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Unformatted text preview: 12. 15. 17. In my perusal of a zip code directmy, 1 found no 00000, nor did I ﬁnd any zip codes with four zeros, a fact
which was not obvious. Thus possible X values are 2, 3, 4, 5 (and not 0 or 1). As examples, X: 5 for the
outcome 15213, X = 4 for the outcome 44014, and X = 3 for 90022. No. In the experiment in which a coin is tossed repeatedly until a H results, let Y: 1 if the experiment
terminates with at most 5 tosses and Y = 0 otherwise. The sample space is inﬁnite, yet Yhas only two
possible values. See the back of the book for another example. The least possible value of Y is 3; all possible values of Yare 3, 4, 5, 6, .... Y= 3: SSS; Y= 4: FSSS; Y= 5: FFSSS, SFSSS; Y = 6: SSFSSS, SFFSSS, FSFSSS, FFFSSS; Y= 7: SSFFSSS, SFSFSSS, SFFFSSS FSSFSSS, FSFFSSS, FFSFSSS, FFFFSSS a. Since there are 50 seats, the flight will accommodate all ticketed passengers who show up as long as
there are no more than 50. P(Yg 50) = .05 + .10 + .12 + .14 + .25 + .17 = .83. b. This is the complement ofpart a: P(Y> 50)=1—P(YS 50) = 1 — .83 = .17. c. If you‘re the ﬁrst standby passenger, you need no more than 49 people to show up (so that there’s space left for you). P075 49) = .05 + .10 + .12 + .14 + .25 = .66. On the other hand, ifyou’re third on
the standby list, you need no more than 47 people to show up (so that, even with the two standby passengers ahead ofyou, there’s still room). P(Y<_Z 47) = .05 + .10 + .12 = .27. a. (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5) b. Xcan only take on the values 0, 1, 2.p(0) =P(X= 0) =P({(3,4) (3,5) (4,5)}) = 31'10 = .3;
p(2) = P(X= 2) =P({(1,2)}) = 1f10 = .1;p(1) = P(X=1)=1— [p(0) +p(2)] = .60; and otherwisep(x)
= 0. c. F(0) =P(X£ 0) = P(X= 0) = .30;
F(1) =P(Xs 1) = P(X= 0 or 1) = .30 + .60 = .90,
F(2) =P(X£ 2) = 1.
Therefore, the complete cdf of X is 0 x<0
.30 0£x<1 .90 1£x<2
1 2£x a 19(2) = P(Y= 2) = P(ﬁrst 2 battelies are acceptable) = P(AA) = (.9)(.9) = .81. a b. p(3) = P(Y= 3) = P(UAA orAUA) = (.1)(.9)2 + (.1)(.9)2 = 2[(.1)(.9)2] = .162. c. The ﬁfth battery must be an A, and exactly one of the ﬁrst four must also be anA.
Thus, p(5) = P(AUUUA or UA WA or UUA UA 01' UUUAA) = 4[(.1)3(.9)2] = .00324. 11. 1901) = P(theytll is am! and so is exactly one ofthe ﬁrsty — 1) = (v — 1)(.1)rH(.9)2, fory = 2, 3, 4, 5, .. .. 21. 29. 33. 36. a. First, 1 + 1.1x > 1 for all x = 1, .. ., 9, so log(1 + lfx) ) 0. Next, check that the probabilities sum to 1: 9 9 ' Zlogm{1+1fx) = 2 log“, = logIn + log“,    + logm using properties of logs,
F1 F1 I this equals logm [%><%><    x?) = log10(10) = 1. b. Using the formula p{x) = log10(1 + 10:) gives the following values: p{1) = .301, 12(2) = .176, p(3) =
.125,p{4) = .097,p(5) = .079, J17(6) = 067,110) = .058,p(8) = .051,p(9) = .046. The distribution
speciﬁed by Benford’s Law is n_ot uniform on these nine digits; rather, lower digits (such as 1 and 2)
are much more likely to be the lead digit of a number than higher digits (such as 8 and 9). c. The jumps in F(x)occ1u' at 0, . .. _. 8. We display the c1unulative probabilities here: 17(1) = .301, F0!) =
477,178)=.602,F(4)=.699,F(5)=.778,F1{6)=.845,F(7)=.903,F(8)=.954,F(9)= 1. So, F(x) =
0 forx< 1;F(x) = .301 for 1 3x<2;
F00 = .47? for 2 5 x < 3; etc. a. P(X:I 3) =F(3) = .602: P(X2 5)=1—P(X< 5)=1—P(X5 4) = 1 —F(4) = 1 — .699 = .301. a. EOE) = prﬁc) = 111.05) + 2(.10) + 411.35) + 8(.40) + 16(.10) = 6.45 GB.
3111 b. V(X) = 20: — ,u)2p(x)={1— 6.45)2(.05) + (2 —6.45)2(.10)+ ... + (16 —6.45)2(.10)= 15.6475.
all: c. 6: "iron = 15.6475 = 3.956 GB. a. E(X2)=Zx’p(x)—12(.05) : 22(10) : 42(35) : 82(40) : 162(.10)=57.25.Using the shortcut
aux formula, mo — E00) ,02 — 57.25 (6.45)]2 — 15.6475. 51. EEK?) = 2x2 p(x) = 02(1—p)+ 12(1)) =p.
x=0 b I’(X)=EIz‘ifg)—[E{JY)]2 =P [P]2=P(1P) c. E0759) = 0T9(1 —p) + 119(1)) =p. In fact, E(X') = p for any nonnegative power 10. You have to be careful here: if $0 damage is incurred, then there’s no deductible for the insured driver to
pay? Heres one approach: let MK) = the amount paid by the insurance company on an accident claim,
which is $0 for a “no damage” event and $500 less than actual damages (X— 500) otherwise. The pmf of
3190 looks like this: Based on the pmf, the average payout across these types of accidents is 1301(ij = OLE) + 5001;. 1) +
4500(08) + 9500(02) = $600. If the insurance company charged $600 per client, they’d break even (a bad ideal). To have an expected proﬁt of $100 — that is, to have a mean proﬁt of $100 per client — they
should charge $600 + $100 = $700. 3'8. 31. 46. 57. (113.5) = $235, while E[h(X)] =E[i] = p(x) = 1 i1 = $403. X x=1 x=1 x 6 F1 x
So you expect to win more if you gamble. Note: In general, if Mac) is concave up then E[h(XJ] ) h{E(X)), while the opposite is tme if h(x) is concave
down. Use the hint: WaX+ b): E[((aX+b) E(aX 1 b))2] — Z[ax 1 b E(aX 1 b)]2p{x) = ZIGJH b—{aﬂ + 33121005) = Z[M—aﬂ]zp{x) = “ZZG’C ﬂfpix) = anliXl 3
a. b(3;3,.35) = [3](.35)3(.65)5 = .279.
a
b. b(5;8,.6) = [5](.6)5{.4)3= .279.
c. P(3 EXE 5) = b(3;7,.6) + 5019.5) +b(5;7_..6) = .745. 11. 13(13):) =1—P(X= 0)=1—[:](.1)°(.9)9 =1—(.9)9 = .613. LetXbe the number of “seconds,” sz~ Bi11{6, .10).
n _ '6 1 5
a. P(X= 1) = p‘n — p)" I = 1 {. 1) (.9) = .3543.
x
6 6
b. P(X2 2) = 1 — [P{X= 0) + P(X= 1)] = 1 — [[0](.1)°(.9)6 +{1](.1)‘(.9)5] = 1 — [.5314 + .3543] =
.1143.
1:. Either 4 or 5 goblets must be selected.
'4
Select 4 goblets with zero defects: P(X= 0) = [0](.1)°(.9)4 = .6561 .
4
Select 4 goblets, one of which has a defect, and the 5‘'1 is good: I}. DI (9)3 : x .9 = .26244 So, the desired probability is .6561 + .26244 = .91854. 52. Let): be the number of students who want a new copy, so X ~ Bin(n = 25, p = .3). a. b. d. E(X) = up = 25(3) = 2.5 and SD{X) = Japﬂ—p) = .[25(.3)(.7) = 2.29. Two standard deviations from the mean converts to 7.5 :: 2(2.29) = 2.92 & 12.08. For X to be more
than two standard deviations from the means requires X 4 2.92 or X 3 12.08. Since X must be a non negative integer, P(X< 2.92 or X? 12.08) = 1 —P{2.92 5X5 12.08) = 1—P(3 5X5 12) =
12 F3 If X i> 15, then more people want new copies than the bookstore canies. At the other end, though,
there are 25 —X students wanting used copies; if 25 —X 2> 151 then there aren’t enough used copies to
meet demand. The inequality 25 —X2> 15 is the same as X < 10, so the bookstore can’t meet demand if eitherX 2> 15 orX< 10. A1125 students get the type they want iﬁ 10 5 X5 15:
15 P(10£X:1 15) = Z[;:](.3)‘(.7)25“ = .1890. x=10 The bookstore sells X new books and 25 —X used books, so total revenue from these 25 sales is given
by hp!) = 100(X) + 70(25 —X) = 30X+ 1150. Using linearityx’rescaljng properties, expected revenue
equals E(h(X)) — E{30Xt 1250) — 30p! : 17'50 — 30(75) + 1150 = $19?5. ...
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This note was uploaded on 01/20/2012 for the course STAT 511 taught by Professor Bud during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 BUD
 Statistics

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