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20061ee131A_1_131a9sol_06

20061ee131A_1_131a9sol_06 - EE 131A 3.84 3.86 Homework#9...

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Unformatted text preview: EE 131A 3.84 3.86 Homework #9 Solution (0 - €)2/€ 10.5. 10.5 17.36 6.88 20.02 5.36 44.02 1.93 6.88 6.88 D = 130.33 Expected, (0 —— €)2/8 9.01 9.43 10.74 » 7.81 77.41 0.27 9.43 9.93 ’ 83.26 Observed N k Obs. Expected 0 0 10.5 1 0 10.5 2 24 10.5 3 2 10.5 4 25 10.5 5 3 10.5 6 32 ' 10.5 7 15 10.5 8 2 10.5 9 2 10.5 105 Obs. 2 24 105/8 .3 2 105/8 4 25 .105/8 5 3 105/8 6 32 105/8 7 15 105/8 8 2 105/8 9 2 105/8 105 k 1 25 2 6 3 19 4 16 5 10 6 20 # degrees of freedom = 9 1% significance level => 21.7 D2 > 21.7 => Reject hypothesis ' that #’s are unif. dist. in {0,1,...,9} # degrees of freedom = 9 1% significance level => 21.7 D2 > 21.7 => Reject hypothesis that #’s are" ’ unif. dist. in {0,1,...,9} Expected mk (N;c — mk)2/mk 16 16 16 16 16 16 81/16 100/16 9/16 0/16 36/16 16/16 D2 = 2m. —— mgr/m,c = 242/16 = 15.125 > 11.07 I: => Reject hypothesis. Winter 2006 K. Yao 2.84 Three tosses of a fair coin result in eight equiprobable outcomes: 000 —> 0 100 ——> 4 001 ——> 1 101 —~> 5 010 —+ 2 101 011 _} 3 111 } —> No output a) P[a number is output in step 1] = 1 ~— P[no output] ‘ _ _ Z _ 3 — 8 “ 4 b) Let A; = {output number i} i: 0, ...,5 and B = {a number is output in step 1} then P[A1- D B] P[binary string corresponds to i] PlAilBl = —P‘B— = _*—‘*§——*~—— [ l 4 Z 6 c) Suppose we want an urn experiment with N equiprobable outcomes. Let n be the smallest integer such that 2” Z N. We can simulate the urn experiment by tossing a fair coin n times and outputting a number when the binary string for the numbers 0, ..., N —— 1 occur and not outputting a number otherwise. 5.15 P ”iv—£9 — ‘2 a] : P[|N(t) — At] 2 at] S W by Chebyshev Inq. At /\ l l 1 1 5.17 M100 = mug + + X100) = @5100 1+2+m 6 ,u -— 8[X]= 6 + 23.5 1 a; = E(1+ 32 + 32 + 42 + 52 + 62) — (3.5)2 = 2.91667 P[300 < 5100 < 400] = [3 < % < 4] Pl‘-5 < M100 - 3.5 < .5] PllMioo — .35] < .5] 2% . 1_-100(%)2 =(18832 ll IV 5.22 The relevant parameters are n = 1000, m = np 2: 500, 02 :: npq = 250. The Central Limit Theorem then gives: 400—5003N~m£600—500 V 250 0 V 250 HwogNg6m]= P[ 22 Qp6amy—Qw3%g=1—2Qw3mq = 1—2Murw) P[500 s N s 550] Q(0) — Q(3.162) = %— 7.300%) 22 526 fl&] VARmd n8[X¢]=n-1=n nai=n~1z=n I Assuming Sn approximately Gaussian: Pwn>wy=Pfigf>JiifiJ~Q(w_")zaw From Table 3.4 => n — 2.3263\/fl— 15 = 0 => n = 27.04 => #y 28 pens A ...
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