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**Unformatted text preview: **41'. 43. 5'0. X ~ Eiﬂ{6, .10} n. _ .5.
a- Pix= n= [Jtmxa—p)" * =[ 1](.1}1(.9}5 = .3543 11- PKX21J=l-[P{X=UJ+P{X=1}]- _
mee.weleeewP{X = 1}: .3543,aed P{X= s)=[:J(. I)“ (.9115 = .5314.
HeneeP(xe 2): 1 — [.3543+.5314]=.1143 I c. Either 4 or 5 geblets must be selected
1} Select4 goblets with zero defects: P0: = n} = [:](.1)“{.9)4 = .6561.
ii) Select 4 geblets, one of which has a defect, and the 5th is good: [[:](.1)1(.9)3]x.9 = 25244 So the desired probability is .6561 + 26244 = .91354 Let S = comes to a cumplete stop, so p = .25 , n = E! a. PIEX E 15] = B(ﬁ;2ﬂ,.25] = .Tr'Eﬁ b. P(X = ti] = hﬁﬁﬁﬂ,.2ﬂ) = Bﬂﬁﬂﬂ,.25} — Bfﬁ;2ﬂ,.25} = .TEﬁ — .ﬁl? = .lISB
c. PIE-{2(3):l—P[X£5}=l—B(5;20,.25)=1—.61?=.333 d. EOE} = [2ﬂ]{.25} = 5. We expect 5 of the next 2D to step. X~Bh1(lﬂ, .150}
a. P-(X eeJ=1—P{xs 5}: 1 —B(5;2ﬂ,.6ﬂ}= 1 — .351: .533 b- EJEXJ=ns=ilﬂII(-ﬁ)=ﬁ; VDQ=np(l—pl=(lﬂK-61I(-4)=2.4;
ox: 1.55 E{X)i5,=(4.45,155).
We desire P{5 £XE?}=P{XE?}—P{XE4}=.333 —.lﬁﬁ= .56? c. HEEXSTJ=P{X5?]—P(X£EJ=.333—.012=.321 75.
a. P{X 5 3] = F{s;5} = .932 b. P(X = s} = F{3;5) — Etta) = .ﬂﬁS e. mem=1—P{x5s}=ses d. H5 5 x. 5 3) = H35} — F{4;5} = .492 e. P(5 5 x 5 3) = F05} — 151:5) = .SﬁT—.ﬁlﬁ=.251 b. P{X22]=l—P[X£ 1)=1—F(1;2}=1—.932=.ols e. PI[15t doesn’t r1 2nd doesn't} = PI:1=it doesn’t} -P[2’1'i doesn't} a. For a two hour period the parameter of the distribution is 11 = (4)12} = E, b. For a 30 minute period, it = (43(5) = 2, so P{X=ﬂ)=F(ﬂ,2) = .135 73. a_ P{X=l}=F{l;2)—F(ﬂ;1}=-932'-319=-1ﬁ3
= (.319){519) = .671
33.
so p0: = m} = F(1ﬂ;3} — FEES) = 1399-
; E{X)=Iu:=2
1. 1 2]”
II 5 5
xdx=112 3. Graph of ﬁx) = .093?5{4 — :2) I15 {Li 2
2 3
1:. Pp: H1) = L .093?5(4 — x2 )dx = .ﬂ93?5(41 3%)] = .5
U 1
c. pm a: x: 1) = LﬂQSTS-{r—l —x? )dx = .58?5 .5
d. Fiﬁ—.5 ﬂaws): 1—P(-_5 5X5 .5]: 1 — L.D93?5(4— fjdx
=1 — .36?2=_ﬁ323 a. ﬁﬂxﬁmx: fie—“Wax: —e"‘2”92t = 0—H) = 1 Elm 2m: 2 2
1:. malt-0F L: J‘"{I;I5‘)-::I'x=Iﬂ Giza" ”‘9 dx
2W
=—+e"‘j”'5‘”]tl m—.1353+1=.364? P(X *4 20D) = PIX 5 20(1) Rs 364?, since 1; is continuous.
mama}: 1 magma .1353 alumni: 1“”
e = L DD
(1. H1m5x52n0)= mf(x;9)dx=— m, 2.4?12 ﬂ. me}O,P{X5x)=
r f(y;9}dy = ”ElEE—ylugitﬁ = _g‘J’2’292t =1_e—x11‘292
4: e ...

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- Fall '04
- Eisenstein