I calculate the unemployment rate as (In Labor Force Employed)/In Labor Force

Transcription

1 Introduction to the Prctice of Sttistics Fifth Edition Moore, McCbe Section 4.5 Homework Answers to 98, 99, 100,102, 103,105, 107, 109,110, 111, 112, 113 Working. In the lnguge of government sttistics, you re "in the lbor force" if you re vilble for work nd either working or ctively seeking work. The unemployment rte is the proportion of the lbor force (not of the entire popultion) who re unemployed. The tble is for people ged 25 yers or older. The tble is in thousnds of people find the unemployment rte for people with ech level of eduction. How does the unemployment rte chnge with eduction? Explin crefully why your results show tht level of eduction nd being employed re not independent. Highest Eduction Totl Popultion In Lbor Force Employed Did not finish high school 28,021 12,623 11,552 High School but no college 59,844 38,210 36,249 Less thn bchelors degree 46,777 33,928 32,429 College Grdute 51,568 40,414 39,250 Totl 186, , ,480 I clculte the unemployment rte s (In Lbor Force Employed)/In Lbor Force Unemployment Highest Eduction In Lbor Force Employed Rte Employed Rte Did not finish high school 12,623 11, High School but no college 38,210 36, Less thn bchelors degree 33,928 32, College Grdute 40,414 39, Totl 125, , Notice tht the unemployment rte chnges s we look t different eduction ctegories, tht is the vlue of the unemployment rte depends on which ctegory of eduction you re considering. Thus, eduction level nd the unemployment rte re not independent.

2 Highest Eduction Totl Popultion In Lbor Force Employed Did not finish high school 28,021 12,623 11,552 High School but no college 59,844 38,210 36,249 Less thn bchelors degree 46,777 33,928 32,429 College Grdute 51,568 40,414 39,250 Totl 186, , , () Wht is the probbility tht rndomly chosen person 25 yers of ge or older is in the lbor force? P(in lbor force) = 125, , (b) If you know tht the person chosen is college grdute, wht is the conditionl probbility tht he or she is in the lbor force? P(in lbor force college grdute) = (c) Are the events "in the lbor force" nd "college grdute" independent? How do you know? t- No, since P(in lbor force college grdute) P(in lbor force); see pge 322.

3 4.100 You know tht person is employed. Wht is the conditionl probbility tht he or she is college grdute? You know tht second person is college grdute. Wht is the conditionl probbility tht he or she is employed? P(college grdute employed) = =.3285 P(employed college grdute) = = This is the nswer if you only consider college grdutes tht re ctegorized in being in the lbor force. P(employed college grdute) = = This is the nswer if you consider the entire popultion of college grdutes The probbility tht rndomly chosen student t the University of New Hrmony is womn is 0.6. The probbility tht the student is studying eduction is The conditionl probbility tht the student is womn, given tht the student is studying eduction, is 0.8. Wht is the conditionl probbility tht the student is studying eduction, given tht she is womn? P(womn) = 0.6, P(study eduction) = P(womn study eduction) = 0.8. P(study eduction womn) =? Notice tht the probbility we wnt is the opposite of the conditionl probbility given. This would normlly signl for me to crete tree digrm, but, when I tried I relize there is informtion tht is missing not llowing me to finish the tree. So, next I look to see if formul will show me the wy. Here is the formul for the conditionl probbility P(A B) = P(A nd B) P(B) this trnsfers to P(study eduction womn) = = P(study eduction AND womn) P(womn) P(study eduction)p(womn study eduction) P(womn) = (0.15)(0.8) 0.6

4 4.103 As explined in Exercise 4.60 (pge 305), spelling errors in text cn be either nonword errors or word errors. Nonword errors mke up 25% of ll errors. A humn proofreder will ctch 90% of nonword errors nd 70% of word errors. Wht percent of ll errors will the proofreder ctch? (Drw tree digrm to orgnize the informtion given.) P(nonword) = 0.25, P(ctch nonword) = 0.9 P(ctch word) = Question: P(ctch) P(ctch) = 0.25(0.9) (0.7) = nonword word ctch not ctch ctch not ctch The voters in lrge city re 40% white, 40% blck, nd 20% Hispnic. (Hispnics my be of ny rce in officil sttistics, but in this cse we re speking of politicl blocks.) A blck myorl cndidte nticiptes ttrcting 30% of the white vote, 90% of the blck vote, nd 50% of the Hispnic vote. Drw tree digrm with probbilities for the rce (white, blck, or Hispnic) nd vote (for or ginst the cndidte) of rndomly chosen voter. Wht percent of the overll vote does the cndidte expect to get? P(white ) = 0.4, P(blck) = 0.4, P(Hispnic) = 0.2 P(get vote white) = 0.3, P(get vote blck) = 0.9, P(get vote Hispnic) = 0.5 The question is P(get vote) =? P(get vote) = 0.4(0.3) + 0.4(0.9) + 0.2(0.5) = 0.58

5 4.105 At self-service gs sttion, 40% of the customers pump regulr gs, 35% pump midgrde, nd 25% pump premium gs. Of those who pump regulr, 30% py t lest $20. Of those who pump midgrde, 50% py t lest $20. And of those who pump premium, 60% py t lest '$20. Wht is the probbility tht the next customer pys t lest $20? P(regulr) = 0.4, P(midgrde) = 0.35, P(premium) = P(py $20 regulr) = 0.3 P(py $20 midgrde) = 0.5 P(py $20 premium) = 0.6 The question is P(py $20) =? regulr mid py $20 py < $20 py $20 py < $ P(py $20) = 0.4(0.3) (0.5) (0.6) = prem 0.6 py $ In the setting of Exercise 4.105, wht percent of customers, who py t lest $20, pump premium? (Write this s conditionl probbility.) P(pump premium py $20) = 0.25(0.6) 0.4(0.3) (0.5) (0.6) = Albinism. People with lbinism hve little pigment in their skin, hir, nd eyes. The gene tht governs lbinism hs two forms (clled lleles), which we denote by nd A. Ech person hs pir of these genes, one inherited from ech prent. A child inherits one of ech prent's two lleles, independently with probbility 0.5. Albinism is recessive trit, so person is lbino only if the inherited pir is () Beth's prents re not lbino but she hs n lbino brother. This implies tht both of Beth's prents hve type A. Why? Becuse if Beth s prents re not lbino, but she hs n lbino brother (), then ech prent must hve the llele.

6 (b) Which of the types, A, AA could child of Beth's prents hve? Wht is the probbility of ech type? Since ech type is eqully likely, nd ech is inherited independently of ech other then A P(AA) = 0.5(0.5) = ¼, P(A) = (0.5)(0.5) + (0.5)(0.5) = ½, nd P() = ¼. A A AA A (c) Beth is not lbino. Wht re the conditionl probbilities for Beth's possible genetic types, given this fct? (Use the definition of conditionl probbility.) Notice tht we cn rule out, since Beth is not n Albino. So hving ruled out tht possibility the new probbilities re: P(AA not ) = 1/3. P(A not ) = 2/3. A A AA A A Albinism, continued. Beth knows the probbilities for her genetic types from prt (c) of the previous exercise. She mrries Bob, who is lbino. Bob's genetic type must be. () Wht is the conditionl probbility tht child of Beth nd Bob is non-lbino if Beth hs type A? Wht is the conditionl probbility of non-lbino child if Beth hs type AA? ' BOB BOB Beth A A A Beth A A A A A A P(no child Beth A) = 1/2 P(no child Beth AA) = 1 (b) Beth nd Bob's first child is non-lbino. Wht is the conditionl probbility tht Beth is crrier, type A? I used the informtion from problem 109c to help me nswer this question. Wht I hd to relize determine ws the strting point, which turned out to be Beth s possible genetic mkeup. How did I now this? Becuse prt (), of this problem, delt with the chnces of Bob nd Beth hving n lbino child, depending (given) on the ssumed genetic mke up of Beth. I relized tht the problem, mteril ws llowing me to crete the second brnch of the tree. The question then is P(Beth A no child), the grph is below, nd then I used it to nswer the question by reding the tree. 1/2 A 2/3 1/2 No 1/3 AA 0 1 No

7 P(Beth A no child) = () 1 = Cystic fibrosis. Cystic fibrosis is lung disorder tht often results in deth. It is inherited but cn be inherited only if both prents re crriers of n bnorml gene. In 1989, the CF gene tht is bnorml in crriers of cystic fibrosis ws identified. The probbility tht rndomly chosen person of Europen ncestry crries n bnorml CF gene is 1/25. (The probbility is less in other ethnic groups) The CF20m test detects most but not ll hrmful muttions of the CF gene. The test is positive for 90% of people who re crriers. It is (ignoring humn error) never positive for people who re not crriers. Json tests positive. Wht is the probbility tht he is crrier? P(crries gene) = 1/25 Note tht I m ssuming from this point on tht we re only considering someone from Europen ncestry. 0.9 pos crry gene P(positive test crries gene) = 0.9 1/ neg P(positive test NOT crries gene) = 0 P(crries gene positive test) =? 24/25 /3 Not cg 0 1 neg pos You cn logiclly see tht the nswer is 1, since the test is never positive if you re crrier. Creting tree digrm will llow you to see this s well. P(crries gene positive test) = 1 ( 0.9) ( 0.9 ) + (0) = Cystic fibrosis, continued. Json knows tht he is crrier of cystic fibrosis. His wife, Julinne, hs brother with cystic fibrosis, which mens the probbility is 2/3 tht she is crrier. If Julinne is crrier, ech child she hs with Json hs probbility ¼ of hving cystic fibrosis. Is she is not crrier, her children cn not hve the disese. Json nd Julinne hve one child, who does not hve cystic fibrosis. This informtion reduces the probbility tht Julinne is crrier; given tht she nd Json hve one child who does not hve cystic fibrosis. This problem involves Julinne which my crry the gene for Cystic Fibrosis. The known fcts re tht her husbnd Json is crrier, her brother is lso crrier, nd tht they hve one child who does not hve the cystic fibrosis. P(crry gene) = 2/3 P(cystic F. crry gene) = ¼, P(cystic F. not crry gene) = 0

8 The question is given tht their child does not hve cystic fibrosis, wht is she is still crrier? P(crry gene not cystic F.) = () = Musculr dystrophy. Musculr dystrophy is n incurble muscle-wsting disese. The most common nd serious type, clled DMD, is cused by sex-linked recessive muttion. Specificlly: women cn be crriers but do not get the disese; son of crrier hs probbility 0.5 of hving DMD; dughter hs probbility 0.5 of being crrier. As mny s 1/3 of DMD cses, however, re due to spontneous muttions in sons of mothers who re not crriers. Toni hs one son, who hs DMD In the bsence of other informtion, the probbility is 1/3 tht the son is the victim of spontneous muttion nd 2/3 tht Toni is crrier. There is screening test clled the CK test tht is positive with probbility 0.7 if womn is crrier nd with probbility 0.1 if she is not. Toni s CK test is positive. Wht is the probbility tht she is crrier? P(son hs DMD mom crrier) = 0.5 P(dughter crrier of DMD mom crrier) = 0.5 P(son hs DMD mom not crrier) = Toni hs one son who hs DMD P(CK test positive mom is crrier) = 0.7 P(CK test is positive mom not crrier) = 0.1 P(Toni crrier test positive) =? After I wrote ll this down, nd I begn constructing the tree, I relized tht I did not need much of the informtion tht ws given. P(Toni crrier test positive) = 2 ( 0.7 ) (0.7) ( ) =