Number of women 0.15

Transcription

1 . Grouped Data (a Mdponts Trmester (months Number o women Relatve Frequency Densty.5 [0, /400 = /3 = [3, /400 = /3 = [6, /400 = /3 =

2 (b To calculate the proporton contaned n [, 4, rst notce that the nterval [, 3 comprses two-thrds o the rst class nterval [0, 3. Smlarly, the nterval [3, 4 comprses one-thrd o the second class nterval [3, 6. Thereore, together, the sample proporton n the nterval [, 4 s equal to ( (0.5 = (See gure, let. 3 3 (c Quartles Frst, n order to splt the total area nto equal halves o 0.5 each, the medan must clearly be located n the rst class nterval [0, 3 n such a way that t dvdes the correspondng rectangle o 0.6 area nto let and rght areas o 0.5 and 0., respectvely. Snce 0.5 s ve-sths o 0.6, the nterval [0, 3 must also be dvded nto ths same proporton. Thus, the medan must be equal to (3,.e., Q =.5 mos. (See gure, rght. Snce the nterval [0,.5 contans eactly 0.5 o the sample, t thereore ollows that the rst quartle must dvde ths nterval nto equal halves at ts mdpont,.e., Q =.5 mos. Fnally, snce the last class rectangle already contans 0.5 o the sample, Q 3 = 6 mos Mean = 400 usng mdponts o class ntervals [(.5(40 + ( 4.5(60 + ( 7.5 (00] = 3.45 months Varance s = 399 [( (40 + ( (60 + ( (00]

3 . Cty resdent Bob has been nvted to three partes on the east sde on the same nght, but cannot mae up hs mnd about whch party to attend, so he decdes to let random chance mae the selecton. At each subway staton, he smply boards the rst eastbound tran that arrves. The probablty that each tran ndependently arrves rst at ts correspondng staton s tabulated below. (Assume only one tran per trac as labeled n the map, n the drectons ndcated by the arrows. Tree Dagram Tran Frst Arrval Prob A 0.0 B 0.30 Bob A B C D E I J C 0.50 D 0.50 E 0.50 F 0.0 F G 0.30 G H 0.60 H I 0.0 J 0.80 (a Lst all o the ordered outcomes n the sample space or the eperment Bob goes to a party. As a generc eample, n order to get to a partcular party, Bob must rst tae Tran X, ollowed by Tran Y, then the correspondng outcome would be the ordered par (X, Y. For three trans, the outcome would have the orm o an ordered trple (X, Y, Z, etc. (5 pts S Party Party Party 3 { A, ( B, D, ( B, E, I, ( C, F, I, ( B, E, J, ( C, F, J, ( C, G, ( C, H} (b Calculate the probablty that Bob wll attend Party : P( A 0.0, and va ndependence, PBD (, (0.3( , (3 pts P( B, E, I (0.3(0.5( and P( C, F, I (0.5(0.( Thereore P(Party = = Party : Va ndependence, P( B, E, J (0.3(0.5(0.8 0., (3 pts P( C, F, J (0.5(0.( and P( C, G (0.5( Thereore P(Party = = 0.3. Party 3: Va ndependence, P( C, H (0.5( ( pts PROBLEM 3 CONTINUES ON NEXT PAGE

4 (c Calculate the probablty that Bob attends Party, gven that he rst taes a B tran. (3 pts P(Party B = P( Party B P( B, D P( B, E, I P( B P( B (d Suppose one o Bob s rends ndependently uses the same method to attend one o the three partes. What s the probablty that they wll both end up at the same party? ( pts Frst, P( Bob goes to Party Bob s rend goes to Party = P(Party P(Party va ndependence, = (0.39 = 0.5 rom (b. Lewse, P(Both go to Party = (0.3 = and P(Both go to Party 3 = (0.3 = Thus, P( Both go to Party Both go to Party Both go to Party 3 = = (e What s the probablty that they wll both end up at the same party, usng eactly the same subway route (.e., ordered outcome o trans? ( pts Frst, va ndependence, we have P( Both go to Party usng the same route = P( A P( A P( B, D P( B, D P( B, E, I P( B, E, I P( C, F, I P( C, F, I = (0.0 (0.5 (0.03 ( Lewse, P( Both go to Party usng the same route = P( B, E, J P( B, E, J P( C, F, J P( C, F, J P( C, G P( C, G = (0. (0.04 ( And nally, P( Both go to Party 3 usng the same route = P( C, H P( C, H ( Thus, P( Both go to same party usng the same route = = 0.9.

5 3. Dene events: A = Adult B = Male P(A = 0.5 P(B = 0.5 P(A B = 0.6 (a Snce ,.e., P(A P(A B, t ollows that events A and B are not statstcally ndependent. (b For any two events A and B, recall that P(A B = P(A B P(B, so P( Man = P(A B = (0.6(0.5 = 0.5 (5%. (c For any two events A and B, recall that P(A B = P(A + P(B P(A B, so that P(A B = = 0.6; thereore P( Grl = P(A c B c = 0.6 = 0.4 (40%. PB ( A (d Usng (b, P( Man, gven Adult = P(B A = = 0.5 = 0.3 (30%. PA ( 0.50 (e Venn dagram: From the gven, we now that P(A = 0.5, and rom (b, P(A B = 0.5, so t ollows that P( Woman = P(A B c = = Lewse, rom the gven, we now that P(B = 0.5, so t ollows that P( Boy = P(A c B = = 0.0. And nally, rom (c, P( Grl = P(A c B c = 0.40, as shown. A = Adult B = Male

6 Event C = Whte A and C are statstcally ndependent P(A C = P(A P(C B and C are statstcally ndependent P(B C = P(B P(C P(C = 0. P(A B C = 0.05 ( From above, P( Whte Adult = P(A C = P(A P(C = (0.5(0. = 0.0 (0% (g From above, P( Whte Male = P(B C = P(B P(C = (0.5(0. = 0.05 (5% (h From above, P( Whte Man = P(A B C = P(A B C P(C = (0.05(0. = 0.0 (% ( P( Whte, gven Man = P(C A B = PC ( A B PA ( B = 0.0 va (h 0.5 va (b = (6.67% (j Venn dagram: From (h, the trple ntersecton has probablty P(A B C = 0.0. Subtractng ths rom (b: P(A B = 0.5, (: P(A C = 0.0, and (g: P(B C = 0.05 leaves P( Nonwhte Man = 0.4, P( Whte Woman = 0.09, and P( Whte Boy = 0.04, respectvely. Subtractng these values rom nown probabltes P(A = 0.5, P(B = 0.5, and P(C = 0. (all gven above yelds P( Nonwhte Woman = 0.5 ( = 0.6, P( Nonwhte Boy = 0.5 ( = 0.06, and P( Whte Grl = 0. ( = Fnally, subtractng ths last value rom P( Grl = 0.4 (va (c leaves P( Nonwhte Grl = 0.34, as shown. A = Adult B = Male C = Whte

7 4. Consder a random sample o n values, arranged n the requency chart shown below, 3 3 n (a The sample mean can be calculated va the ormula n. Usng ths, provde a ormal algebrac proo that the sum o the devatons rom the mean s always equal to 0. That s, prove the ollowng statement: (6 pts ( 0 Proo: ( ( n n n 0 (b The sample varance can be calculated va the ormula provde a ormal algebrac proo the alternate or shortcut ormula: ( n s. Usng ths, (9 pts ( n s n Hnt: ( a b a ab b Proo: ( ( n n s n n ( n n n n n n