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stat400sln5 - 3.4—4(a f:z)[email protected]—K 1 z:1,2,3 365 E g 1 1 ...

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Unformatted text preview: 3.4—4 (a) f(:z:)= (@f—K 1 ) z:1,2,3,..., 365 E g 1 1 § i (b) [J = T = 365, 1 3—65 g 364 1 ! 02 = —§§§.§ = 132,860, ; 1 3 a = 364.500; J I 400 1 1 364 ‘ 364 299 > P(X < 300) _ 1 — (E) 795591 ‘ 3.4-10 (a) Negative binomial with r 2 10,12 2 0.6 so ’ NJ- 10 10(0.40) = — = 1 . 7, 2 z = 11.111, = 3.333; " 0.60 6 66 ‘7 ((160)2 " 1 . , (b) P(X : 16) : ( 95) (0.60)1°(0.40)'3 = 0.12406 . Ai;:”m X"e"‘ N «A W ,. A N 3.5-4 _ 3 1! = 2! i 'e-*,\(/\—6) = 0 Thus P(X = 4) = 0.285 — 0.151 = 0.134. 3.5-6 A = (1)(50/100) = 0.5, so P(X = 0) = e-0-5/0! = 0.607.} g 3.5-8 np : 1000(0.005) :- 5; (‘9 P(X S 1) “0.040; 1 .(b) PM : i1? 5? Q: TX 5 6) ‘ (X S 3) w 0762 41265 = 06497 4.1—2 (a) (i) f0”x3/4d:c : 1 "" “m ‘ c4/16 : 1 ' c = 2; (ii) 1%) = 1:00 f(t)dt a ‘ 1, fix} HI) , 2.0 15 1.5 1.0 ‘ 1.0 05 0.4 0.8 1.2 1.6 2.0 0.4 0.8 1.2 1.8 2.0 Figure 4.1—2: (a) Continuous distribution p.d.f. and c.d.f. r< 7,, W (b) (i) [few/1mm = 1 I ‘ 63/8 = 1 ' E c = 2; i (a) F(:c) = ffwfgtwt = ff2(3/16)t2dt 1 t3 , = [1—4-2 _ £3 1 g ’ _ 15 2’ l ‘ 0, —oo<z<—2, \ 3 1 F(z)= ¥6+§, —2g:x:<2, Fa) :- (b)’:’Co’_Ix1t;1‘1"1fiouskdistribfifion p.d.f. and c.d.f. (c) (i) /1%dx = 1 o :12 20:1 “7 1 = k dt f. m z = M] = ., D 0, —oo < a: < 0, F05): x/E, 0<z<1, 17 1 < I < m fix} Fa) 2.0 240 1.5, x 1.5 1.0 0.5 0.2 0.4 0.6 0.8 1 12" —o.2 ‘ v‘ 032 0,4 Figure 4.1—2: (c) Continuous distribution p.d.f. and c.d.f. ‘4.1—4 (a) fizzy“) = /2£:dz , ‘ 0 0.8 07?: 1.0 (C) (b) M=E(X) : /(i)z3dz 02 = VHI(X) II N m w /“\ ale» \_/ at E II no Cal“ 8 0! L—J I N M II I + l I #21500 = fid" 0 2 x 1 A x/5 - [)de _ [ms/2r 1 3 0‘5’ 1 2 02=V X :/(_1 L ad) 0 z 3 2 audit ...
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