Homework#7_Solutions

Homework#7_Solutions - ENGINEERING PROBABILITY Spring 2012...

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Unformatted text preview: ENGINEERING PROBABILITY Spring 2012 Homework #7 _ Solutions Topic Discrete random variables & Probability distribution functions & Expected values 32. 34. 37. 315. 316. All probabilities are greater than or equal to zero and sum to one. a) P(X 1)=P(X=1)=0.5714 317. Probabilities are nonnegative and sum to one. a) P(X = 4) = 9/25 b) P(X 1) = 1/25 + 3/25 = 4/25 c) P(2 X < 4) = 5/25 + 7/25 = 12/25 d) P(X > 10) = 1 b) P(X>1)= 1P(X=1)=10.5714=0.4286 c) P(2<X<6)=P(X=3)=0.1429 d) P(X1 or X>1)= P(X=1)+ P(X=2)+P(X=3)=1 All probabilities are greater than or equal to zero and sum to one. a) P(X 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(1 X 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X 1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1/2 The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is 0,12,... , The range of X is 0,12,3,4,5 , The range of X is 0,12,...,50 , Assignment Due Next Class 3-2,4,7,15, 16, 17, 33, 39, 40, 49, 50 1 333. x 2 0, 1 / 8 2 x 1 3 / 8 1 x 0 F ( x) where 0 x 1 5 / 8 7 / 8 1 x 2 1 2 x a) P(X 1.25) = 7/8 b) P(X 2.2) = 1 c) P(1.1 < X 1) = 7/8 1/8 = 3/4 d) P(X > 0) = 1 P(X 0) = 1 5/8 = 3/8 f X (2) 1 / 8 f X (1) 2 / 8 f X ( 0) 2 / 8 f X (1) 2 / 8 f X ( 2) 1 / 8 339. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.5, f(3) = 0.5 a) P(X 3) = 1 340. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.7, f(4) = 0.2, f(7) = 0.1 a) P(X 4) = 0.9 349. Determine E(X) and V(X) for random variable in exercise 315 b) P(X > 7) = 0 c) P(X 5) = 0.9 d) P(X>4) = 0.1 e) P(X2) = 0.7 b) P(X 2) = 0.5 c) P(1 X 2) = P(X=1) = 0.5 d) P(X>2) = 1 P(X2) = 0.5 . E ( X ) 2 f (2) 1 f (1) 0 f (0) 1 f (1) 2 f (2) 2(1 / 8) 1( 2 / 8) 0( 2 / 8) 1( 2 / 8) 2(1 / 8) 0 2 V ( X ) 2 2 f (2) 12 f (1) 0 2 f (0) 12 f (1) 2 2 f (2) 2 4(1 / 8) 1(2 / 8) 0(2 / 8) 1(2 / 8) 4(1 / 8) 0 2 1.5 350. Determine E(X) and V(X) for random variable in exercise 316 E ( X ) 1 f (1) 2 f (2) 3 f (3) 1(0.5714286) 2(0.2857143) 3(0.1428571) 1.571429 V ( X ) 12 f (1) 22 f (2) 32 f (3) 2 1.428571 3 ...
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This note was uploaded on 02/27/2012 for the course ISE 540:210 taught by Professor Susanalbin during the Spring '12 term at Rutgers.

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