This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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. ...
View
Full
Document
This note was uploaded on 02/20/2010 for the course STAT 3022 taught by Professor Yanpingqu during the Spring '10 term at University of Minnesota Duluth.
 Spring '10
 YanpingQu

Click to edit the document details