Assignment 2 solution (unstarred)

# Assignment 2 solution (unstarred) - STATO301 Elementary...

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Unformatted text preview: STATO301 Elementary Statistical Methods Semester 1 2009/2010 Assignment 2 solution 1 un-starred Questions) (21) LetX be the length of a certain kind of item. We have X ~ N(8.5,0.022) P(X > 8.545) : P[X—,u > 8.545 4.5] 0' 0.02 = P(Z > 2.25) = 0.5—0.4878 = 0.0122 (b) LetX be the length of the sardines received by a cannery. We have X ~ N(11.02,0.52) (i) P(X <10) :P(X—,u <10—1102) 0' 0.5 = P(Z < —2.04) = 0.5 — 0.4793 = 0.0207 (ii) P(10.5 < X <12) =P(10.5—11.02 < X—,u <12—1102] 0.5 0' 0.5 = P(—1.04 < Z < 1.96) = 0.3508 + 0.4750 = 0.8258 (c) LetX be the demand for cash in any ordinary weekend. We have X ~ N(125000 ,15000 2) P(X < c)2 0.85 :> P X‘“ <————C~125000 20.85 0' 15000 c — 125000 :> P Z < 15000 )2 0.85 (Look up 0.35 in Table B, it gives 1.036) . c—125000 ” 15000 0 2140540 Hence, 1406 \$100 bank notes must be kept in the machine. 21.036 (d) P(,u—ko‘<x<,u+2k0')=0.8 3HW<Z<£2§£15208 0' 0' :>P(—k<Z<2k)=O.8 P(—k < Z < 2k) 0.8185 0.0185 0.8002 0.0002 0.7963 -0.0037 0.79825 -0.00175 Take k = 0.95 LetXbe the. number of couples get maimed using the dating service. We have X m Bi(125,11.25) E(;X)=125 :4 0.25 = 31.25 and Var(X)=125 x 0.25 x 0.75 = 23.4375 By NABT, .X can be appraximated by X ~ N(31..25,23.4375) (1') P(X :2 20) = Pm; 219.5) = PF; —,11 31.95—31.25] ’— 0' 11/234375 = 13(2 2 4.43) = 0.5 + 0.4925 = 0.9925 (ii) PLX g 37) = 13(3) 2:315) = Pf; ~11 g 315—3125} 0' «1,123.43 5 = 13(25 1.29) = 0.5 + (3.4015 = 0.9015 (b) (d) (e) LetX be the number of cracked marbles in the sample. We have X ~ Bi(60,0.085) E(X) = 60 x 0.085 = 5.1 By PABT, X can be approximated by X ~ Po(5.1) P(X s 4) = P(X = 0)+P(X =1)+P(X = 2)+P(X = 3)+P(X = 4) : e‘5-‘(5.1)° + e'“(5.1)1 + (“(5.1)2 + e‘5‘1(5.1)3 + e"“(5.1)“ 0! 1! 2! 3! 4! = 0.4231 According to Binomial probabilities table of n=12 and p=0.30, the most probable value of r that will occur is 3 and the probability is 0.2397. Given np = 350 and np(1— p): 105 ——————"p(1_p)=£:p=0.7 andn=3_5—9=500 mp 350 0.7 (1) 2k=3+k:>k:3 (2) By C: =C” 2k=15~(3+k):k=4 n—r 9 The sum at the end of the 6th year will be \$10000 x e6”11 2 \$19347 .9233 (a) The resulting mean, ,u = w = 310 ohms 2+1 The resulting standard deviation, a = \/[%(252 +3002)+%(202 +3302)]—310 = 4/750 = 27.38610th (b) LetX be the resistance of an item in the combined batch. We have X ~ N(310,750) P(X<,u—20'ORX>,u+20') =P(X<,u—20')+P(X>,u+20') =P(X—,u <y—20—y]+P(X—,u >,u+20'—,u) a o- o- 0' = P(Z < 2)+ P(Z > 2) = 2x P(Z > 2) = 2><(0.5—0.4772) = 0.0456 (a) (i) Let X be the number of persons going to watch the FB Cup (Final) football match. We have X ~ N(15000 .2000 2) P(X < 15440) : P£X—,u < 15440 —15000 J 0' 2000 = P(Z < 0.22) = 0.5 + 0.0871 = 0.5871 (ii) In order to make a proﬁt of at least \$90000, there must be at 90000 50 P(X > 17240) : P[X—,u > 17240 45000] 0' 2000 = P(Z > 1.12) = 0.5—0.3686 = 0.1314 least 15440 + = 17240 people going to watch the match. (b) Let X be the number of times of getting an even number. we have X ~ Bi(27,0.62) P(15 s X s 22) = P(X =15)+ P(X = 16)+ + P(X = 21)+ P(X = 22) = of; (0.62)” (0.38)12 + C1267 (0.62)16 (0.38)” + + 022,7 (0.62)21 (0.3 8)6 + C2227 (0.62)22 (0.3 8)5 =0.8054 Alternative method: Since [1] np = 27x 0.62 = 16.74 > 5 [2] n(1— p) = 27 x (1—0.62) = 10.26 > 5 [3] p=0.62:>0.1<p<0.9 By NABT, Xcan be approximated by X0 ~ N(16.74,6.3612) P(15 sX s 22) =P(14.5£Xc \$22.5) :P(14.5—16.74 < Xe ——,u 22.5—16.74] 3 76.3612 “ 0‘ 06.3612 = P(— 0.89 3 Z s 2.28) = 0.3133 + 0.4887 = 0.8020 (0.8054 is the exact probability while 0.8020 is only an approximation! ! ll!) (e) Let X be the diameter (in mm) of the lid and Y be the diameter (in mm) ofthe base. We have X ~ N(35.5,0.042) and Y ~ N(35.42,0.032) :5 X—Y ~ N(35.5 435420.042 + 0.032): N(O.O8.052) P(0.02 < X~Y < 0.15) 2 P 002-008 < Z < 0.15—0.08) 0.05 0.05 = P(—1.2 < Z < 1.4) = 0.3849+ 0.4192 0.8041 (a) Let X be the number of persons: who WiIE experience the symptoms of jet lag. We have X m Bi(8030.72) 15(3): 80 x- 0.72 = 516 and Var(X)= 80 x 0.72 x 0,28 =16.128 Bar NABT,Xca11 be approximatéd by X; m N (516,1 6.128) P-(X g 50) = Pg; 1: 59.5) = 1363f; —,.u 50.5 — 5?.6) )ﬁ‘k a _ Jim = 13(2 3 —1 ,77) =U.5—O,4616 =0.0334 (b) LetX be the number of persons who will reply. We have X ~ Bi(92,0.075) P(X > 5) = 1 — P(X s 5) =1—P(X = 0)—P(Xé1)—P(X = 2)—P(X = 3)—P(X = 4)—P(X = 5) = 1 — 032 (0.075)0 (0.925)92 — C192(0.075)1(0.925)91 ~ c? (0.075)2 (0.925)90 — C392 (0.075)3 (0.925)” ~ 0:2 (0075)“ (0.925)88 — C592 (0.075)5 (0.925)87 = 0.6961 Alternative method: E(X)= 92 x 0.075 = 6.9 By PABT, Xcan be approximated by X ~ Po(6.9) P(X > 5) = 1 — P(X g 5) =1~P(X = O)—P(X=1)—P(X = 2)—P(X§3)—P(X =4)—P(X =5) e‘6‘9(6.9)3 (0.696] is the exact probability while 0.6863 is only an approximation! /! l !) (c) P(A beats B) = P(A gets 4 correct in 4 questions, B gets 0 or 1 correct in 2 questions) + P(A gets 3 correct in 4 questions, B gets 0 correct in 2 questions) = C3 (025)“ (0.75)0 x [C§(0.25)°(0.75)2 + 012(025)‘ (0.75)] + C34(0.25)3(0.75)‘ X1C02(0.25)0(0.75)2J = 0.0300 (d) ,u = 20x 0.25 = 20 0'2 = 20x0.25><0.75 = 15 10. Number of tickets bought: C68 = 28 If each ticket costs P , total stake: 28P Consider the awards for the lSt prize is A, 2nd prize is Az ,. . .,7th prize is A7 Number of winnin_ tickets Winnin_ numbers 6 regular numbers 5 regular numbers+ lextra 5 regular numbers ' 4 regular numbers+ 1 extra C]1 x C j x C13 _ 3 - 4 regular numbers C; x C: x 023 = 3 - 3 regular numbers+ 1 extra 011 x €34 >< C23 = 12 - 3 regular numbers C3, x C: x C: = 4 Counting the number of winning tickets Consider three groups of number Group 1: {1}, Group 2: {234,5} and Group 3: {6,7,8} ix. :3; 4g; CA 7 For example, The number of winning tickets for 4th prize = Number of selecting {1} from group 1 x Number of selecting {234,5} from group 2 x Number of selecting any one number from group 3 = C11 XC: ><C13 =3 The number of winning tickets for 5th prize = Number of selecting 0 number from group 1 x Number of selecting 4 numbers from group 2 x Number of selecting any 2 numbers from group 3 = C; x C: x C23 = 3 Total win = 3.44 + 3A5 +12A6 + 4A7 — 28P 3A4 +3A5 +12216 +4A7 —28P Each person’s net win 2 5 11. I 0.1465 8.1954 (3.1954 _m Prob. Of sreliing r - books 0-9133 Number of books Profit 1 Loss stocked .2 I! 0.0?33 —a-m———_ ___-_- ___-_— ___—_- Number of banks stocked £x necteé ?rof"t 3 335159 ¥ 8.01 0‘356 . T388528 6.966795 Hence the bookstore shouici stock 3 copies per issue and the comesponding expected pmﬁt will be 3?.3680‘ ...
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