SolHW3_S06

SolHW3_S06 - 5 a 1.0 5! o : 5’4’3'2'l 4.4x . 0= ~ -...

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Unformatted text preview: 5 a 1.0 5! o : 5’4’3'2'l 4.4x . 0= ~ - =—- - =—— ' =' a I,“ (0)“)(3) Maul) l_5_4_3_2_l(ixooz43) 00243 = 5 l 5—1: 5! = PU) mung) ~IE4!(.7)'(.3)‘ .02335 _ 5 3 5.2_ 5! z _ p(2)— [Jon (.3) — EU) (.3)’ — .1323 : (7)13)“ = i (.7)’(.3)l = .3087 Pa)=[ J 3‘31 12(4)= 5 (7)13)“ = i (may = seals 4 ' ' 4m ' ' ' = 5 5 L5=i s = L 9(5) (5)67) (.3) 5m (.7)(.3)" .sto7 b. u=np= 5(.7)= 3.5 a= 1|[Irpq = .i5(.7)(.3 = l.0247 c. #3: 26 :3 3.5 3: 20.0247) :3 ( | .4506, 5.5494) 3 o H, 3! u , 3-2-l 3 4.5 _ = . . = -— . . = . =. 0 a 17(0) (0)0)(7) omtmv) I31.1mm) 343 - I _ 3! 2. pm— (3)0)“ — BMW) —.441 5! J n_ fit” (.7) —.027 4.62 a. Let: = number ofdemocratic regimes that allow a free press in 50 trials. For this problem,p = .8. p = 5(1) = up = 50(3) = 40 We would expect 40 democratic regimes out ofthe 50 to have a free press. a= .[npfl — p = 50(.8)(.2) = 2.828 We would expect most observations to fall within 2 standard deviations ofthe mean: u i 20: 40 i 2(2.828) = 40 i 5.656 2 (34.344, 45.656) We would expect to see anywhere between 35 and 4S democratic regimes to have a free press out ofa sample ofSO. b. Let: = number ofnon-democr-atic regimes that allow a free press in 50 trials. For this ‘ problem, p = .l. ‘ i y=E(x)=np=50(.l)=5 We would expect 5 non—democratic regimes out of the 50 to have a free press. a= .lnpa—p) = 50(,l)(.9) =2.121 We would expect most observations to fall within 2 standard deviations of the mean: piZd: 5 i 2(2.|2l):> 5 i424] 2 (0.757, 9.243) We would expect to see anywhere between I to 9 non-democratic regimes to have a free press out ofa sample of50. ' 4.78 From Table ll, Appendix A, with n = 25 andp = .05, l, p(0) = P(x S O) = .277 j Ml)=P(.rsl)—P(x$0)=,642—.277=.165 . p(2)=P(xS 2) —P(xs l)= .873 — .642 = .23l l= ,u= up = 25(.05) = [.25 -_4_ -'-”=, 7 [7(0) ’1 0' e 28 | «I15 p(l)= ['25: I.25e‘”’— 358 N v II | | N N a». 4.86 5.4 From Exercise 5.3,](x) = .05 b. P(x 5 3) = .010 using Table III, Appendix A with l= IO. Yes. The probability ofobserving 3 or fewer crimes in a year if the mean is still ID is extremely small. This is evidence that the Crime Watch group has been effective in this neighborhood. (l05x530) P(|05x525)=(25 -— |0)(.05)= .75 H20 <x < 30) = (30 — 20x05) = .5 P(x 2 25) = (30 — 25x05) = .25 ms 10)=(10— 10x.05)=0 ms 25)= (25 7 10x.05)= .75 P(20.5 SK 5 25.5) = (25.5 — 20.5)(.05) = .25 5.14 Let .r = length oflime a bus is late. Then x is a uniform random variable with probability distribution: 5.22 l Jix1= [E b. P(xz 19)=(2o— 19). (—J C. 0+20 _ fl: _ (051520) 0 otherwise |0 2 =— =.05 20 20 It would be doubtful that the director’s claim is true, since the probability ofthe being more than l9 minutes late is so small. Using Table [VI Appendix A: 2.. P(z> L46) = .5 P(0 <25 L46) = .5 - .4279 = .072] P(.r < -|.56)= .5 — P(—l.56 5 z < 0) = .5 — .4406 = .0594 P(.67 $25 2.4l) =P(0 <25 2.4l)—P(0 <z< .67) = .4920 — .2486 = .2434 P(—l.96 s .- <—.33) =P(—l.96s:<0)-P(—.33 52<0) =.4750—.1293 = .3457 P(22 0) = .5 P(—2.JJ <1 < 1.50) = P(—2.33 <2 < 0) + P(0 <2 < 1.50) = .4901 + .4332 = .9233 P(z 2 —2.33) = P(-2.33 s 25 0) + P(z 2 0) .4901 + .5000 .9901 P(z < 2.33) = P(z 5 0) + P(0 5 z 5 2.33) = .5000 + .490l = .990l "/ //// 1 - 12- 5.28 a. P(10$x$12)=l’[o Hst 2”] =P(-0.50$z$0.50) . =AI+AZ J .I =.l915 +.l9|5=.3830 izs 2 2 — — I b. P(6st10)=P{6 H '0 I] = P(—2.50 s z s —o.50) . =r(—2.505250) 490505250) =.4933-.19|5=.3023 525 c. P(lssxs16)=P['3‘” '6‘”) =P(|.OOSzSZ.50) u” x =l’(0 st 2.50) —P(0SzS 1.00) = .4938— .3413 = .1525 d. I’(7.8 Sx S 12.6) = (7.8—11325126—11] 2 2 = P(—1.60 5 2 S 0.80) =A, +A: = .4452 + .288I = .7333 e. P(x213.24)= {egg—1'1] =P(221.12) k L12 =A:=.5 —A. = .5000 - .3686= .1314 f. P(x27.62)= 11(22 752‘”) 2 = P( z 2 -1.69) =A| +142 = .4545 + .5000 = .9545 5.30 a. P(x2xn)=.5:P[zzx";30] ' 1 5 i = P(x 2 zo) = .5 I a 10 = 0 = I" :50 a 5 3x0=8(0)+30=30 b. P(x<xo)=.025: P(z< x";30] i = P(z < za) = .025 m A. = .5 — .025 = .4750 1 Looking up the area .4750 in Table IV gives to = 1.96. I. o Since 20 is to the lefl of0, 20 = —l.96. x0 —30 Za=—1.96= :xo= 8(—1.96) +30= 14.32 c. P(x>xa)=.10: P(z> x";30] = P(1 > 20) = -|0 A. =.5 — .10 = .4000 Looking up the area .4000 in Table 1V gives 20 = 1.28. — o z.,=|.2zz=""x3 =xo=8(l.28)+30=40.24 d. P(x>.\'o)= .95 = I’(z> ’2”) = I’(z > za) = .95 AI = .95 - .50 = .4500 Looking up the area .4500 in Table IV gives 20 = 1.645. Since 20 is to the lefl of0, 29 = —l.645. xfl—30 za=—1.645= 3x0=8(—1.645)+30=16.84 5.40 a. Using Table IV, Appendix A, with ,u = 24.l and a= 6.30, 20 —24.l P 220 = 2—~ (x ) P(z 6.30 = .2422 + .5 = .7422 J = P(z 2 —.65) = P(—.65 s z s 0) + .5 b. P(xS 10.5)= P(zsL‘EE—‘L—l] =P(zS—2.16)=-.5 —P(—2.16SzS0) =.5—.4846=.0l54 c. No. The probability of having a cardiac patient who participates regularly in sports or exercise with a maximum oxygen uptake of 10.5 or smaller is very small at = .0l54). it is very unlikely that this patient participates regularly in sports or exercise. 5.44 a. Let x = gestation length. Using Table IV, Appendix A, 275.5—280 276.5—280 —— < z < —— P(275.5 <x<276.5)= P( J=P(—.23 <z<—.|8) 20 20 = .09I0— .07l4 = .0196. b. Using Table IV, Appendix A, P(258.5 <x<259.5) = M<z <M] =p(-|.os <t<—l.03) 20 20 3 = .3599 — .3485 = .0] l4. C. Using Table IV, Appendix A, . — . —28 P(254.5 <x < 255.5) = P(~———254 :0 280 < z < ————255 :0 0] = P(—l .28 <2 < —l.23) = .3997 — .3907 = .0090. e. If births are independentI then P(baby l is 4 days early A baby 2 is 2| days early ('1 baby 3 is 25 days early) = P(baby l is 4 days early) P(baby 2 is 2] days early) P(baby 3 is 25 days early) = .0l96(.0114)(.0090)= .00000201. 5.66 When n is large, there often are not Binomial Tables for finding probabilities. lfn is large, then finding the probabilities ofevents by using the formulas can be quite tedious and time consuming. The normal approximation gives very good estimates of the true probabilities. 5.70 a. Using Table 1!, P(x s H) = .345 .u = "P = 25(5) = l2.5. a= $ng = 25(.5)(.5) = 2.5 Using the normal approximation, (l|+.5)—12.5 P(xSll)= P(zS 25 J = P(zS —.40) = .5 — .l554 = .3446 b. UsingTable |l,P(x2 l6)= l —-P(xS l5)= l — .885 =.l l5 Using the normal approximation, P(x2l6)=P[22“6—_'25)5:1—23]=P(22L2)=.5 —.3849= .l l5l (from Table lV, Appendix A) c. Using Table ll, P(8 S x S I6) = P(x S 16) — P(x S 7) = .946 — .022 = .924 Using the nonnal approximation, my“). 2.5 2.5 = p(—2.o s z s 1.6) = .4772 + .4452 = .9224 (from Table lV, Appendix A) 5.78 Let x = number of parents who condone spanking in ISO trials. Then x is a binomial random variable with n = 150 andp = .6. ,u = up = l50(.6) = 90 a: m =./150(.6x.4) =J3_6 =6 P(xszo)=p[zs£‘”:i9]=P(zs—i l.58)=0 ...
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SolHW3_S06 - 5 a 1.0 5! o : 5’4’3'2'l 4.4x . 0= ~ -...

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