chapter 4

Prévia do material em texto

CHAPTER<! 
00 I"' 4·>. a) P(l <X)= J e- xdx = (- e- x) 
1 
= e- 1 = 0.3679 
I 
15 2 . 
b) P(l < X < 2.5) = J e- xdx = (-e-x)~ -> = e- 1- e-2-5 = 0.2858 
I I 
J 
c) P(X = 3) = J e- xdx = O 
'J 
00 11>0 0 P(x < X) = J ,- x dx = (- e- x) x = , - x = 0.1 0 
X 
T~.en.x = - ln(0 .10) = ~,3. 
X IX &l P(Xsx) = J e- xdx = (- e-x) =l- e- x = O.lO 
0 0 
T~.en.x = -ln(0.9) = 0 .1054 
0 0 
4 ·:). a) P(X < 0) = J 0.5cosxdx = (0.5sinx)l = 0- (- 0.5) = 0.5 
- -:r/ 2 
- o/2 
- -:r/ 4 / 4 
bl P(X < - -rrl 4) = J 0.5cosxclx = (0.5sinx)l-" =-0.3536 - (- 0.5) = 0. 1464 
- ':r/2 
- -:r/ 2 
<:r/4 
J [. /4 c) P( - -rr I 4 < X < -rr 14) = 0.5 cosxdx = (0.5 sin x) = 0.3536 - ( - 0.3536) = 0. 7072 <:r/ 4 
- -:r/4 
<:r/2 12 
dl P(X > - -rr I 4) = J 0.5 cosxdx = (0.5 sin x >I" = 0.5 - ( - 0.3536) = 0.8536 
- -:r/ 4 
- -:r/4 
X 
•l P(X < x) = J 0.5cosxdx = (0.5sinx)lx = (0.5sinx) - (- 0.5) = 0.95 
- ':r/2 
- -:r/2 
T~.en.sinx = O.<).and x = l .n (}.l) radians. 
4 4 2 42 "2 Jx x - J 4·5- 8) P(X < 4) = - dx=- = = 0.4375.~>=,..fx(xl=Oforxq 
A 8 16 16 
' 3 
4 4 
Jx x2 42 - 32 b) P(X< 4)= - dx=- = = 0.437) bilca,..fx(xl=Oforx >; 
' 8 16 16 
' 3 
5 x x2[ 52 - 42 
c) P(4<X<5) = J - dx=- = = 0.5625 
4 8 16 16 
4.5 2 .5 2 2 
J X X 4.5 - J . -- 0.-IO'J 1 dl P(X< 4.5) = - dx=- =---38 163 16 
•l P(X> 4.5)+P(X<3.5) = - d,- + - dx = + - = . + J. J = 0.5 J
s x J3.S,. x2[ x2 ;.> 52 - 4 52 '52 - ,2 
4.5 8 3 8 16 .5 16 J 16 16 
4·7. a) l{o <X)=O,s.bys)mmetry. 
I I 
b) P(0.5<X) = J 1.5x2dx = 0.5x31 = 0.5- 0.0625= 0.4375 
0.5 
0.5 
0.5 0.5 
cl P(- 0.5sXs0.5) = J 1.5x2dx = 0.5x31 = 0.125 
- 0.5 - 0.5 
d) P(X < -:l)=O 
e) P(X < o or X>- o,s) = 1 
0 P(x <X) = J 1.5x2dx = 0.5x31: = 0.5 - 0.5x3 = 0.05 
X 
Theo.x = O:~.;s 
50.25 50.25 
4-9. 8) P(X >50) = J 1.0dx = b t = 0.5 
50 
50.25 Sil.25 
b) P(X>x) = 0.90= J 2.0dx = 2xlx =100.5- 2x 
X 
Then,~= <)9.6andx = 49.8 
4-11. a) P{X < 2.2,5 or X> 2.7.;) = P(X < ~.:!.5)- P{X > 2.7.;) boollnse the two t'\'ttll$ate mutuaJiyexclusiw:. n.en. 
2.8 
P(X < :l.2..j) = 0 and P(X > :L75) = J 2dx = 2{0 .05) = 0 .10 
2.15 
b) If the probability d-ensity function isoeotcri!dst :l,s.; mcters, thenfx(x) = :l for :l.3 < x < :l.Sandall rods "'ill meet specificatioll$. 
4·lj. a) P(X < 2.8) = P{X s -2.8) bee!! use X iss cnntin.-ollS random variable. 
Tr.en. P(X < ~.8) = A,i.S) = <U{~.S) = o..;6. 
b) RX> 1-5) = 1-RX ~ 1-5) = l-O.i ( lS) = 0 .7 
c) RX < -~) = Fx{-i) = o 
d) IV<>6)=l-FX(6)=0 
4·17. NowJ{x) = o.:; 006Xfot- rt/2 < x < st/~and 
X 
Fx (x) = J 0.5cos udu = (0.5sinx)lx = 0.5sinx + 0.5 
- .;rf2 
- .;rf2, 
O,x5 - 1T 12 
Tlten. 
Fx(x) = 0.5sin x + 0.5, - -rr / 2 5 x < -rr / 2 . 
I , x ?.T. / 1 
0 ., ,x< J 
foro< x. Th.en. 
, .2 - 9 
' 3< x <5. 
16 ' -
l,x> 5 
X 
4·~l . Now.ftx)=2fori<.J<X < 2.8and F (x) = JUly= 2x -4.6 
for i,l < x < 2.8. TM:n. 
0, ~· <~ 2.3 
F (x) = 2x -4.6, 2.3s x <2.8. 
1, 2.Ssx 
2.3 
RX > ~.7) = 1-FV<:;. i .7) = 1- R:2.7) = 1-0.8 = O.i becatse X is a continuous random \"ariable. 
4·:!;. 
4·'3· F(>')-J 0.5xdx = 0.5x 2 = 0.25x2 foro <x < , ,Thtn. X [ 
0 2 
0. x<O 
O<x<2. 
I. 
/(x)- 0.2. 0< .r< 4 
0.04. 4 < x< 9 
. , . 
H7. £(X)= J 0.25xdx = 0.25: -2 
0 - 0 
4·29. 
I 4 I 
£(X)= J 1.5x3dx = 1.5 ·: -0 
- 1 -1 
I I 
I"(X) = JL5x3(x - 0)2dr =J.5J x 4dx 
- 1 -1 
X; L 
= 1.5- = 0.6 
5 I 
4-3 1. 8 2 [ - ']
8 
J 3(8x - x ) x ' Jx E (X) = x dx = --- = (16 -12) - 0 = 4 0 256 32 1024 0 
V(X) = J <x -4/> x- x dx = J - x +_:_-~+ >X x 8 '(8 2) 8 l 3 • 3 , I_ 2 _ r 
0 256 0 2 56 16 16 2 
8 
V(X)=[-Jx5 +Jx' _ 5x3 +3x2 ] = [ - 384+ 192 -160+48]= 3.2 
. 1280 64 16 4 5 
0 
4-35. a) 
1!20 600 1120 E(X) = x-'-'-'-dx = 600 1n x =109.39 
x2- 100 100 
V(X) = J (x -I 09.39)2 600 dx = 600J 1- 2(109.39) +(I 09.39)2 dx 
100 x2 JOO x x2 
= 600(x - 2 18.781nx -109.392x - 1) = 33.19 1
120 
100 
00 
• ·J7. •> E(X)= J x iOe-iO(x- S)dx. 
s -to(r-sl 
Using integNtion bypsrts "'ith tt = x aodd11 = 10e dx, we obtain 
l
oo ~ -IO{x- 5) r 
£(X) = - xe-IO(x- 5) + J , -IO(x- 5)dx = 5- e = 5. 1. 
5 5 I 0 5 
«> 
Now, Jl (X) = J (x - 5.J)2 J Oe-IO(x- 5) d\· , Using th.e integtation byparts,-.itb 11 = (x-.; .1)2 and:dv = lOt-lO{X-s). \l.'tobtaio 
5 ·I"' "" V (X) = - (x - 5.1)2 e -lO(x - S) + 2 J (x - 5. 1)e -IO(x- 5) dx. From tf!.cdefini tionof~X)theintegral abo\'t' istt."OOgni7.«i to equal o. 
' 5 5 Therefore. V{X) = {5-.;.1) = 0 .01. 
06 "" 
b) P(X > 5. 1) = J I Oe -iO(x- 5)dx = - e -IO{x- 5)1 = e -10{5.1-5) = 0.3679 
5 ~ I .I 
4-39· a) E.OO = {- 1 • l)/ 2 = o. 
2 
V(X) = (l- (-l)) 1/J,andox=O.;n 
12 
X I X 
b) P(- x < X < x) = J - dt = 0.511 = 0.5(2x)= x 
2 - x 
- x 
Therefore. x slt.onld eqooJ o .90. 
c) 
F(x) = 
0, 
0.5x +0.5, 
I' 
x < -1 
-l ::; x< l 
I ~ x 
4-41. a) TltedistributionofXisj(x) = u) for i'L9.S < x < 1.0.5. Noh·, 
0, x <0.95 
Fx(x) = I Ox - 9.5, 0.95~ x< 1.05 
I' 1.05 ~X 
b) JV<>l.02) = 1- JV( Sl.02) = l- Fx(l.02) = 0~ 
<:) lt P(X > >:) = 0.91), th.eo 1- HX) = 0 .90 and F{X) = 0 .10. Th.erefote,lOX- 9..; = 0 .10 and>: = 0.¢ . 
(1.05- 0.95)2 
d) E.{X) = (1.0.5 • 0.9.:;)/ 2 = l .OO and V(X) = = 0.00083 
12 
4-43. a) ThedistribntionofXisj(x) = 100 for o.~o.so < x < o.:Hso. Therefore. 
0, x <0.2050 
F(x) = IOOx- 20.50, 0.2050~x< 0.2 1 50. 
I, 0.2 1 50 ~ x 
b) RX > 0 .212.5) = l- 1\0.212.5) = l-l t00(0.2li.S)-20..;o J = 0 .2.5 
<:) lfP(X > x) = 0 .10, then 1- ftX) = 0 .10 and F{X) = 0 .91). Therefore,lOOX - 20..;0 = 0.90 and>:= 0 .2140. 
? (0.2150- 0.2050t -6 • 
d) E.{X)= (O.iO.SO -0.21.50)/ 2 =0.2.100¥tnandV(X)= ? =8j$Xl0 Jlffi 
I_ 
4--45. (a) l.ct X be t~.etime(in minutes) bef\,·~narri\'<11 and 8:;}0 am. 
I j(x) = - . f or 0 s x!: 91). 
90 X 
SotlteCOFis}{x) = - .joJ'O ::x s 9Q. 
90 • 
(b) f;(X)=4.S.Var(X)=90 / 12- =67.5 
(c) Thet'\l:nt isananh'al in f.t!.i! intervaJs 8:.:;0 -9:00am or <)::.oo -9:30am or 9:.50 - 10:00 sm so t~.at IJ!.e probabili ty= j0/ 90 = 1/ :.}. 
(d) Similarly, flt.e e\'ent is anarri\'<IJ in tl'.e iotervsls8:30 -8:40am or 9:00 - 9:10 sm or 9:30 -9:40am so that t~.o: probabili ty= :J0/90 = t/:.J. 
4-47. {a) tet X be th.earrivaJ time(io mim:.tes)after 9:00am. 
(120- ol 
l'{X) = =1200andax=.')4JJ4. 
12 
b) Wewomt to determine the probability the mess.sgeatri\'es inanyofth.c follo\.>o;ng inteNa!s: 
9:05- 9:1.; amor 9:35- 9:45am or 10:05- 10:1samor 10:35 -10:45 am. n-£prob3bilityofthise>.l:nt is40/l~O = t/ j.. 
c) We want todd ennine tfo.e probabilityt~.e mess.ageatri\'CS inanyoftf!.£ follo\.>oiQg. inter\'als: 
9:15-9:30 amor 9:45- 10:00 a m or 10:15- tO:jO sm or 10:4.;- u :oo a m. Tlt.e probabilityoftbis~tisGoft'J.O = l/ 2 
4-49. a) P('l < 1;.32) = 0 .91)6.;8 
b) P{l < j .O) = 0 .9()86.; 
<:) P{l > 145) = 1-0.92647 = O.Oi.J.5j. 
d) P(Z >- 2.1.;) = p(Z < ~Ll.;) = 0 .9&$1!! 
c) F(- :!,34 < z < 1.76) = P{l < 1.76)- P(Z> 1,34) = 0 .9.;116 
4 -.;1. a) P(Z< 1.28) = 0.90 
b) P(Z < O) =O.S 
c) (f P(Z > z) = 0 .1. then P(Z < z) = 0 :!)1) and z = 1.18 
d) lfP(Z > 1.) = 0 .9. then P(Z < 1.) = 0 .10 andz = -1.28 
d F(- 1.::!4 < Z < z) = P('L < z)- P('L < -1.::!4) 
= P{Z< 7.) - 0 .10749. 
Th.erefore. P{Z < l) = o.B ... 0 .10749 = 0 .9Q749andt. = 1,33 
4·.53· a) PCX< l.))= P(Z < (lJ-10)/~) 
= P(Z< 1.)) 
=0.9.)319 
b) P(X>9)= >- P(X< 9) 
= 1- P{l.-< (9-10)/1) 
= 1- FfL< -o..;) 
=0.69146. 
[6-10 14-10] cl P(6 <X<>4l = P l <2< 2 
= F(- 2<Z< 2) 
= P('L< 2)- P(Z <- 2)1 
= 0.9.;45 
[2-10 4-10] dl P(' <X«l = P 2 < 2 < 2 
= P(-4 <Z< -J) 
= 1'('/. < - j ) - I'V- < - 4) 
= O.OOlj.!! 
e) P(- 2<X<8) =P(X<8) -P(X < -2) 
= p[2 < 8 -2 1 0 ]- P [2 < -2~ 1 0] 
=P<Z<-l) -P(Z < -6) 
= 0 .1.;866 
4·55· a) I'(X< u) • P[z < I I~ 5] 
• P(7. < J.,S) 
•0.93319 
[ 0- 5) b) I'()(>O)•P Z> - 4
-
• P(1. > .. u.s) 
• l - P(7. < .. 1.2,$)• •.89435 
[3-5 7- 5] <) I'U<X<7l• P --<Z <--4 4 
• P(-O.S < Z< O.S) 
• P(1. < 0._11) .. fl(Z<- O,S) 
• 0 ,18:192. 
d) PH < X<9l • P <Z<--[- 2- 5 9- 5] 4 4 
• N.- J.7.S < z < J) 
• P("l. < 11 ... P(Z <- ,.,su 
• o.Bcn:tB 
['-5 8- 5) t) P(><X<SI • P =-;--<Z< - 4-
• F(·0.15<Z<o.;s) 
• m < o.:s)- P(Z < - o.;s) 
• 0.$4674 
[ 
6250- 6000] 4•57· •I P(X<6250)= P Z< 100 
• P(7. < 1.S) 
"" 0.993?9 
b) P<.58oo <X (5900) = P < Z < .:.c:...:..:,-:7c:..:.~ [
5800- 6000 5900- 6000] 
I 00 100 
• P(· 2<Z<-J) 
•l'tZ<->)- PCl.<-2) 
'"0.J3$Cn 
[ ·-6000] ·-6000 <l P(X > X) • P Z> · =0.95_,.,..,., .... -· -:-::::-- • -1.65andx=;S:ls 100 100 
4'59- (a)14(2)•0.0218 
(b) Lt1Xdmcy<U.6PH. 
X- ~(129,14 ) 
[
100-129] RX <ton) • <.() 
14 
= 4><-1.07141 = o.ott;t6 
_, 
{<:) <D (0.9,5)Jit14 •1:t9= 1,51.02:80 
9.;9fiof 1~Al$UI"&trle& will be fini&I>.OO. with.io 152.02:8 mim:.tes. 
(d) 1C)Q>>1!1:l.O:.t8ao Uw: w lumcot s~hs1!.rgcrics is \oerysmall (leu ih.nn,s9{.), 
4~>1 . a) P(X > of>:l_) = > ----P[z 0.62- 0.5] 0.05 
=PfL> :.!4) 
=l - ffL<24) 
= 0 .0082 
b) P(04 7 <X< 0 .6.)) = . < < P [ 
0.4 7- 0.5 z ..:..:0 . .::63'-:---:-0..:..:.5 ] 
0.05 0.05 
= P(- 0 .6 <Z< 2.6) 
= P(Z < 2.6) - P(Z < - 0 .6) 
= 0 .99.;j4- 0 .21425 
P[z x - 0.5 ] c) ~«l= P(X < x) = < = 0.90. 0.05 
Therefore. x - O.S = 1.28andx = o ..;64. 
0.05 
4-63. a) If~> >') =0.99\).then P Z > = 0.999 [ 12- '' ] 0. I 
12 - JL 
Tltcrefore. = - j .09 aod11 = l :l$09 
0. I 
b) lfP(X> » ) =0.99\).thco P Z > = 0.999 [ 12 - .u·] 0.05 
12 - JL 
Therefore. =- ;3.09aod;~ = 12.1545 
0.05 
[ 70- 60] 4-65. a ) ~> 70l = P l > 4 
=l-W.< ::!..;) 
= 1-0.99379 = O.OOM.l 
b) ~qs) = P[z < 58 ~ 60] 
= RZ<-O,s) 
= 0.:)08538 
c) 1.ooo.ooo h)1es·s bits/b)1e = s.ooo.ooo hits 
8,000,000 bits 
60,000 bits/sec I 33.33 secoods 
4-672 l.etXdenote theheigJtt. 
X - N(l4 1. 0.01 ) 
<al P(X > 1.42) =I- P(X < 1.42) = 1- q,[ l .42- 1.4 1 ]=I- q,(l) = 0. 1587 
0.0 I 
-I 
{b) <I> (0 .0S)KO.Ol- 14l=l ,.l9$6 
!cl P(l .39 < X < I .43) = of>[ 1.43- 1.4 I]- of>[ 1.39 - I .4 l l = .P(2) - .P(- 2) = 0.9545 
0.01 0.01 
4.£>!). a) P!XqOOO)= p l <==-,-=:. [ 
5000- 7000] 
600 
= P(Z < - .3..'.13) = 0 .0004,3 
bl P!X>xl =0.9s.T~.cr.rou. P[z > x~~OOO] = 0.95 and 
Cons.eqnently.x = 6016. 
X 7000 = _ (.64 
600 
cl P(X > 7000)= P Z > = P(Z > 0) = 0.5 [ 
7000 - 7000 l 
600 
P(t.J!.r« laserso)X!toltingafter 7000 hours) = {l/'J} = 1j8 
4-71. al P(X > 13) = P Z > = P(Z > 2) = 0.022 75 [ 13 -12] . 0.5 
[ 
13-1 2] b) lfP!Xo3)=o.m .w.eo P Z < 
0 
= 0.999 
Therefore.!/ a= .J.i>9aod =a 1/ .).09 = 0,3:24 
[ 13-p] c) II P!Xo3)=o.m .w.eo P Z < , = 0.999 0.) 
13- .u 
nnefote, =,3.09Sod,!l=ll455 
0.5 
4·7,3. From tho:sltapeoftft.e normal cut\'C the probability is maxim ius (or an intet\'al S)'lllmetricaixmt the mean. 
Tttercl'orea = 2.1-5 witll probability= 0 .19']4. The standard a'e\iationdoes not affect tt.ec~.oioe:ofintm'l'l l. 
[ 
100 - 50.9] 4 ·15· a) P(X >lOO) = P Z> 
25 
= P(Z >l.9()4)=0.0::!48 
[ 
25- 50.9] b) P(X < 2,5) = p Z < 
25 
= P{"/. < - 1:0.}6) = 0 .1501 
-1 
c) P(X > :.:) = o.o.;. tlten<l> {0:9,s)x~·.;0 .9 = 1.f*l9x2.5-50.9 = 92.M1.) 
4·77. a ) F..(X) = 200(04) = 80. V{X) = 200(04X0.6) = 48and0' X = .J48 
Thco, P(X 5 70) ~ P[z 5 70~80 ]= P(Z 5 - 1.37) = 0.0853 
b) 
c) 
[ 
70.5- so 89.5- so] , PrO<X<90) ~P f4i <25 f4i = P(-I.J7<251.37) 
= 0.91466-0.08534= 0.8293 
Pr9.S< X 5 80.5)~ P[79~SO < Z < 80~80 ]= P(- 0.072 17 <2 5 0.072 17) 
= 0.0575 
X - 64 
4·79 z = 0)6; X - 64 isapproxim.stelyN(O,l) 8 
[ 72-64] <•l P(X > 72) = 1- P(X ~ 72) = 1- P Z ~ 8 
= 1-IV-~1) = 1-o.&u,l = o.1,s87 
lf a oontinl!.ityoorrection '"ert:nsed tlt.t: fol!o,,;ng result is obtained. 
P(X > 72) = P(X > 73) ~ P[z 2 73 - 0~5 - 64] 
'"P(Z > 1.06) = 1- 0.855428 = 0. 1446 
{b) o ,s 
lf a oontinl!.ityoorrection '"ert:tl.SOO tlt.t: follo"'ing res-ult is obtained. 
P(X < 64) = P(X < 63) ~ P[z ~ 63+ 0~5 - 64]= P(Z ~ - 0.06) = 0.4761 
[68- 64] [60-64] (c) K60 <X<68)=~<6S)-~s 60) = of> S -of> S 
= ¢(O,s)-<li(-O,s) = O.j.8::!9 
If a oontinliityoorro::tionwere us.ed the following re.sult is obtained. 
P(60 < X< 68) = P(61 ~X ~ 68)'" p[ 61- 0~5 - 64 < z < 68+ 0~5- 64] 
~ P(- 0.44 < Z <0.56)= 0.3823 
4-Bl. tct X d~to: tt.e nu.mb.."t of poople,,;t~.a disability in tlt.O!samp!e. 
X - 81N(lOOO,O.l})3) 
Z = X -1000x0.193 X -193 
J 193(1-0. 193) 12.4800 is approximately t\(0.1) 
!al P(X > 200) = 1- P(X ~ 200) = 1- P(X ~ 200+ 0.5) = 1- .P[2005 - 193]= 1- <!>(0.6) = 0.2743 
12.48 
!bl P(180< X < 300) = P(181 <X ~ 299) = q,[ 299.5-193]- <I>[ 180.5 -193] 
12.48 12.48 
4-83. td X denote the number of original component$ that fai l during tlt.e-..sefullifeof the prodnct. Then, X is a binomial random \"ariablewith p = 0 .001 and n = .:;ooo. 
Also. E.{X) = 5 MO {0.001) = 5 aod V(X) = 5000(0.001X0.999) = 4.99.;. 
P(X > 10)'" P[z ?:: r-5] = P(Z ?:: 2.01) = 1- P(Z < 2.01) = 1- 0.978 = 0.022 
4.995 
4-85. l.et Xd:eoote the: number of p.'lrticles in JO <'Jn 
2 
of dust. n.en. X i.sa Poisson random variab!e,.,;tn A = tO{looo)-= to.ooo.Aiso, E.(X) =A= 10 4 and V(X) = i. = 10 4 
P(X > 10000) = 1- P(X < 10000) ~ I- P[z ~ I OO~O- I OOOO]~ I- P(Z < 0)~ 0.5 
I 0000 
If a oontinu.ityoorrection \,·en USi'd: tl:e follo,..-ing tf$u.J t isobtaiood. 
P(X > 10000) = P(X ?:: I 0001) ~ P[z > LOOOI ~ O.S - 1 OOOO ]~ P(Z > 0) ~ 0.5 
10000 
4-87. l.d Xdeootet~.e rmmbcrofhits to a website. Tben.Xisa Poisson random \'ariablewitha mesnof ,\ = 10.000 hits perday.Aiso. E(X) = ..\ = 10.000 = V{X). 
a) 
P(X> 20000)= I- P(X ~ 20000)= I- P[ z ~ 20000 - IOOOO ] 
0J10000 
= I- P(Z ~ I 00)"" I- I = 0 
lf a oontinuityoorr-ection ,,·ere toed lite follo,-ring result is obtained. 
. -
P(X> 20000)= P(X ?:: 2000 1)~ P[Z ?:: 20000- 0.5 - IOOOO ] 
.JI 0000 
= P(Z ?:: 99.995)"' I- I= 0 
bl P(X <9900) = P(X ~ 9899) = P[z ~ 98j-1 OOOO l = P(Z ~ - 1.0 I) = 0. I 562 
10000 
lf a oontin"ityoon-ectionwereoed the following result is obtained. 
P(X < 9900) = P(X ~ 9899) ~ P[z < 9899 + O.S- 1 O, OOO] "' P(Z?:: - 1.0 I) = 0. I 562 
J10ooo 
[ 
x - 10000] x - 10000 
c) lf'P(X>x) =O.Ol.f.l!.enP Z > .J = O.OI.n.erdore. = 2.33aodx= lj.jOO. 
10000 10000 
d) let Xdeootethenu.mbetofh.its to a website. Tlten.Xisa Poisson r.aodom \'ariable,ritha of mean 10.000 per day. F..(X) =A= 10.000 aod V{X) = 10.000. 
P(X > 10200) - P[z?:: 10200 - 1 OOOO]= P(Z ?:: 2) = I- P(Z < 2) 
.J10ooo 
= I- 0.97725 = 0.022 75 
If a oontim<.ityoorrflction is USi'd we obtain tlt.e follo,'ling result 
P(X > I 0, 200) ~ P [z?:: 1 0•2j.5-1 O, OOO] = P(Z ?:: 2.005) = I- P(Z < 2.005) 
I 0,000 
that approximatelyeq-u.aJs the NS-dt '"if.l!.out theoontin~<.ityoorrtction. 
n.e expected number of <f.ttys v.ith more tft..an u),200 hits is{O .O~J.;)•j6.; = 8JO dsys per ~~r 
e) tct Y dmoto: tf!.e nu.mb.."t of days per )'e.'lr ,.,.;t~over lO,:iOO f!.its to a ,,·ebsite. 
TJ!..en. Y is a binomial 101ndom \'ariablewith n = j6.;aod p = 0.0~27.; . 
E(\') = 8..}0 and V(Y) = j6.;{0 .01~7.;XO:l>77i.5) = 8.18 
P(Y > I 5) ~ P[z > 1 5j-8·30]= P(Z ?:: 2.56) =I- P(Z < 2.56) 
8.28 
=I- 0.9948 = 0.0052 
4-89. Witlt lO .,S~)O astJ!.ma incidents in children ina ::n-month period. th-en mean nu.mb.._r of incidents per month is to.;ooj~l = .;oo. l.et X denotes Poisson modom \"ariablev.itha mean of .;~)0 
per month. Also. E(X) =A= .;oo = V(X). 
a) Using a oontinu.ityoorrection. th.i!: follo,,ing ti$-dt is obtained. 
P(X > 550) ~ P[ Z 2 550+~ 500 l = P(Z > 2.2584) = 1- 0.9880= 0.012 
P(X > 550) = P[z 2 55~00 ]= P(Z 2 2.2361) = 1- P(Z <2.2361) = 1- 0.9873= 0.0127 
b) Using a a mtincityoorrcction, tbe follo,\ing resclt is obtained. 
P(450< X < 550) = P(451 <X< 549)= p[450.5- 500 < z < 549.5- 500] 
- ' - - J 5oo - - J5oo 
= P(Z s 2.2137)- P(Z s - 2.2137) = 0.9866- 0.0134 = 0.9732 
P(450< X <550)= P(X < 550) - P(X < 450) = P[z < ss;-500]- P[z s 45t 500] 
500 500 
= P(Z s 2.236 1) - P(Z s - 2.2361) = 0.9873- 0.0127 = 0.9746 
c) Jt\' s x) = 0.9.; 
x = .p- 1(0.95)x v'5oo + 500 = 536.78 
Secoon+S 
d) The Poisson distribution would not be appropriate bcc.sl!SO: the rate of e>.'tntsshou.ld beoonsttlnt for a Poisson distribution. 
4-91. lfF..(:X)=lO,tl'.en..\ ::0.1 
00 00 
al P(X> 10) = J o. le- 0·" dx =-e-O.tx l
10 
= , -I = 0.3679 
10 
00 
b) P(X>20) =-e- O. tx l = e-2 = 0.1353 
20 
JO 
c) P(X < 30) =-e- O.Ixlo = 1- e- 3 = 0.9502 
X X 
d) P(X < x) = Jo. ~«-0· 1'dt =-e-0·"1 0 = 1- e-O.tx = 0.95andx =,9.96 
0 
4-93. l.et X denote th-e time u.ntil the first cou.nt. Tr.en. X is an exponential 1<1ndom \'8tiab!e,~it!\ ,\ = ~ oo.:.nts per minute. 
00 00 
a) P(X > 0.5) = J 1e- 2xdx = - e- 2xl = e- 1 = 0.3679 
0.5 
0.5 
116 116 
bl P[x < ~~] = J 1e-2'dx =-e- 2x l 
0 
= 1- e- 1!3 = 0.283 5 
0 
2 
c) P(I < X < 2) =-e-2'1
1 
= e- 2 - e- 4 = 0.1170 
4-9'.5· Lct X denote the time t:ntil the first call. n.en. X iso:pooential and A = I = _I_ calls/ minute. 
E(X) 15 
"" "" 
a) P(X > 30) = J-1 e -T.dx = - e -f; = e-2 = 0. 1353 
30 
15 JO 
b) The probability of a t l~st one call ina tO·mitmte intrn"SJ «!,lUlls one minu.s the probsbilityof 1.eroea1Js ina U)·mioute iotenosl and tf!.st is P{X > 10) 
"" P(X > I 0) = - e - -;; = e - 2'J = 0.5 134 
10 
n.erefore. t.l>.eanswer is l- 0 ,5l34 = O.t~8G6.Altcrnati\'dy, therea.ui:StOO: probabilityisf:<luaJ to P(X < 10) = 0486& 
10 
P 5 X 0 - · - 1/3 - 2'J 0 203 c) ( < < I ) = - e u = e - e = . 1 
5 
X 
d) P<>: o:) = 0:90 aocl P(X <X) = - e -t, = 1-e- x/ IS = 0.90. Therefore,>:= .34-54 minutes. 
0 
4-97. tet X d-enote tf!.e time to fai lt:re(in f!.ours)offans ioa personal oomp1:ter. Then. X is an o:ponential random \ 'Sriable and A = l/ E(X) = 0 .0003. 
a) 
"' 
"' P(X > 10,000) = J O.OOOJe- xo.ocoJdx = _ , - xo.oooJI = , -J = 0.0498 
10.000 10.000 
1.000 1.000 
bl P(X < 7,000) = J O.OOOJe- xo.oooJd,· = - e- xo.oooJI = 1- e-2·1 = 0.8775 
0 0 
4-99. tet X d-enote tf!.e timenntiJ the arrival of a taxi. n.en. X is an exponential tana'om \'3riable,-rithA = l/ l£(>.1 = 0 .1 a rrh>alsjminnte. 
<" 00 
a) P(X > 60)= Jo. ~e-0· "dx =-e-0· 1t0 = , - • = 0.0025 
&.1 
10 10 
b) P(X < 10)= J o. J,,- o.lxdx =-e- 0·"1
0 
= 1- e- 1 = 0.632 1 
0 
00 .. 
c) P(X > x) = J o. ~e-0· 1'di = _ , - o .. . l X = . - O.Ix = 0.1 and X= 23.03 minmes. 
X 
d) P{X < >:) = 0 .9 implies tf!.st P(X > x) = 0.1. Therefore. this answer is tf!.esame as part c). 
X 
e) P(X < x) = - e- 0·1'1 
0 
= 1- , -0.1> = 0.5 and x = 6.93 minmes. 
4·101. tet X deootet lw! m:mber ofit'JS(!Cf fragments per gr<~m . Tlten 
X-IU/{!44/~l 
a) ::!2..5/ 144 = 15.fu..5 
b) 
- Ad 
P (X = 0) = .:.." .,..:.".:... 
0! 
7 . -6 
C) (0.1629) : JXlO 
14.4x28.35 
e 225 = 0.1629 
4-103. tet Xdcootethe lifetime of an assembly. Tltcn. X isano:pooeotiaJ NIOOom \ "ariable,,itttA = Jj l-;(X) = l/401) fil ilute$ per lt.our. 
100 100 
a) P(X < 100)= J 4~0 e-x1400dx =-e-x14001 0 = 1- e- 0·25 = 0.22 12 
0 
00 
b) P(X > 500) = - e -x/400I = e - 5I• = 0.2865 
500 
c) From the memoryless propertyofth.ef'Jipoo.."'tltial. t.t.isanswer is th.est~meas pan a .• P(X < 100) = 0.2::!:1~L 
d) tet U denote tl".e nll.mber o1 ass.embliesoll.t of 10 thst filii bclOre 100 ~.ours. By the memoryless property of a Poisson prooi:SS, U f!.asa binomial distribll.tioo,vit~ n = 10 and p = 0.2211 
{from part {a)). Tlwn. 
P(U ?. I)= 1- P(U = 0)= 1-[ ~O ]0.2212°(1- 0.2212)10 = 0.9179 
e) l.d v deootetlw! m:mber of a.ss.emh!ieso~>t oflo ths t fail before Boo ~.ours. Tltcn. vis a binomial tilodom vsriablewitll n = u) aDd p = P(X < Boo). wlt.erf: X denotes t~ lifetime of an 
assembly. 
800 800 
Now, P(X < 800)= J -1-e- x 1400 dx =-e- xi400 = 1- , -2 = 0.8647 
0 400 0 
Therefore, P(V = I 0) = [: ~ ]0.864710 (1- 0.8647)0 = 0.2337 
4·105. l.et X deoote th.c rmmher of calls in jO minutes. B«:ause the time bet\<.·eencaJis isanf'.Xpoo.."'tltial tilndom variable. X is a Poisson 1<1ndom variable,-ri th.A = l/ 1-;(X) = 0 .1 calls per minute 
: JC.'t!ls per jO minutes. 
a) 
b) 
P(X> 3)= 1- P(X<3) =1-
P(X = 0) = -3 ... 0 e .> = 0.04979 
0! 
- 3 ... 0 - 331 - 3 ... 2 - 3 ... 3 
e J + <' + e J + e .> 
0 ! I! 2! 3! 
c) l.ct Y deoote the time bml.·eencalls in minutes. Then. JlY ~ x) = 0 .01 and 
00 "' 
= 0.3528 
P(Y> x ) = J 0. Je- 0·1'dt = - e- 0.1' x = "- O.lx. Therefore, , - o.1x = 0.01 and x = 46.05 mimnes. 
X 
00 00 
d) P(Y > 120) = J o. le- O.lydv=-e- 0·1Y = e- 12 = 6.14x i0- 6 • 
120 
120 
c) Because thee11Us area Poisson process. the numbers of caJis in disjoint inten>alsare independent. Ftom Exercise4-90 p;~rt b). the prob.!lbi!ityof no calls inooe:-hslt ~.oru ise -3 = 0.04979. 
Therefore. t!teanswe:r is [e-"3 r = e-12 = 6. )4 X J0-6 .A!temath>ely, liN! answ-er is tf!..e probability of oocaJis in 1\~·o ~.ours. From part d) of this exercise, this ise -l.2. 
() Beca.:sea Poisson process is memoryless. probabilities do not depend on ,,·lti:llt.er or oot inten'3lsareoons.ec:uti\'e. Tlt.;:refore. parts d) and e) have tltesameanswer. 
00 00 
P(X > B) = J!,.,-•'9dx = _ , - xte = , -• = 0.3679 
e e 
e 
"" b) P(X > 28) = - e- x/9 = e- 2 = 0.1353 
29 
"" c) P(X > 38) = - e- x;e = e- 3 = 0.0498 
39 
d) n.eresultsdonot<iependonO. 
4·109. Xisanexpooential random \'ariablewithp = 3-5 days. 
2 
a) P(X < 2) = J _2_.. - x!l . S dx = 1- e -213.5 = 0.435 
0 3.5 
00 
b) P(X>7) = J~-xt3.5dx = e-m5 = 0. 135 
? 3.) 
-X/f!.S 
c) RX>x)=0.9and:JV(>x)=e =0.9 
Tlbetefore.x =- .'),sln(0.9) = Ojf>9. 
d) From tbe JackofmemorypropertyP{X < 10 I X> j) = P(X < 7)andfrom p.llrt(b)ti'.i.sf.'qllals 1-0.13,5 = 0.86.;. 
Sectlon4-9 
4 ·lll. a) f{6) = .s!=l:!O 
b) r [i)= l r[.1)= JJ.r[.l) = l.,.v2 = 1.32934 
2 2 2 22 2 4 
c) r[9)= lr[l)= li3.ir[.l)= I05.,. •n = 11.63 17 
2 2 2 2 2 2 2 2 16 
4-113. a) Th.e time until the tent"' call is an Erlsng random variable \'ittoA =.;calls per mirmte and: r = 10. 
b) E(X) = l0/ 5 = 2 mint:t.-s. V{X) = 10/'.15 = 04 minctc.s. 
c) Bcca~a Poisson process is memol)1e.ss. th.emeantimeis l/5 = 0 .2 minute.; or 1~seoonds 
tetY deoote tlt.enu.mberofcalls in one minute. Th.en, Y is a Poisson Nlodom variablewitlls calls per min~>tc. 
, -ss• 
d ) P(Y= 4)= 0.1755 
4! 
e) 
- SsO 
P(Y > 2) = I- P(Y < 2) = 1- e 
0
; 
-ss• e . 
1 ! 
- ss? e -
2! 
0.8754 
l.et W deoote the number of one minute inten<~l.soct of 10 tha t oootaio more than::!. calls. Broiltse th.ec.allsarea Poisson process. W is a binomial tana'om \'3 riable,,i t~. n = 10 and p = 0 .87.;4. 
n.<rclore. P( W = I 0) = [: ~ ]0.875410(1- 0.8754)0 = 0.2643 
4·ll5. l.et X <teoote the time bctween failure.sofa laser. X isClo:ponentiaJ ""itha mean of ::;.:;.ooo. 
a) Expected time until tl:eseoond (ai!l!.re: E(X) = J' 1 A= :!. 10.00004 = 50.000 ltom·s. 
b) N = oooffaiJures io50000 hollJ'S 
£(N) = 50000 2 
25000 
? - ? 2)' P(N ~ 1)= '£/ "( = 0.6767 
k l k = O • 
4·U7. l.et X denote the nnmber o( bits t:ntil fiveerrorsocct:r. n.en. X lt.ssan Erlang. distribt:tion,,ith r = 5 and A= 10 -s error per bit. 
a) 
b) 
E(X) = ~= 5x 105 bits. 
.A 
V(X) = .!_= 5x 1010 and u x = ) sx 1010 = 223607 bits. 
.>-2 
c) l.et Y deoote IJ!£ number o( errors in u) 5 bits. Then. Y is a Poisson rancf.om variable,,ilhA = 1110 5 = u) - 5 error per bit = ! error per 10 5 bits. 
P(Y ;,:: 3) =I- P(Y < 2)= I- e +£..___+ e - = 0.0803 ! -110 -Ill -11?1 0! I! 2! 
4-119. a ) let X denote IJ!.e number of customers that arri\'e in 10 milllit<S. Tlt.en. X iss Poisson random \'Sriab!e,~ith 
A = 0.:!-anh-als per mint:te =:! arrh'lls per 10 minntes. 
P(X>J)=I- P(X<J) =I- e -- + " -- + " -- + " = 0. 1429 [ 
_??O _??I _??2 -2231 
- 0! I! 2! 3! 
b) let Y deootet.J!.<' m:.mber of customersIJ!.atatri\'t in 15 mint:tes. Then. Y is a Poisson rancf.om variable,,ilhA = 3atrivals per 15 minutes. 
P(Y ;::: 5) =1- P(Y~4) =1- e + " + " -' + " -' + " -' = 0.1847 [ 
- 330 - 331 - 3·2 - 3·,3 - 3· 4] 
0! I! 2! 3! 4! 
t'O oo Ar xr-1 f! - >.x oo r-1 _ y 
4·121. J f(x;A,r)dx = J X . lety= At'.th-entludntegralis JA Y e !!f_. From tlt.eddinitionofr(•·).thisintcgr.aJisrccognizedtoequall. 
o o r (r) o r (r) • 
4-1~. l.et X deoote the number of patientsarri\'ttt t the emergencydq>artment. n.en. X h.ssa Poissondistribt:tion '~iti!:A = 6-5 patients per tt.ou.r. 
a ) E(X)=riA = 101 6-5 = 1-539~.ou.r. 
b) l.etY deootetf!.e ncmberofpatients thstarrh't'in :!.0 minl!.tes. n£n. Y is a Poisson random \>ariablewithA. = 6-5/3 = 2.1667 a nh>als per :!.0 minl!.tes. Thee\'ent tltat tfte thirdanh'31 o:ceeds 
:!.0 minutes istqliivalent to tlteC\>er.t that t!t.creare t\\'OOT fewer arrivals in :!.0 minutes. Tttereforc. 
[
. - 2.16672 16670 . -2.16672 16671 . - 2.1661? 166721 P(Y~ 1) = ' · + ' · +' -· = 0.63 17 
0! I! 21 
Tile solution mayalso be obtained from tl:e rf:S-..!t that tlte time until tl:f! thirdatri\'31 follows a gamma distribl!.tioowitn •· = jaod i. = 6-5 a rri\'lJs per hour. n.e probability isobtaioed by integra ting. the 
probtlbilitydi!OS:ityfunction (rom :!.0 mir.11.tes to infinity. 
S«:Uon+IO 
£(XJ = 10or [1 + -1 ]= IOOx5!= 12,ooo 
0.2 
V(X) = I002 r[ I+ 0~2 ]- I 002 [r[ I+ O~J2 = 3.61 x I 010 
4 ·12'7. l f X isa Wei bull Ntndom \ 'Sriable \'lith U = 1 and 6 = 1000. thedistribli.tionof X is theo:poo.-:ntia! distribt:tion \~i th). = 0.()01. 
/(x)= [-1 ][-x ]0 j ,:oo)' Jbrx >O 
1000 1000 
= 0.00 lc - O.OOix for x > 0 
TJ>.emeaoofX is F.(X) = l/ A= 1000. 
4·>'9 a) £(X) = or[ I +~)= 900f (l+ I /3) = 900[(4 / 3) = 900(0.89298). = 803.68 hours 
b) V(X) =o2r[l +~]-o2 r[l +_~f = 9002 r[l +~] - 9002 [r[l + ~w 
= 9002(0.90274) - 900\0.89298)2 = 853 I 9.64 hours2 
c) [
500J 
P(X < 500) = Px(500) =I- e 900 = 0. I 576 
4-1 31. a) B= 2, l\ = 500 
b) 
c) 
£(X)= 5oor [ 1 + ~] = 500f(L5J 
= 500(0.5)[(0.5) = 250.{; = 443. I I 
= 443. I I hours 
V(X) = 5002[(1 +1)- 5002[r[ I+ ~w 
= 5oo2 r (2) - soo2[f(L5)f = 53650.5 
[
250)' 
P(X < 250) = F(250) = I- e 500 = I - 0. 7788 = 0.22 I 2 
4- I 33. o2r[ I+ .! ]= Var(X) + (EX)2 = I 0.3 + 4.92 = 34.3 I 
or[ I + .~]= E(X)= 10.3 
Rcqdresa numuicaJ solntion to th.esi!: two tqiUitiom. 
b) P(X > 6oOOi X> 3000) = P(X > 6000, X> 3000) = P(X > 6000) 
P(X > 3000) P(X > 3000) 
1- F (6000) " - (&JOCV4000)' 
= 1 _ ~ (3000) = -e --(.,-30.,-0CV.,-4.,-00.,-0)7' 0. I 054 0.5698 0. I 850 
c) If it is an exponential distribution, then 13 = I and 
I- F (6000) e- (.oocv•ooo) 0.223 I 
= X = = 0.4724 
1- Fx (3000) e - CJOOCV4000J 0.4724 
For the Wei bull distriblltioo{\\i t~ IS= ::!) tJ!.ere is oo Jsckofmcmorypropertyso th£tttheans\~·ers to parts (a)and (b)di fter \''h.ereas tt.eywould be t~.esame if an exponential distrib~ttion ,,·ere assumed. 
From part (b), f.t!.e probability of surviva l ~~mel 6000 lt.ours.gi\'eot.ede-.ict hasalr-eadys-.:rvh't'd3000 ~.ot:tS, is lower than the probability of survival be)-ond$000 ~.ours from tbe start time. 
-{3s00/2000) 
4·1,37. a ) Jt\"> j.SOO) =l-FX (lsOO) =e =0.0468 
b) Themesnoft~.is Weibull distrib.:tioni.s (2000) 0.5,J; = 1772.45. 
r ( it is an o:poo.."'Otia I distrihl:ti_1is~MW'~~n then 
ft\'> .',1.500) = 1-F.>.U;OO) = e = O.ljSH. 
c) Tlte probability tf!.at f.t!~ lifetimeo:ceOOs j.SOO ~.ours is greater unrli!T tJ!.eo:pone:otiaJ distribution than under this Wei bull distribution model . 
• 4·lJ9. a) Xisa logoonn.al distributionwitl\0=- :!and(l) =9. 
P(500 < X < I 000) = P(500 < ew < I 000) = P(ln(500) < W < ln(l 000)) 
= .P[ In( I O~O) + 2]- .P[ In( SO~)+ 2 ] = .P(2.97) - .P(2.74) = 0.0016 
rr [ln(x)+? ] b) P(X < x) = P(e :o; x) = P(JV < IIl(x)) = .P 
3 
- = 0.1 
ln(x) + 2 28(3)- 2 
- 1.28 x = e-1. = 0.0029 
3 
a) P(X > I 0) = P(e"' > I 0) = P(W > In( I 0)) = 1- q,[ ln(I0)
1
-
0
·
5] 
= 1- q,( l.80) = 1-0.96407 = 0.03593 
w [ln(x) - 0.5] , b) P(X 5. x) = P(e < x)= P(W < ln(x)) = .P 
1 
= 0.)0 
b) 
c) 
ln(x) - 0·5 0 x = ,o( J)+0.5 = 1.6 5 seconds 
I 
P(X > I 0000)= P(ew > I 0000) = P(W > In( I 0000)) = 1- q,[ In (I OOOO) - 8.4056] 
1.2686 
= l-q,(0.63) = 1- 0.7357 = 0.2643 
P(X > x) = P(e > x )= P(W > ln(x)) = q, = 0.1 w [ln(x) - 8.4056] 
1.2686 
ln(x) - 8.4056 - 1.28 x = e -1 100( 1.2686)+8.4056 = 88 1.65 hours 
1.2686 
2 X 2 
4·145· tet X· N(lt, o ). then Y = e folio\~ 'Sa lognonnal distrib~tion,.,rifl!. meanpaod \'arianoeo . By definition. 
Fy(y) = P( Y :S: y)= P(ex < y) = P(X < logy)= Fx(log y) = q,[ log:- f£ J. 
s;""" Y = ,x sod X- N(l'.a\ wecansltow tl>at}~ (Y) = _!_ fx(log y) 
v 
8P (y) 8F .(logy) 1 . 1 I j~o'(."")' 
Finally,fy(y) = Y = ' =-fx(log y )=-· .Jh' l · . 
8y 8y y Y u 2rr 
b) P(X <8) = P(ew < 8) = P(W < ln(8)) = .P[l n(S) - 1.5] = .P(l .4486) = 0.9263 
0.4 
c) IV<< o) = o for tl:e lognonnaJ distriba:tion.lftl:edistrihl'-tion is normal. th.en 
P(X < 0) = P(Z < O- 4·855) = 0.008 . 
../4.0898 
Bec!luscwaiting times cannot be o.-:gath't the nonnal distribt:.tiongmeN~tessome modeling error. 
section + u 
4-1 49. a) P(X <0.25) = 
0
f.
25 
r(<> + .B) )x0 - 1(1- x)P-I 
0 
r (a)r (,B) 
b) 
c) 
= j r(3.5) )xu = (2.5)(1.5)(0.5).[; .,.25 025 = 0.2525 = 0.03 13 
0 r (2.5)r(l) ( l.5X0.5)0); 2.5 0 
o.•5 r f3 
P(0.25<X < 0.75) = J (<> + ))x0- 1(1- x)"- 1 
o.25 r(a)r (,B) 
o•5 r us 
= J r(3.5) ).,.u = (2.5Xl.5X0.5)\Irr x25 = 0.7 52s _ 0.252.5 = 0.4559 
025 r (2 .S)r (l) (t.sxo.s}J,. 2.5 o.2s 
E(x "' 2.5 0 7 ' p.= ) = = = . 14> 
<> + {3 2.5+ l 
"2 = V(X) = o,(3 = _ _::2.:.:_5_ 
(a+,8)2 (a+{3+ 1) (3.5)2(4.5) 
0.0454 
4-1 51. a) <> -1 Mode = ---
a +.B- 1 
p.= E(X) = --=." -
<> + 13 
2 
----= 0.8333 
3 + 1.4 - 2 
3 
--='---= 0.6818 
3+ 1.4 
,,Z = V(X) = . (\{3 
(<> + .Bi(a+ i3 + I) 
4
'
2 
= 0.0402 
(4.4)2 (5.4) 
b) Mode= <>-I 9 0.63 16 = 
<> +.B- 2 10+ 6.25- 2 
f' = E(X ) = <> 10 0.6154 
<> + 13 10 + 6.25 
2 
= V(X) = <.¥3 62.5 0.0137 (1 
(a + .B)2(<> + /3 + I) (16.25f(J7 .25) 
c) Botltth.e meananch-sriance from part a)aregreatcr than for part b). 
4·1$3. l.d X <ieootc tl:coompletion proportionoft~.e maximum time. Tl:co:erciseoonsid.e:rs tl:c proportion'.!/ '.!.!) = 0.8 
I 
P(X>0.8)= J r(a + /3)_,.•-' (1 -x)~-c 
0.8 r (a)f(/3) 
I 
= J' r (S) x(l - x)2 = (4)(J)f(J) [x2 - 2x3 + x•] = 12(0.0833- 0.08 11) = 0.0272 
0.8 f(2)f(3) f(2)r(J) 2 3 4 0.8 
4-155. a) P(X < 40) = P[z < 40 ~ 35] 
= P(Z < 2.5) 
= 0.99379 
[ 30- 35] b) P(X < 30) = P Z < 2 
= P(Z <- 2.5) 
= 0.0062 1 
0.621% are scrapped 
4-1 57. a) P(X > 90.3) + P(X < 89.7) 
= P[z > 90.3 - 90.2 ] + P[z < 89.7- 90.2 ] 
0.1 0.1 
= P(Z > I)+ P(Z <- 5) 
= 1- P(Z < I)+ P(Z <- 5) 
= I - 0.84134 + 0 
= 0. 15866. 
Tlterefore. tJt.eans"'·cr is 1).15866. 
b) n.e proress me.'lnsltould be set at th.ee.."tlter of the specifications: t~.at is. at#= <)0.0. 
c) P(89.7 < X < 90.3)= p [89.7 - 90 <Z < 90.3 - 90] 
0.1 0.1 
= P(- 3 < Z < 3)=0.9973. 
Tite yield is 100*0.9973 = 99.73%. 
d) P(89.7 < X < 90.3) = ? [89.7- 90 <Z < 90.3 - 90] 
0.1 0.1 
= P(- 3 < Z < 3) 
= 0.9973. 
P(X = 10) = (0.9973 )10 = 0.9733 
e) tet Ytep«'SCnt the numberofeastSoutofthesampleofu) th£ttare f>ct\•·ren~.7 and<)()~ mi. n .en Y follo\•-sa bioomiaJ distribntion"'itlln = u) andp = 0 .9973. Thu.s, E(}') = 9.973or u). 
4·159· E{X) = 1000{0.:/) = :l-00 and V{X) = l000(0.2X0.8) = 160 
a) P(X > 225) = P(X '2 226)~ 1- P[z < 225~200 l = 1-P(Z < 2.02) = 1-0.9783 = 0.0217 
b) P(175 <X< 225) ""' P[ 174.5- 200 < Z < 225.5- 200] = P(- 2.02< Z < 2.02) 
--- ..JI60 -- JJ60 --
= 0.97 83 - 0.021 7 = .9566 
c) lf P(X> x) = O.OI, then P[Z>xJ!toO]= O.OI. 
Therefore, 'Ji¥o0 = 2.33 and x = 229.5 
160 
4-161. let X denote the nt<.mber of calls inS ~.ou.rs.Because tlto: time b.."'\\.ox:ncalls is an exponential Nodom \oariab!e. tlw! m:mber of calls in;} hol!rs isa Poisson hlodom \oariab!e. t\o'"· th.e mean 
time bef\~·eeneaJis is o,s ~.ours and A= 1/0s = ~calls per tt.ou.r = 6calls in$ ~.ol!.rs. 
P(X > 4) = I- P(X < 3) = I- e + e + e + e = 0.8488 
1
- 6 6o - 6 6, - 662 - 663 I 
- - 0! I! 2! 31 
4-16;}. tetXdeoote the lifctimc. 
a) £(XJ = 70or (l + ~)= 620.4 
2 2 2 
'?/ I' (X) =too 1l 2l-700 l ll>~ll 
= 700 {1)-7oo (o.::;,s;r) = u)_;,t54:9 
c) [
620.4)' 
P(X > 620.4) = e 700 = 0.4559 
a) 
2.5 [ 2 ]2.5 
P(X<2.5) = J (O.Sx -l)dx = 0.5 '
2 
- x = 0.0625 
2 2 
b) 
4 ? 4 
P(X>J) = J (O.Sx - l)dx = 0.5x- - x = 0.75 
2 J J 
c) 
3.5 2 3.5 
P(2.5 <X < 3.5)= f <O.Sx - l)cb· = O.SL- x = 0.5 2 25 
2.5 
d) 
X 2 X ? 
F(x )= J (O.St - l )dt = 0.5-1 - t = L - x + I. Then, 
2 2 2 4 
0, x <2 
F(x) = x2 --x+ l, 
4 
25x < 4 
I . 4 5 x 
e) 4 • 2 4 ( ) E(X) = J x(0.5x -l)dx = 0.5L - L = 11- 8- i-2 = .!Q 
2 J 22 3 3 3 
4 4 
V(X) = J (x - 'f J\0.5x - l)dx = J(x2 - 2°x+ 100](0.5x- l)dx 
2 2 3 9 
'[ ' l 4 4 = J 0.5xJ_ .L!.x2+l!.Q.x - 100 dx = L - ll,,.l+ 55xz - IOOx 
2 3 9 9 89 9 92 
= 0.2222 
a) 
b) Lct W deoote the ncmber ofCPUs that fa il "'it.f!.in tfte next tllreeyears. Tlteo. W is a binomial random \'Sriable "'ith n = 10 and p = 0.$93.5 {from E.xercisc4·1.)0 ). Then . 
• S+1.i~l a) Fiod tlte'llh~3fOJM(,) giveo tft.:. t F.{X) =so andV{X) = 4000 
so =e 41)00 =e {e -1) 
8 6t 1 21 2 . .[; 
letx =e andy=e tft.eo{l )SO = X Yand{~)4000= x y{J,'-l)= x y - x y. 
50 
Squ.are{l ) forx X = .{; andsubs.tih!te into(~) 
50 2 50 
[ ]
2 [ ]2 4000 = [; y - [; y = 2500(y - I) 
y = 2.6 
50 
st:bstiruteyba<:k in to(J)andsolvtfor x X = ~ = 3 1. 
• " 2.6 
0= l otll) =j~aod61 = ln{2.6) =0.91) 
b) P(X < 150) = P(ew < 150) = P(W < ln(ISO)) = <I>[ ln(ISO) - 3.43] 
0.98 
= <P( I.61) = 0.946301 
4-171. ld X d-enote t~.d.eight of a plant. 
a) 
b) 
c) 
P(X > 2.25) = P[z > 2·2~~ 25] = P(Z> - 0.5) = I- P(Z < - 0.5) = 0.69 15 
P(2.0 < X < 3.0) = P[2·0 - _2·5 < Z < J .O- 2·5] =P(- I < Z < I) = 0.683 
0.) 0.5 
P(X > x) = 0.90= P[Z >x - ~.5 ] = 0.90and x- ~.5 = - 1.28. 
0.) 0.) 
a) [ 
5.5- 5] P(X > 5.5) = P Z > = P(Z > 2.5) = 0. 0062 
0.2 
b) P(4.5 < X < 5.5)= P [4·5- 5 < Z < 5·5 - 5] = P (- 2.5 < Z < 2 .5) = 0.9876 
0.2 0.2 
n neforc. th-e proportion ll'.at do not moo specific.'ltions is 
1- P(4,s < X< .;,s) = 0 .0 12-. 
b) If P(X < x) = 0.95, then P Z > -- = 0.95. Therefore, -- = 1.65 and x = 5.33. [ x-5] x - 5 0.2 0.2 
4·!7.5· {(P{0.002 - X< X< 0 .00:!.- X), t.f!..eo P(- X/0.0004 < Z < xj0.00CJ4) = 0 .99Jj.. n .erefote,xj0.0004 = j and:x = 0 .0(H2.. n .esp!lcific.'ltionsare from 0.0008toO.OOj.~. 
a) 
b) 
5800 X J 5800) 
P(X < 5800) = J - 1 - e ?OOOdx = I- c -l7000 = 0.5633 
0 7000 
00 X X 
P(X > x) = J 7~00 e ?OOOdx = 0.9 Therefore, e 7000 = 0.9 and x = - 70001n(0.9)= 737.5 hours. 
X 
4-179. a) Using tl:e normal approximation to t~.e bioomial ,.,;tn n = .:;o•j6.j6 = 64.800.and p = O.OO()l wt: Ita\'€:: E(X) = 64800(0.0001) = 648 
P(X > I) "" p[ X - np > 0.5 - 6.48 l 
- - Jnp(l- p) J64800(0.000I):0.9999) 
= P(Z >- 2.35 ) =I- 0.0094 = 0.9906 
b) 
P(X > 4) "' p[ X - np > 3.5- 6.48 l 
- - )np(l- p) - )64800(0.0001X0.9999) 
= P(Z '2 - 1.1 7) = I- 0. I 2 I 0 = 0.8790 
4-181. a) P{X > x) implie; that there are r - 1 or lessoollllt$ in an intenoaJ of length x. l.et \'denote the number of ooc.nts in an intc:nosl of!engt~x. T~""ll. Y is a Poisson ~<~ndom \ 'arisble ,.,; ,~ 
psrameter Ax. 
r-1 
Then.P(X> x) = P(Y 5 r - I) = L "-.lx (~)'. 
1=0 
1'-1 
b) P(X < x) =I- L "-.lx (~? 
f=O 
c) 
,._, I," I' ,._, I >.x\')' (:..,·)'-' 
f, (x) = ..iL F (x) = ;.., -xx"'r:::.L_- , -"'"' Afr:::.L= >.e-.lx.L._L-~ * X ~ ., ~ . , ( I) ' i=O I . f= O I . r - . 
• • • 4·l8j. a) Q-uality loss= £k{X- m) = kE(X- m) = ko , byth.edcllnitionofth.e,'tlriance. 
• • b) Qusl;tyk==~-m) =kEIX-#"#-m) 
= kEI(X- "! "(jl- m) •" o(jl- m)(X- #ll 
:d:.E(X- p) - A"(p- m) - !!J.'f#l-m}}-;{X- p). 
Tlbe last term oqu:alszero by2t!tedefiniti~nofthe me.'ln. 
Therefore.qu.slity loss=ko .o.k(p:-m) . 
I P(t < X < t +t ) 4·18..;. P( X < t 1 + t., X > t 1) = 1 1 2 from thedelinitionotoooditionaJ probability. Now, 
- P(X > t1) 
t1+•! ',+s~ 
P(t
1
< X < t
1
+ t
2
) = J >..e-"'dx =-e- >.x 
1 
= e- /.1• - e- '-(1,- l,) 
' ', 
P(X >t
1
) = - e- >.x 
I, 
- /.1 
= e * 
- AI 
1- e ' = P(X < t2) • 
	1
	2
	3
	4
	5
	6
	7
	8
	9
	10
	11
	12
	13
	14
	15
	16
	17
	18
	19
	20
	21
	22
	23
	24
	25
	26
	27