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HW_7 - 4.113 Eaeh unemployment rate is computed as Did ttet...

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Unformatted text preview: 4.113. Eaeh unemployment rate is computed as Did ttet finish H3 1 — Efl ' 11.11343 11.31.11 = £119“ 011 1.1113 right. (Alternatively. subtract 1.1112! HSJIID college 1 _ 35:39 i [”35 13 number employed from the number In the labor 3mg mnege 1 — i—dites é ””442 force, then divide that difl’erenee by the number 39:30 _ is the labor force.) Because these tstes {probabili— “0"“? firm“: ‘ tut—m = ”“233 1.1125} are different+ education level and being employed are not independent. 4.122. The given probabilities ste P(W} = 13.152, PHI) = div, and P(W |E) = 113. By the multiplication tttle. Pie stte 1v}: PUE} Pitv | E) = (13.111211113) = 0.135. Thetefere. this | iv} = 4—1? $33)” = % '= 0.2194. 4.124. With E. M. and D teprflenting the three kinds ofdegl'ees+ and W meaning the degree recipient vim a woman, we have been given P13} 2 121.73, P111”): 121.21, P113} 2 121.1316, P11? | B} = 1143, P(W Ht!) 2 1142, PUFF | D} = 1129. Therefore.we find P(W} = P(W andB] + P(W andM} + 1’1me} 2 P1B)P(W |B}+ P1M1P(W|M}+P(D)P(W | D) =1].4515. so PBaJliW PBPWB . . P13|W}=¥=%=%=fljfifi4 4.1215. {11] If Beth is Act, then the first table on the right givfi A a A A the {equally likely} allele eettthittstietts for a ehile. so P(ehild a isnon—albinolEefl'lis Act] :51me is AAJhen ssthe ‘1 A“ “a A“ 41' seeenti table shows. theit child will definitely he type Act {she non-albino}. so P[ehild is non—albino |Eeth is AA}=1.{b]lMe have Piehile is sen-slhittel = Piehile sesseBeut sel+Piehile sesseeeut AA) = Ptueut Aa)P(ehild lie |Beth se)+ PfBeth ssipiehile es |Eeth as) _31 l _2 —3'2+3'1—3- thetefete. PfBEfl'l is Act |ehild is Act) = % :1. 4.123. Let C be the event that Julianne is a carrier. and let D be the event that Jason's and Julianne‘s ehild hm the disease. We have been given P113] = %. Pt‘IDICl = 1—. and PiflIC") = 1]. Therefore. P111”) 2 P1C)P(D"|C) + P1C"}P13”|C"} = (§)(3‘)+(%)113=i+i=itm P1C|D“l=%=§. 5.14. {a} X, the number of auetion site visitors. is B(15,115). {bl Symmetry tells us that this must be 115, which is confirmed in Table C: P1X 3 3} = 111964 + 121.152? + 11139115 + 0341;: + 13.131139 + 1111132 + 11121115 2 121.5. 5.22. For E. p, = 0.49 and tr 2 Jpfl — fife i 0.01563. As 15 is approximately Normally distributed with this mean and standard deviation. we find P110445 {.- fi 4:. 0.52.] i P(—l.91 a: Z «c: 1191} £03433 {Exact calculation gives 0.9442.) 5.24. When n = 3130 the distribution of ii is approximately Notmal with mean 0.49 and standard deviation 0.02336 {nearly twice that in Exercise 5.22). When n = 51200, the standard deviation drops to 0.0]?0‘? {leg than half m big m in Exercise 5.22]. Therefore. a = 31]]: P{0.4fi 4:: g? {0.52) i P(—l.04 4: Z <1.04)é0.?016 n=5l]]]: P{0.46<:fi{0.52)éP(—4.24<Z{4.24)é1 Larger samples give a better probability that :3 will be close to the true proportion p. {Exact calculation of the first probability gives 0.?014, but this more accurate answer does not change our conclusion.) 5.3. {3] fl- : {IEIIJHCIJS} : ml] and. 0|" 2 Cflflfiflflfly cam-ti?" $225 = 15 studmts. {11] PHI 2 951} i Table Software Table Software PL? 2 3.4) = com. {1:} was it = 1300, Hunt le Normal Normal For a 951) '= FILE 2 454} = 0.9332. Dther ”-9332 ”-9379 ”-9413 ”-941? answers are shown in the table on the right. 530. {a} Mhas abinomial distribution with n = 30 and p = 0.6. so PI‘IM = 20} = @jmnfflmxtm s 0.1152. {b} Pflst woman is the 4m call) = {0.fi)3(0.4} = 0.0364. ...
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