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Unformatted text preview: r . (a) All are equally likely; the prnhﬂhility is f3it. (b) Because I3 sluts are red, the
pmbabilitv of a red is Hard) = g ‘= 0.4?4. (c) There are 12 winning slots. so thin a
Iumn bet) = ﬁ i 0.316. I133. {11] There an: 10'4 = 10.000 pussihlc PINS {0000 through 9999)? (b) The probabilin that
a PIN has no 05 is 0.9“ (because then: are 9“ PINS that can he made from the nine nonzero
digits]. so the probability of at least one 0 is l — 0.9“ = 0.3439. *If we assume that PINs cannot have leading Us. then there are only 9000 ptissthe codes (1000—9999). and the probability of at least one 0 is 1 _ 9% = 0.271. 4.35. If we assume that each site is independent of the others (and that they can be considered
as a random sample from the collection 01' sites referenced in scientiﬁc Journals). then Hall seven are still good) = 0.13.1r1 i Item. 4.62. {a} P1X 3 0.351 = 0.65. (11) Pt): =0.351= 0. {c} P1035 e X e 1.35; =
H035 < X <11: 0.65. (d) P10.l5 5 X s 0.25m0.s 5 X 50.91: 0.1 +0.1: 0.2.
{e} Pine: [0.3 5 X i: 0311: 1— 1310.3 5 X «5 0.?}=1— 0.4 = 0.5. l. (1) Histogram on the right. (bi "At least one noanrd
emf“ is the event “X 3 1"(dr “X > 0“}. P[X =_ 1) =
1— PM = 0) = 0.9. {cl “X 5 2" is “no mere than two 0.3
mnwnrd cn'nrs.” or "fewer than three nonword ermrs." “'2
P(X52)=U.T=PI:X=U}+P(X=l)+P(X=2] m I
=0.1+0.3+0.3 I 
P{X«:2}=0.4=P(X=UJ+P[X=l}=ﬂ.l+0.3 00
ﬂ 1 2 3 Il 463. {at} The height shhuld he % since the area under . ,
the curve must he I. The density curve is at the right. (h) P0” 51.5}: 3—5 : 0.75.10 P(0.6 q: r e. 131: 4%”. I
% =0.55. {d} PC? 3 0.91 = % = 0.55, a 0.5 ‘l 1.5 2 4.109. (31"The vehicle is alight I‘— pm) _ 031— —PM—r)‘_0 59
"Wk = A i PM 3 = 069 PIS} :02: J ethane 31 :00: Pretender =0. (1)) “The vehicle is an imported pm“
' l= 0.?8 PEA and. B: 211.33 P ‘ r 2
car‘ = Aand B. To ﬁnd this I ' (A and B l o prnbahility. note that we have been given P(8"} = 0.18 and PM‘ and 3"‘1 = 0.55. F
this we can determine that 73% — 55% = 23% of vehicles snld were domestic cars—that
PM and 3‘12023—50 PM and B} = PIA)  PMand B“) = 0.31— 0.23 = 0.08. Note: The tabfe shown here summarizes all the
mutton (bold). r we can determine from the given 1' 4.111. See also the solution to Exercise 4 109 e ' ' ' ' '
{a} PU“  B} = ﬁrmm m _ M L   SPCCIally the table of probabilities given there. _ I Pm] — _ “In — 0.6364. {b} The events A‘ and B are nor independent;
if they were, PM‘  B) wedld be the same as PM") = 0.69. HI. With E = {LBW +0.21”. we have me = Uﬁnw + 0.2,uy =i]16% and:
r on = (0.3m)? + {0.2m}? +2pwr{0.3my){ﬂ.20rl a 15.929193: 4.136. (a) The. mean is m; = (INDIA) + (1.5){D2J + (2}(0.2) + {41(0J) + {101ml} = 2.5
million dollars. The variance is: ' a; = (1' — 2.5)2(0.4) + (1.5 — 2.512(02} + {2 — 2.519102;
+ (4 — 2.532(0.” + no — 2.3%.”
= :r'
The standard deviation is ox = ﬂ 2 2.6453 million dollars. {h} m, = mm , m =
0.9,!” — 0.2 = 2.05 million dollars. and cry = Golgxruz = 0.9ax =' 2.3312 million dollars 1.126. Tonya's score standardizes to z = W '= 1.4067, while Jermaine's score _ lg . T . .
corresponds to z = w 2 LS. Jarmalne s scare 15 hlgher. ...
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 Spring '10
 YanpingQu
 Standard Deviation, Variance, Probability theory, alight I‘— pm, pussihlc PINS, ptissthe codes

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