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3+; __(‘ULrt’u/m ___0!_U€W1M_ _CW @11 a particular ﬁrm 10% ofexecutives are “empty suits” (have no knowledge ofthc product and are generally unhelpful, but
specialize in handshaking, backslapping, and golfing). Let X = the number of empty suits in a random sample M3. 3) What are the possible values ofX and their respective probabilities? X Probabilitv
0 “g _ —
(3‘) (115111 , .72a ‘f 1;:
l 1? L L l l
(“M (‘31 1’3 1 077
aﬂf'ﬂalli ‘017 3 .ool (214190?“ .001 h) Draw 21 Probability Histogram oﬂthis random variable X 1:) What is the standard deviation ofthis random variable X? .{—\ hrj): Kﬂlfﬁ >=r=nP/A‘)\lm=
@ ormal Approximation to the Binomial (section 512). Consider the ﬁrm 1n question 1. Again assume 10% of 1ts cxecuthes
are erupt} suits, but now let X= the n11111bc1 ofcmpty suits in. random samples of 100. Mt l’l‘ If): 900/) l ~ /0
G“ 11 1 i H ‘ ' 3
a) Use the Normal approximation to the Binomial to ﬁnd the probability that X 15 greater or GKetplialltloC g and lgthan or equal to
‘ 5 (v1 ithout using the continuity correction — u ithout adjusting/or the fact that Binomial is dis rete, while normal 13 :mitinuous), P(S<X<IS). K‘ .5) p 5 ~ )0 I 8 ~ 1% l3{s“éxéls) <2<ﬁt§1 3
" 13 % < 2 < 53 = [3 (241.57} 13(2< ~L67) 1) Now use the Normal approximation to the Binomial to ﬁnd the probability that X is greater or equal to 4.5 and less than or
quul to l5. 5 (using the eontimStV; co1rection) P(415 <X 15 5). 1;;10<Z<’5:”’0 P{l83 <Z<l§>b
p{z<12g) Pf? 4231] Fleet/303% ) Use the normal approximation to ﬁnd the probability that X — 10, (with continuity correction  this is necessary to ﬁnd an
stimate of the probability of one value) 1 e ,P(9. 5<)( 10 5). 1': '5—1’042< “13—“ .~ 73(17.<2<17 Q: ’3)
5'0
1p(2<17)—p[2l<—1/L_\@
171:0 ‘ .5674" HRH bl : ‘4525— 01/7; f} 1) “QB below shows number of‘ziuto insurance customers in each category. Rezili7c total number of customers = 2000; and
nivers ﬁle a claim after an accident that requires repairs.
Filed Claim Did Not File Claim Female: 100 800 4? 190
Male: 290“ __ __9_90_ __ __~ ______1_1l{_£_)
3m) 17 W 20120 300
)Randomly SClt c1 3 customer. The unconditional probability they ﬁled a claim — / (x E0
”800 U "‘
)Thc conditional probabilitv P(Filed Claim [ Female) L03 1 f _ i D
' ' 900 .lH ‘ l tl/D l Are gender and likelihood ofﬂling 21 c laim independent) for these customer’s'. AM) Pﬂlarwll‘S‘ .lll *6 p (Clo/MA:
IRandomly select a customer. The unconditional probability the» me female '— 9 Ob : ill 3 ‘ 2/ gj’b 10 ’35 Theconditionalproba‘bilityP(Fe1nalelFiled Claim): 195 K 1 23 ; 33 90 ll“
3 ' , Supp iletes are tested for use ot'bztnned hormones. The test may be either nosrtive or not positive (negative), and the
tlete may be an actual user or nonuser. Assume the test correctly identiﬁes a user 99% of the time robabili ositive given
3199); gives false positives 5% of the time (probabilitV positive given non user' —— .05); and that 10% ofthe athletes actually 2 users (unconditional probability 0 r0 . = 10) What' IS probabili that an athlete who tests positive is actualh a user? I In: 1 be done using Bayes’ Rule on pan .1 +l LU pﬂ/ll
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MHU ) E( M) (126‘) mlutitw) ”’qu
ml 6 i i 697?
1( U\ 4.qu Wt 1M Pﬂ/3; JD __ :
‘\ fathom f’it’llfto l l’ (0?\((l0\ ‘leU// ossible values for event B? @vem A occurs with probability 0.3. If A and B ar.then what is the ran Zita") M41910 ‘ ﬂ PQUé‘i
wat Whig) age/9) lhe true percent return 0' lemmon stock of two companies is below. X=Percent return A__' Probability Y=Percent return B_Co Probability
3% .5 1% .3
4% .3 4% .3
% .2 7% .4 1) Find the mean ofthe random 7v7{iable X @ variance ' 6 i and standard deviation____ ['2’ : humidiﬁer” CSlé 371 4138* 332416375 “3* *7 “”1 335" 1) Find the mean of the random variable Y @ variance_ 6; 2. i and standard deviation_ Vbl 2 :6.
26
(3363 (Mil {£167} = 4.; 946—le 4.3 611.52 t.H(7—4,a)l= 30277 ;) Assume the returns are independent so the true linear correlation between X and Y, pxy = 0. 3; (Z i b _ 6 2  _____
What is the average return (in %) r a portfolio of S 0:. A_Co and 50% B_Co? 1 371+ ,s‘ 1/133
I.) Censider the percent‘return ofthe portfolio in c above. Find the variance< i ‘ 706‘ 2nd standard deviation
G73" ,slttf m1 of i 2 [r] OXi/giﬂ/
$56” i‘ .2§(6.Z)] ‘ 115251.).9523— ': {(703— )a   large population, 59/11 ofthe people have the ﬂu. If two people are chosen at random, what is the probability that they 0th have the flu? (write your answer to at he; ‘ ' 131 places) 029%, ) Suppose a factor) has 400 workers and 5% ofthem (20) have the ﬂu. If each person can be picked only once what is the
obabilib. that two people chosen at random both have the ﬂu? (write vour answci to at least 6 decimal places) 340 {iii 3&0 ‘ ico2%\ {HP/5 HUD (Wt («:qu @Fllp a fair coin and consider the numb
vertical axis, and number ofﬂips on   n
Nonnal or Binomial 01' @Flip a fair coin. Consider the number . 1138 required for the ﬁrst he ' e probability ot'gening the ﬁrst head on
he 66' toss (i.e., 5 tails in a row, followed by a head)? (1556  '0 [562.5— ' 5 required for the first head. If We graph probability of ﬁrst head on the
’hat type of distribution is this?
or Exponential probab' '
A, ' "u r V Consider tossing a coin. Draw the probability that the ﬁﬂead is on the Xth toss to the right (up to X = 5)
Z . a What is the probability that the first Head is
on the second toss? 8: 5’2 , 2 (f ‘ number of toss 0 get the ﬁrst “head” 1 V ' 3 < < this sum = —.
$918 Geometric series can be written; 3.0%; + If 0 a: 1, l ~1 1 _ a
. . . 7 = 9 ‘
What is the sum of the geometric series if a =.2. l‘ “”3 .1 . "l r
' ‘7 _ /7""~ _L l 2
m: 12 ...
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 Spring '07
 Guggenberger

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