Two random variable X and Y have the joint PDF given by: J;(x³y+ 1) ,0
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Fundamentals of Applied Probability and Random Processes
![Question# 1:
Two random variable X and Y have the joint PDF given by:
Si (x³y+ 1) ,0 <x< 2, 0 < y < 1
,0 <x< 2,0 < y<1
fxy(x, y) = { 4
,otherwise
a) What are the marginal PDFS of X and Y?
b) Are X and Y independent?
c) Find P[x < 1,y < 1]?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19a3223a-1d7c-4291-91d1-494278bfcd76%2Ff969a7fe-25fd-44af-ad4b-e008ed2f1470%2F8jod5k_processed.png&w=3840&q=75)
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- Question 10 The joint probability density function of X and Y is given byQUESTION 11 The probability density function, f(x), for any continuous random variable X, represents: a. all possible values that X will assume within some interval a ≤ x ≤ b. b. the probability that X takes on a specific value x. c. the height of the density function at x. d. None of these choices.Question 1 : Suppose that the probability density function (p.d.f.) of the life (in weeks) of a certain part is f(x) = 3 x 2 (400)3 , 0 ≤ x < 400. (a) Compute the probability the a certain part will fail in less than 200 weeks. (b) Compute the mean lifetime of a part and the standard deviation of the lifetime of a part. (c) To decrease the probability in part (a), four independent parts are placed in parallel. So all must fail, if the system fails. Let Y = max{X1, X2, X3, X4} denote the lifetime of such a system, where Xi denotes the lifetime of the ith component. Show that fY (y) = 12 y 11 (400)12 , y > 0. Hint : First construct FY (y) = P(Y ≤ y), by noticing that {Y ≤ y} = {X1 ≤ y} ∩ {X2 ≤ y} ∩ {X3 ≤ y} ∩ {X4 ≤ y}. (d) Determine P(Y ≤ 200) and compare it to the answer in part (a)
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