STAT 349 - Study Guide Exercise Solutions Ch 3: Ch 4:

STAT 349 - Study Guide Exercise SolutionsCh 3:5. (a) P (X = i|Y = 3) = ( 3)( 6 (i 3−i)/9 )3 ; (b) E[X|Y = 1] =538. (a) E[X] = 6; (b) E[X|Y = 1] = 7;(c) E[X|Y = 5] = 1( 1 5 ) + 2( 4 5 )( 1 5 ) + 3( 4 5 )2 ( 1 5 ) + 4( 4 5 )3 ( 1 5 ) + 6( 4 5 )4 ( 1 6 ) + 7( 4 5 )4 ( 5 6 )( 1 6 ) + . . .21. (a)X = ∑ Ni=1 T i ; (b)E[N] = 3; (c)E[T N ] = 2; (d)E[X|N = n] = 4(n−1)+2 = 4n−2;(e) E[X] = 4E[N] − 2 = 1025. (a) E[W ] = np + np 2 ; (b) E[W ] = 1 + p( 1 p ) = 237. (a) E[X] = 3; (b) V ar(X) = 3.106742. (a) E[X] = 5.5; (b) V ar(X) = 38.544. V ar = E[N]V ar(X) + E[X] 2 V ar(N) = 3333355. (a) E[X] = 5.5; (b) P (you are nth)= E( 1 N) where N ∼Poisson(10)Ch 4:3. P rrr,rrr =P ddd,ddd =0.8, P rrd,rdd =P rdr,drr =P rdd,ddd =P drr,rrr =P drd,rdd =P ddr,drr = 0.6,P rrd,rdr =P rdr,drd =P rdd,ddr =P drr,rrd =P drd,rdr =P ddr,drd = 0.4, P rrr,rrd =P ddd,ddr = 0.25. E[X 3 ] = 1 4 (47/108) + 1 4 (11/27) + 1 2 (13/36)10. this probability is 1 − P 3 0,218. (a) the proportion is π 1 = 5/9; (b) P 4 1,2 = 0.444419. r 0 = 1/4, r 1 = 3/20, r 2 = 3/20, r 3 = 9/20, so r 0 + r 1 = 2/529. π 1 = 6/17, π 2 = 7/17, π 3 = 4/1752. π A = 3/7, π B = 4/7, so E[X] = 3 7 (0.6 · 6 + 0.4 · 12) + 4 7(0.3 · 12 + 0.7 · 8) = 62/753.23 (326.4) + 1 3 (200π 1 1 4 + 250π 2 1 4 + 400π 3 1 4 + 600π 4 1 4 )1

Ch 5:4. (a) 0; (b) 1/27; (c) 1/46.(λ1λ 1 +λ 2) 2+() 2 λ2λ 1 +λ 2( )9. 1 − e −λ1t + e −λ 1t λ 1λ 1 +λ 215. E[T ] = 200100 + 20099 + 20098 + 20097 + 20096V ar[T ] = ( 200100 )2 + ( 20099 )2 + ( 20098 )2 + ( 20097 )2 + ( 20096 )241.λ 1λ 1 +λ 242. (a) E[S 4 ] = 4 λ ; (b) E[S 4|N(1) = 2] = 1 + 2 λ; (c) E[N(4) − N(2)|N(1) = 3] = 2λ50. (a) E[X] = E[7T ] = 7/2; (b) V ar[X] = 7E[T ] + 49V ar[T ] = 91/1260. (a) 1/9; (b) 5/9Ch 6:4. N(t) = # of customers in station at time t= birth-death process with λ n = λα n , µ n = µ6. E[T 0 ] = 1 λ , E[T 1] = 12λ + µ 2λ E[T 0], E[T 2 ] = 13λ + 2µ3λ E[T 1], E[T 3 ] = 14λ + 3µ4λ E[T 2],E[T 4 ] = 15λ + 4µ5λ E[T 3]; (a) E[T 0 ] + E[T 1 ] + E[T 2 ] + E[T 3 ]; (b) E[T 2 ] + E[T 3 ] + E[T 4 ]13. π 1 = 3 4 π 0, π 2 = 3 4 π 1 = 3 24 π0 , so π 0 = 1637(a) π 1 + 2π 2 = 3037 ; (b) 1 − π 2 = 28(c) now ˜π 0 = 6497 , so 1 − ˜π 2 = 888897, so improvement adds λ(97 − 2837 ) = 0.4514. π 1 = 5 3 π 0, π 2 = 5 3 π 1, π 3 = 5 3 π 2, so π 0 = 27272(a) 1 − π 0 = 245272 ; (b) π 3 = 12527215. π 1 = 3 2 π 0, π 2 = 3 4 π 1, π 3 = 3 4 π 2, so π 0 = 32143(a) 1−π 3 = 116143 ; (b) ˜π 0 = 64175 , 1− ˜π 3 = 148175(18. π i = λ µ iπ 0 , so π 0 = 1 + ∑ )k λi=1 µ i(a) π i ; (b) π 023. π 1 = 12 5 π 0, π 2 = 4 5 π 1, π 3 = 410 π 2, so π 0 = 2501522(a) π 1 + 2π 2 + 3π 3 ; (b) π 2 + π 337 ;2