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Homework Solutions 03

# Homework Solutions 03 - 1 Lecture Plan Experiments Outcomes...

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Solutions to Homework #3 (W4150 ‘Intro to Probability and Statistics’, S04) Sec. 3.3. (1) X: discrete Y: continuous M: continuous N: discrete P: discrete Q: continuous Sec.3.3. (3) Elements of S w HHH 3 HHT 1 HTH 1 THH 1 HTT -1 THT -1 TTH -1 TTT -3 Sec. 3.3. (5) (a) Since 1 = Σ 3 x=0 c(x 2 + 4) = 30c ° c = 1/30; (b) Since 1 = Σ 2 x=0 c(C 2 x )(C 3 3-x ) = c(1 + 6 + 3) = 10c ° c = 1/10. Sec. 3.3. (6) (a) P(X>200) = dx x ± + 200 3 ) 100 ( 000 , 20 = + - 200 2 ) 100 ( 000 , 10 x = 1/9; (b) P(80 < X <120) = dx x ± + 120 80 3 ) 100 ( 000 , 20 = 120 80 2 ) 100 ( 000 , 10 + - x = .102 Sec. 3.3. (12) (a) P(T = 5) = F(5) – F(4) = ¾ - ½ = ¼ ; (b) P(T >3) = 1 - F(3) = 1 – ½ = ½ ; (c) P(1.4 < T < 6) = F(6) – F(1.4) = ¾ - ¼ = ½. Sec. 3.3. (14a) (a) P(X < 12 min) = P(X < .2 hours) = F(.2) = 1 – e -1.6 = .7981 Sec. 3.4. (1) (a) ² 3 x=1 ² 3 y=1 f(x,y) = c { ² 3 x=1 ² 3 y=1 xy} = 36c = 1 ° c = 1/36;

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(b) ² x ² y f(x,y) = c { ² x ² y | x-y | = 15c = 1 ° c = 1/15. Sec.3.4. (2) Let’s calculate f(x,y) for every possible pair (x,y): y x 0 1 2 3 0 0 1/30 2/30 3/30 1 1/30 2/30
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