mont4e_sm_ch03_supplemental - 3-118. a) Let X denote the...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Supplemental Exercises 3-119. 4 1 3 1 8 3 3 1 4 1 3 1 8 1 ) ( = + + = X E , 0104 . 0 4 1 3 1 8 3 3 1 4 1 3 1 8 1 ) ( 2 2 2 2 = + + = X V 3-120. a) 3681 . 0 ) 999 . 0 ( 001 . 0 1 1000 ) 1 ( 999 1 = = = X P () () () 999 . 0 ) 999 . 0 )( 001 . 0 ( 1000 ) ( 1 ) 001 . 0 ( 1000 ) ( ) d 9198 . 0 999 . 0 001 . 0 2 1000 999 . 0 001 . 0 1 1000 999 . 0 001 . 0 0 1000 ) 2 ( ) c 6319 . 0 999 . 0 001 . 0 0 1000 1 ) 0 ( 1 ) 1 ( ) b 998 2 999 1 1000 0 999 0 = = = = = + + = = = = = X V X E X P X P X P 3-121. a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np. b) () () () 112 . 0 9 . 0 1 . 0 2 50 9 . 0 1 . 0 1 50 9 . 0 1 . 0 0 50 ) 2 ( 48 2 49 1 50 0 = + + = X P c) 48 0 50 1 49 10 51 . 4 9 . 0 1 . 0 50 50 9 . 0 1 . 0 49 50 ) 49 ( × = + = X P 3-122. (a)Binomial distribution, p=0.01, n=12. (b) P(X>1)=1-P(X 1)= 1- - =0.0062 12 0 ) 1 ( 0 12 p p 14 1 ) 1 ( 1 12 p p (c) μ =E(X)= np =12*0.01 = 0.12 V(X)=np(1-p) = 0.1188 σ = ) ( X V = 0.3447 3-123. (a) 12 (0.5) 0.000244 = (b) = 0.2256 66 12 (0.5) (0.5) C 6 (c) 4189 . 0 ) 5 . 0 ( ) 5 . 0 ( ) 5 . 0 ( ) 5 . 0 ( 6 6 12 6 7 5 12 5 = + C C 3-124. (a) Binomial distribution, n=100, p=0.01. (b) P(X 1) = 0.634 (c) P(X 2)= 0.264 3-24
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(d) μ=E(X)= np=100*0.01=1 V(X)=np(1-p) = 0.99 σ = ) ( X V =0.995 (e) Let p d = P(X 2)= 0.264, Y: # of messages requires two or more packets be resent. Y is binomial distributed with n=10, p m =p d *(1/10)=0.0264 P(Y 1) = 0.235 3-125 Let X denote the number of mornings needed to obtain a green light. Then X is a geometric random variable with p = 0.20. a) P(X = 4) = (1-0.2) 3 0.2= 0.1024 b) By independence, (0.8) 10 = 0.1074. (Also, P(X > 10) = 0.1074) 3-126 Let X denote the number of attempts needed to obtain a calibration that conforms to specifications. Then, X is geometric with p = 0.6. P(X 3) = P(X=1) + P(X=2) + P(X=3) = 0.6 + 0.4(0.6) + 0.4 2 (0.6) = 0.936. 3-127. Let X denote the number of fills needed to detect three underweight packages. Then X is a negative binomial random variable with p = 0.001 and r = 3. a) E(X) = 3/0.001 = 3000 b) V(X) = [3(0.999)/0.001 2 ] = 2997000. Therefore, σ X = 1731.18 3-128. Geometric with p=0.1 (a) f(x)=(1-p) x-1 p=0.9 (x-1) 0.1 (b) P(X=5) = 0.9 4 *0.1=0.0656 (c) μ=E(X)= 1/p=10 (d) P(X 10)=0.651 3-129. (a) λ =6*0.5=3.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.

Page1 / 6

mont4e_sm_ch03_supplemental - 3-118. a) Let X denote the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online