Assignments_5_Solitions[1]

Assignments_5_Solitions[1] - Assignment 5 SOLUTIONS Ch 16:...

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Unformatted text preview: Assignment 5 SOLUTIONS Ch 16: # 2, 4, 10, 12, 26 Ch 17: #2, 20, 22, 24, 28 2. Expected value. a) it = E(Y)= C(02) + 1(0.4)1+ 2(0.4) = 1.2 b) [u = E(Y)= 100(0.1) + 200(02) + 300(05) + 400(02) = 280 4. You bet! a) } Win $100 $50 P(amount won) % b) ,u = .E (amount won) = MODE?) + $50[%) + gig-z) z $23.61 c) Answers may vary. In the long run, the expected payoff of this game is $23.61 per play. Any amount less than $23.61 would be a reasonable amount to pay in order to play. Your decision should depend on how long you intend to play. If you are only going to play a few times, you should risk less. ' 10. Variation 2. a) l 02 = Var(Y) = (0 — 1.2)2(0.2) + (1 — 1.2)2(0.4) + (2— 1.2)2(0.4) = 0.56 a=SD(Y)=W=Wz0.75 L b) 02 = Var(Y) = (100 — 280)2(0.1) + (200 — 280)2(0.2) + (300 — 280)2(0.5) + (400 — 280)2(0.2) = 7600 0 = SD(Y) = W = W x 87.18 12. The die. Answers may vary slightly (due to rounding of the mean) 02 = Var(Won) = (100 — 23.61)2[%) + (50 — 23.61)2(§6—] + (0 — 23.61)2[—:%) z 14564043 o = SD(Won) = JV'ar(Won) z #14504043 = $38.16 26. Random variables. a) I”: 0: C) u: b) ‘u = E(3X) = 3(E(X)) = 3(80) = 240 0' = SD(3X) = 3(SD(X)) = 3(12) = 36 E(2Y + 20) = 2(E(Y)) + 20 = 2(12) + 20 = 44 SD(2Y + 20) = 2(SD(Y)) = 2(3) = 6 d) it: E(X-5Y) = E(X)—5(E(Y)) = 80 — 5(12) 2 20 E(0.25X + Y) = O.25(E(X)) + E(Y) = 025(80) +12 = 32 0 =_ SD(0.25X + Y) = 0.252Var(X) + Var(Y) 0' = SD(X — 5r) = Var(X) + 52Var(Y) = 1/0.252(122) + 32 z 4.24 = 4127’— + 52(32) 4 19.21 2. Bernoulli 2. a) b) C) d) These may be considered Bernoulli trials. There are only two possible outcomes, gettingz 6 and not getting a 6. The probability of getting a 6 is constant at 1/6. The rolls are independent of one another, since the outcome of one die roll doesn’t affect the other rolls These are not Bernoulli trials. There are more than two possible outcomes for eye color. These can be considered Bernoulli trials. There are only two possible outcomes, properly attached buttons and improperly attached buttons. As long as the button problem occurs randomly, the probability of a doll having improperly attached buttons is constant at abot 3%. The trails are not independent, since the total number of dolls is finite, but 37 dolls is probably less than 10% of all dolls. These are not Bernoulli trials. The trials are not independent, since the probability of picking a council member with a particular political affiliation changes depending on WhC V has already been picked. The 10% condition is not met, since the sample of size 4 is more than 10% of the population of 19 people. These may be considered Bernoulli trials There are only two possible outcomes, cheating and not cheating. Assuming that cheating patterns in this school are similar to the patterr in the nation, the probability that a student has cheated is constant, at 74%. The trials are not independent, since the population of all students is finite, but 481 is less than 10% of a: students. 20. Frogs. The frog examinations can be considered Bernoulli trials. There are only two possible outcomes, having the trait and not having the trait. If the frequency of the trait has not changed, and the biologist collects a representative sample of frogs, then the probability of a frog having the trait is constant, at p = 0.125. The trials are not independent since the population of frogs is finite, but 12 frogs is fewer than 10% of all frogs. Since the biologist is collecting 12 frogs, use Bin0m(12,0.125). Let X : the number of frogs with the trait, from n = 12 frogs. a) P(no frogs have the trait) = P(X = 0) :(15)(0.125)0(0.875)12 z 0.201 b) _ ' ' P(at least two frogs) : P(X 2 2) 0R '— P(X>’2> 7‘ 1—? We“ 9:) : a ‘ «_." {1:}, '23: : P(X:2)+P(X:3)+..,+P(X:12) “l L? ’f "43 +53% 0"” ,1 ri—‘QCylmrs-‘a W5 jig-9.125 ~92}: ; r : 12 (0 125m 875)m + 12 0 125 3 12 i r 17 ' $9353 2 _ . 3 < . ) (0.875) +..‘+ ’12 (0.123) «018M 2 0.453 C) P(three or four frogs have trait) : P(X = 3) + P(X : 4) ‘r 2(132)(0.125)3(0.875)9 + (1:)(O.125)4(04875)8 : 0.1.71 d) P(no more than four) = P(X S 4) : P(X : 0) + P(X : 1) + . . . + P(X = 4) :(102)(0.125)°(0.875)12 + (112)(0.125)1(0.875)” + . . . + (142‘)(0.125>4(0.875 x 0.989 22. More arrows. These may be considered Bernoulli trials. There are only two possible outcomes, hitting the bull's-eye and not hitting the bull’s-eye. The probability of hitting the bull’s-eye is given, p = 0.80. The shots are assumed to be independent. Since she will be shooting 200 arrows, use Bin0m(200, 0.80). Let Y = the number of bull’s—eyes in n = 200 shots. E(Y) 2 mp = 200(080) = 160 bull’s—eyes. 50m = mm = «/200(0.80)(0.20) z 5.66 bull's—eyes. Since np = 160 and nq = 40 are both greater than 10, Binom(200,0.80) may be approximated by the Normal model, N (160, 5.66). According to the Normal model, in matches with 200 arrows, she is expected to get between 154.34 and 165.66 bull’s-eyes approximately 68% of the time, between 148.68 and 171.32 bull’s-eyes approximately 95% of the time, and between 143.02 and 176.98 bull’s-eyes approximately 99.7% of the time. Bull's-eyes made in 200 shots. 160 165.66 17132 176.98 99.7% .1 143.02 148.68 154.34 (1) Using Bin0m(200, 0.80): P(at most 140 hits) 2 P(Y S 140) = P(Y:0)+P(Y:1)+...+P(Y=140) =(280)(0.80)0(0.20)200 + (220)(0.80)1(0.20)199 + . . . + (5:3)(0.80)140(0.70)60 x 0.0005 According to the Binomial model, the probability that she makes at most 140 bull’s—eyes out of 200 is approximately 0.0005. Using N (160, 5.66): According to the Normal model, the probability that she Z 2 y *1“ hits at most 140 bull’s—eyes out 14% 160 Bull,s_eyes of 200 is approximately 0.0002. ~ — made 111 Z — 0.0002 200 shots z z —3.534 140 l60 z = — 3.534 P(Y S 200) z P(z < —3.534) x 0.0002 Using either model, it is apparent that it is very unlikely that the archer would hit only 14 ' bull’s~eyes out of 200. 24. Apples. a) A binomial model and a normal model are both appropriate for modeling the number of cider apples that may come from the tree. Let X = the number of cider apples found in the n = 300 apples from the tree. The quality of the apples may be considered Bernoulli trials. There are only two possible outcomes, cider apple or not a cider apple. The probability that an apple must be used for a cider apple is constant, given as p = 0.06. The trials are not independent, since the population of apples is finite, but the apples on the tree are undoubtedly less than 10% of all the apples that the farmer has ever produced, so model with Bin0m(300, 0.06). E(X) : np : 300(006) = 18 cider apples. SD(X) = «Inpq : «l300(0.06)(0.94) : 4.11 cider apples. Since np = 18 and nq = 282 are both greater than 10, Bin0m(300,0.06) may be approximated by the Normal model, N (18, 4.11). b) Using Bin0m(300, 0.06): P(at most 12 cider apples): P(X S 12) : P(X = 0) +."..+ P(X : 12) :(380)(O.06)°(0.94)300 + . . . + (31°20)(0.06>12(0.94)”2 2 0.085 According to the Binomial model the b ' ‘ ' , pro abilit that no from the tree is approximately 0.085. y more than 12 Clder apples come Using N(18, 4.11): Z: x_# P(X312) zP(z<—1.460) z0.072 a . 12 _ 18 Cider apples According to the Normal Z Z 4 11 0.072 Egg/lg: model, the probability that no _ . more than 12 apples out of 300 z —— ~1.460 are cider apples is 12 18 approximately 0.072. z = 1.460 c) It is extremely unlikely that the tree will bear more than 50 cider apples. Using the Norm“ model, N (18, 4.11), 50 cider apples is approximately 7.8 standard deviations above the mean. 28. The euro. Let X = the number of heads after spinning a Belgian euro 11 = 250 times. Using Bin0m(250, 0.5): These may be considered Bernoulli trials. There are only two possible outcomes, heads and tails. The probability that a fair Belgian euro lands heads is p = 0.5. The trials are independent, since the outcome of a spin does not affect other spins. Therefore, Binom (250, 0.5) may be used to model the number of heads after spinning a Belgian euro 250 times. ‘ P(at least 140) : P(X 2 140) ' :P(X:140)+...+P(X=250) =(f:§)(0.5)140(0.5)110 + . . . + (igmsflmsf 2 0.0332 According to the Binomial model, the probability that a fair Belgian euro comes up heads at least 140 times is 0.0332. Using N (125, 7.91): E(X) 2 mp = 250(005) : 125 heads. SD(X) = 1lnpq = 1/250(0.5)(0.5) z 7.91 heads. Since np = 125 and nq = 125 are both greater than 10, Bin0m(250,0.5) may be approximated by the Normal model, N (125, 7.91). P(X 2 140) z P(z > 1.896) z 0.0290 Z: XT‘U Numberof G h32§08s$ilrt1§f 0-0290 According to the Normal _ 140 ‘ 125 model, the probability that a Z — 7.91 fair Belgian euro lands heads 2 at 1.896 125 140 at least 140 out of 250 spins is z = 1396 approximately 0.0290. Since the probability that a fair Belgian euro lands heads at least 140 out of 250 spins is low, it is unlikely that the euro spins fairly. However, the probability is not extremely low, and we aren’t sure of the source of the data, so it might be a good idea to spin it some more. ...
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Assignments_5_Solitions[1] - Assignment 5 SOLUTIONS Ch 16:...

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