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Unformatted text preview: SOLUTIONS TO STAT 310 . 010 Ah
EXAM 0N JULY 27, 2000 ‘1) M R Problem 1. (30 points) The probability that a newborn baby will be a boy is 51.3%. (a) If 100 babies are born in a city, then the chance that at least half of them will be boys is closest to (choose one option and explain): 51.3% 56% 60% 90% 100% (b) If 2,500 babies are born in a city, then the chance that at least half of them are boys is closest to (choose one option): 51.3% 56% 60% 64% 100% boys is closest to (choose one option): 51.3% 56% 60% 64% 90% 100% The mmim X M h Wm (faint/5
is {>011 Mogwffﬁfj P: 05(2):? (65 1% 11:100, PO< 250) = POO £161.55
%1~C§(‘”‘v§ﬁ%‘d=$‘1) = PCPW)
[email protected](~0.Sé) = 6(1),“) =_o__é;_<___0_6_
(b) 1% n= 2600) PM; (250) = P(X>1)2L101.s)
a: 1~©(W) = @032) = M_______O___é_é_
(Q 1( “1101000J P0920000) = P()<>mﬁ<m.s)
act—@052) 6: 110‘7 rel —.—. Problem 2. (25 points) A machine makes sticks of butter whose average weight is 4.0 ounces; the SD of the
weights is 0.05 ounces. There is no trend or pattern in the weights of the sticks produced by the machine. (You may assume that the weights of the different sticks of butter are independent.) There are 4 sticks to a package. (a) A package weighs [6.0 ounce; give or take OJ ounces or so. (Fill in the blanks and explain.) A store plans to buy 100 packages.
(b) The 100 packages will weigh IOO lbs , give or take I OMMCE. or so. (0) Estimate the chance that the store gets 100 pounds ofbutter, to within 2 ounces. Note: One pound of butter equals 16 ounces. Letn= number of Slick; o¥ bulkr. Er k=l)_,,Jn,
let X11 = W3“ 0? slick AW‘XP h, (in ounces)
Ill20m = FLO oz, d=suow< = 0.050;. Problem 3. (10 points)
The Census Bureau is planning to take a random sample amounting to one hundredth of 1% of the population in each state in order to estimate the percentage of the population in that state earning over $50,000 a year. Assume that in each state the Census Bureau will take a simple random sample. Other things being equal: (i) The accuracy to be expected in California (population 33 million) is about the same as the accuracy to be expected in Nevada (population 1.8 million).
The accuracy to be expected in California is quite a bit higher than in Nevada. (iii) The accuracy to be expected in California is quite a bit lower than in Nevada. Choose one option and explain. 069“,le acowm [m Cat/C ‘a £4 W
Li W ‘l’lu cagedcol «Lu/1403 'w Nevada. Problem 4. (10 points) A survey organization takes a simple random sample of 1,500 persons from the residents of a large city. Among these sample persons, 1,035 were renters.
(3) Estimate the fraction of all the residents of the city who are renters. (b) Find a 95% conﬁdence interval for the fraction of the population who are renters. $=%g—%§=o.eq, SEot/ﬁ = n =0.012 0.4201 ‘1? 0.02 in am maxi/male q§°o mfiolwa
imp/mot (far M (Imam '0“ 0§ Matias wattage/1%. 9&1“ 5‘. Lab F1 = wommm 01 (mm Wm 04am P2 = 4Mme 'sé—{ﬁNn aw Mega H
Wax/L3 managaikma/UA. Mg ”m” A Wm“ 03 gage—67m W mfg Wm
{M a 5am d n, = (80 (OM34 gnaaéua/YIA /\
P2 z WWW“ d) '35qu W‘ m WK»:
(«L a Emil/(A «jg/MHZ = {00 Mmg/‘fgéwa/Q 0'4 '7 :0
ll $1 54 WOXi/Makza mm‘rmcuﬂ (P1 )M%)
132— “ Woxmldg .mwd (P2 ) Elk—Pl) {3‘4“ (/3: EA Wax. mmd ([31 P1) '61P +34];le)
/\ A A /\ SE 0% 31‘ P:— : e.(;]1—e.)+ Earl—PL) (5“ MM gzmzwo.) (a3 31 = £3— = 070. 32. = [3‘3 = 0.5% ,, Problem 6. (20 points) The hyacinth macaw is a rare, large, blue South American parrot. We want to estimate the number of hyacinth macaws in a region of the Amazon rain forest. The sizes of animal populations are often estimated by using a capturetag—recapture method. In this population. Suppose that k = 110 macaws are tagged and then released. A week later, a sample ofn : 225 macaws is then selected at random from the same population. Suppose that there are exactly 22 tagged macaws among the 225 birds in the second sample.
(a) Estimate the fraction 6 = k/ N of tagged macaws in the population. (b) Give an approximate 95% conﬁdence interval for 6. (c) Estimate N. ((1) Give an approximate 95% conﬁdence interval for N. (a) 8 = {Faction oi: hggeal birds in HR sew W0( .Sam 9‘»: 22/225 = 0.09%? FTM = 0.0/93
AW“. 95% mgi mud five: 63:: SE /\ (a) e=k/N W {CI k/§= ”Lg—‘25 22. ll I! (5) SE 04 e3 2.
”25 STAT 310 D. MONK/M) (600 9570 Wgwmj Olaf} : [0.0582 0.37%] ) 0.0532 < e < 0.137% m 0.0532 <,§",< 0.1374 iﬂC 0.3m < N < 075%; iac 30/5 Ns mo
9570 w. WEMOLM N: [301 [8‘10]
MUQ
L(6) = {(9936 )‘7((sz9> {(M9) : ﬁe—X X/G ‘ 61):: e—Xz/e ‘ . . %Exn/a
= __1x’6)n<2 q X: (X+X+...+Xn3/e
683 n E ) JW’ 9> O
(MI.(9§= WP “ﬂ— anew) — W ) 6>O
(L) M awﬂood 610111011 0 = £64119) =  Hg". +11%"
kmksoéulwm 9" "“q;1*_x"=71;7 . Wu; MLEis STAT 310 D.MONRAD
7(4) Shaw X1Jle...JXh m wawwé,
w?» —— ($5)2(vm(x‘\+vm(xz\+...+ Maw
vmm = Wm = = Wm —— M": 1/97: Z so van/e3) wanMeg” = :3: (e) émce @ {A wad/x64 amd Vat/($3 —% 0 cm H'QI‘OO e {A 0L Wsingmﬁ mam/tar of 9. J mm gbqgé‘ {gm/46> =2%.msmdw(
wwwmm/wa/ZZJJMSD; A 6
5549 = m (33 m comm me Thaw/m m «A #10th (W
V] Hm ﬂdmdtlm (RI/VIM Q [/3 WOXW“% manta/6 MM MW (9, A" Wo/(Wéa 95%
can/Mm WZZIW/é f” (9 VJ [e3 4315) @ + 235] I H=ZSMAé=7SJMMSE=O.7S‘
w¢ W WOXWQZ: Cigz) CmeVlOMCL
Maw 75 i 1.5”. SOLUTIONS To STAT L100 1). MONRAD
EXAM 0N JULY 27, 2006 Prngm1 I? X ole/MOM #41; Ca/JWOS 3am W1 a
m— doﬂa/x £06 on “red“ 11mm Hm PM”? 01)(15 §61)=§g=73a7 W $03=3€§=J§ (a) 77m CMWO’S Medea! M (M dad/15) W: a
mdoﬂm—ZM‘FPW ”aid" (1;: 0 =(~1) {(1) +(1) WW1) 1 00526
(53 E00 = ( W461) + (7) ?'(x1'(1) = 1
= Van)0< =)E(X‘ = ~32? =V§+° = Te?“ 10 (= 09‘186101:/1:r:) (C) 1;; l=1W23 .606)? 61014016 #aCMWo’s an 60“ W60: 1' an “red" 77m cméno’s
gm an 2%: 25 Eek is X z X1+X2+X3+ + X25 = £00 = EO<A+...+E(X2§3 = 25*,“ 3 ~25 x = Var/moo = VaA()<,3+...+Va/10<zg) = 25dZ
dx=\/'o? =5~d =?—9W0‘. 3am mall Using a“; Can/1a! [W'tz’ 741W we 5.1% P(X < o) 2»! ﬂ‘Efﬁ = 915(—5/6vro‘)
s3 $602435) = 0,376 M is W a 40% drama 741.4% (#4 (MM
{0501 Midway an 25 014430va W—doﬂa/z—édz
on “red“. AMOMJA soé/izm : [1/4 N" Hum/(5M 01‘ 544 [ML 01‘ 2% 25):“‘5645020 wins.
N is M25339. (0‘) 1? X]: X1 +Xz + H. + X10040“ H1014 )
ﬁx = Looqooox/u = MW $52631.“ ,dx =V1040,000'xd = $ 943.61
' (e) RX 2 50,000) x I ~¢—‘’“~5°Z§f “ X) x l— @2635) *  @1435) =:+ 0.9958 Problem 2. (10 points) There are over 50,000 households in a certain city. The average number of persons age l6 and over living in each household is known to be 2.38; the standard deviation is 1.87.
A survey organization plans to take a simple random sample of 400 households, and interview all persons age 16 and over living in the sample households. The total number ofinterviews will be around q 52. , give or take 37 or so. Fill in the blanks and explain. N = mo. cg thM 7/ 50,000, nzsmdiﬁaglze=qoa
Er kzilztaj...)qoo M X=m (msa {éNWMOm/MWS
h hmoéexgimsm [:2 3 WM 50% = 2.38. d = Shot XV.
010% Wit/\Uim: X= X1+Xz+... 4'qu
E00 = mi = 400.238 = 952. SD otX = d'V—rTN‘j’s’adV—FT 2: 37 EQé/em 3(a) /4 = §°o< «new = éXZOZ/Wak
200 —z __ 2_ 40° =
=zegxwze[><]9 26 Ebb/6m L/ (a! 7m Mae/Mood {Mm/Wm is
Me) : {(XMQ) {(ij93 {(99119)
= (3xf/QZ)(SX:/e ) (EMT/93) = 3196. XZZW m2 493” ) r 92x 62X .. 62%. (£2) 59m Ma ”Adi/I000! Hone/10w IS dggasm
Hue MLE is Ma S/maﬂwé pomé/e 9 (Ia/u
Th; rad/\ncéon 97X 9>XZJH 9?%,,,
Show #LQ/‘L 14m S/Vna/fanl possi/E 9 Ua/H/Q
[5 ML mgwé of 74m oésmm/ims: g" = mm{%,LJ *3
”7L4 IN E is #1 lggf 9W s/a/ws/m:
9 z}; : WMX {X’lX .Xn E — laAﬁ/aé O‘F X1JJXZJ.. X". 'J M n M W
L(9)= ;9H(xz;93 Ema) = QZYEXJE(M ,.>/e (M (’an(9) = [email protected](9) + gm»  %:’“ 9>O,
TM UWﬂmod eguakcm O 3 ()me) = 2n _Z__;_<A SWQQ, X; (a gamma oUdM’uleA mm at: 2)
£09) = 009 = 29 So 13$) = ﬁnZe = e. MMé‘ i4 MGM/3M. (cl) Me?) = (2%)Z<vm(x.>+...+vm(xn>> = MW). 2 vmog.) ”(.92 2e? so Ma) = 3.0 (e) Shack 9 (A MM'O‘IM€€( (WA Vm(9}—>
M “900 9 WsislowF 1/)
, /\ e A A
a) Show = ﬁ .W, 3356 = 3%
If n=32 and ’9‘: 7.2, Hm SE = 0.9 (3) (15% CI “For 9 ~ 9 i 23E or [5H ¥>mblam é (a) /P\ = H8Z/IOZS s M] 09> SE = W {A = MW'qfogs = 0.0156 (c) A i 28}: is an apmeWak 95% conﬁdence
WWW {w [3. We, M M or [aw J 0.50] _
no 7 0L
F" 3.1
SE OicX = V‘sﬁ" : vgg ”(“55 = 0.62 40135 (5) Win% n = 25 oéSWm/iomJ am appmxfmak
95% commence mam/6 {W ﬂ is 36 i 2.0éL1SE (We, (we 14m 15" dl‘sﬂh’Eu/im MM 217/ degrees
of“ WW7!) We 32b 5"] i 1.3 461% SOLUTlONS TO STFlT ‘400 D. MONRHD
EXAM ON JULY 26, 2007 Problem 1. (20 points)
The probability that a newborn baby will be a girl is 48.7% (a) If 100 babies are born in a city, then the chance that at least half of them will be
girls is closest to (choose one option): 0% 10% 30% 32% 40% 44% 48.7% 51.3% (b) If 400 babies are born in a City. then the chance that at least hall" of them will be
girls is closest to (choose one option): 0% 10% 30% 40% 44% 48.7% 51.3% (c) If 2,500 babies are born in a city, then the chance that at least half of them will be
girls is closest to (choose one option): 0% 30% 32% 40% 44% 48.7% 51.3% (d) If 15,000 babies are born in a city, then the chances that at least half of them will
be girls is closest to (choose one option): 10% 30% 32% 40% 44% 48.7% 51.3% Tlu number 01: girls X all of n newborn babies
is binomial, 51:01 P) M p= 0.487. (a) It n=too) PO< 250) = P(X>W5) ~
1__ @(M) = l __ @(qqsHm) —.
.— Vn—prrrts: Lima
<E(o.ie) = 00.5456, = 0.4%! (531% n=uoo P(Xzzoo) P(X>m5)
l" @(Jﬂ——_ qqqH'qug l"@(0."l7) = 0.3l92. (1c) 14 n=2500l P(X31250) = P0“ «2%,5) x
1_§(W) = I~©(L28) = 0.1003
(A) It «145000, {>(Xz 7500) = P(X>7qqq,5)ﬁ PMW) = (—95(3.!8) = 0.0007 Problem 2. (10 points) Let 6 > 0 be an unknown parameter and let XI , X2,...,Xn be be 11 independent
continuous random variables, each with the p.d.f. f(x ;(9)= H/x for >6. (a) TM (Mali hood MCHM is
MO): {(9916) fit36) ' " {(99} 9)
29/0“ Xz' Er 6‘99 9599.,“ 659% ((7) Since #10 Mh02; Mann/1 Ls Malawi/Ice; the
HLE ts “:th (Gt/13,246 Paid): M re, esbicboms emX ...es Problem 3. (15 points) (a) Com ute the mean ,u of the continuous distribution with .d.f.
P P
f(x;€) 2—617XU’6W, for O <x <1, where the constant (9 is greater than 0. (b) Given n independent observations from this distribution, Xx, X3,...,Xn , find the method—of—moments estimator 5 of the unknown parameter 6’. (c) Compute 67 if $20.12. ((0 ft =_o§o><f(9<'9)4)< = 5015‘” *1/64“ = e1 00 (in) Soiving {or 9 we 30" 9= (i—ﬂ)//IL . Home by —>’€)/X (a) It home Pm §= 0.88/0JZ = 733 Probie/Wx LN“) 13 O at 1849
HQ =_§o Hagmow = { J _ (5) Y1 how. pot? (See. W $17)
91%) = “I1F’(a>]"“{(a;e> = men/aw {or g2 6 .
(c) E00 zitgram = iii9'73" dz; l HM) If h? >3 “new
E<Y12)= :3 236334; = fnGVgMalgn = 719292 vath)= EM) (E(YeZ—=)) Mg”: a?" 5m EtY) +9 A Y (Y)
(e) Magi—>00 ti; {plan/M «M M Y, is a
musk/mic ash/Maniac of 8‘. Problem 5. (10 points)
Viral infections contracted early during a woman’s pregnancy can be very harmful to the
fetus. One study found that among 202 pregnancies complicated by a first—trimester German measles infection, 86 resulted in the death of the fetus or in major birth defects. (a) Estimate the proportion p of firsttrimester German measles infections that result in the death of the fetus or in major birth defects. (at) 3 = 8é/2OZ = 0.425 “’5 SE at? =‘/————————OL’ZZZC;ZO57L’ a: 0035 $1: 2'<5Eof‘/{S\) or mg i 0.07 is am apmef/mak 95% MGM/Vice Mat/we
for p. Pleem é
(a) Me) = {(x,;e>f(xz,e) Mm?) _ __:1__ (xe)/e._1_ (19)/9H__1_ (1—93/6
‘ e X‘ 6 X2 6 X“ :. 9'” éxmg‘“ Kn 3049M} 9>O. (0 M6) = 41W) + 1392,“ We»
Th (AlzzI'KooA eiuakm Is "'MJMQB = “9‘ " ‘57:: 610(th
Wm M m 50am 9 = "—2," MM.
TM MU: is @ =“T1TZH 6w(X;)
(a) E<é>= —— %Z."E[MX;)] = ~Em0<>I
= —50A(x)§(xe)ac< _._
= [(’m(x§ X4/e]0'+ +1XSO X—[XVedX : [9X1/el=9.
Sma E163 = 9) 9 Is am mbimwﬁ wk/malvroﬁe.
(on W63 = (la)ZZTVMWXA) = %—Vm(6n0<)>. EWXV] = 33%»? mew = 291
 (Inkgmkom 19g park.) rmblcm b, Conﬁngﬂd
vmwm) = EHMX»? ~ (Ewan)a e = "(E—[Mm + Wm MM] Is agpmxink/g havmaX MM WW 6, 24m
aPmeW/an—Q 0! o Wharf/mu EVI‘k’AUZt/Q ﬁre 23 E5 is 2(5509 €53) /\ I? V1= éq MA 9 = Sé we 31% MI; inlmua)
5Q i1.” ...
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 Spring '05
 TBA
 Statistics, Probability

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