\ not\ independent.$