Question

(1) Suppose the pdf of a random variable X is 0, otherwise. (a) Find P(2 < X < 3). (b) Find P(X < 1). (e) Find t such that P(X <t) = (d) After the value of X has been observed, let y be the integer closest to X. Find the PMF of the random variable y U

Answer 1

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Let X be a random van'áble OL f (x) - \. 0 5X23 LIN ; Otherwise a) Pl 24X53) = +(x) dx 2 3 . 12 / 2 x dx. 2 3 2 c 23 [ 22 ] [ 2 4 L ] 2 2 X 5 g x2 0.5556 P ( 2 < X 3) = 0.5556 b) p(x+1) = 2 adx= [2]. g O 0.1111 g TP(x<1) = 0o!!!! c) we wan to find t such that

P ( XL t) = 2 t . 3 3 x dx. = 1 / 2 x2 t=2 [ g 27.4 ta 9/2 t= 9/ to 2:1213 t=201213 After the value of x has been observed Let y be the integer dosest to X. ;P ( 2 40.5) Y= ip(0.5<x<15) 2 ; P ( 1.5<x<2.5) 3 ; P ( X25) I 0.5 2? P(x4 (5) = J 3/8 x dx = 1095=025 g 9 = 1.5 0:0278 1:5 pl 0.5 <X215) - 0:5 C | - 52 - 0:52) / 3 38 xdx= (*) - = 0.2222 pl 45 <x< 2.5) - 2 ã cm) 215- (2-52-1-53) 19 15 0.4444 PC 2.5 2x) 3 ģx dx (32_205²) 19 205 - 0.3056

i. = PC X20-5) 0.0278 pl 0-5 < X< 1:5) = 0·2222 PL 1.5 < X<215) = 0.4444 p(x > 265) = 0.3056 ..pom.f of y is 3 (Y=y) 0 L 2 04444 0.3656 0.2222 Plys 9) 0.0278