2003-Fall-Midterm_s

2003-Fall-Midterm_s - More PastPaper:...

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Unformatted text preview: More PastPaper: http://ihome.ust.hk/~cs_gxx Suggestion solution for ISMT 111 Fall 2003 Mid-term Exam Question 1: [16 Marks] (a) The sample mean, standard deviation and interquartile range are 1.27, 5.52 and 6.63 respectively. (b) The z-score of ’13.8’ is (13.8-1.27)/5.52 = 2.27 > 2. This observation is regarded as an extreme value. (c) As there exists an extreme value in the sample, it may be more appropriate to use median to describe the location of the distribution of the revenue change than mean. The median is 1.30. 15 ∑x i =1 15 ∑x i =1 = 1.27 × 10 + 1.20 × 5 = 18.7 i 2 i = 290.01 + [4.78 2 × 4 + 1.2 2 × 5] = 388.60 1 18.7 2 [388.60 − = 5.11. 14 15 Combined sample standard deviation = Question 2: [18 Marks] (a) Let X be the number of defective items in the sample. X~Hypergeometric(N=20,A=5,n=6) and E(X)=n(A/N)=6(5/20)=1.5 Then, (b) Pr(X≤1)=Pr(X=0)+Pr(X=1) =(C50C156/C206)+(C51C155/C206)=0.1291+0.3873=0.5164 or 0.5165 (c) Given the first item of the sample was found to be defective, X-1 ~ Hypergeometric(N=19,A=4,n=5). The probability that the shipment is accepted is given by Pr{X-1=0|First item of the sample is defective}= C40C155/C195=0.2583 or, more directly, k/ ~c s_ (d) Given there is one defective item in the first four items of the sample, X-1 ~Hypergeometric(N=16,A=4,n=2). The probability that the specific shipment is accepted is given by Pr{X-1=0|1 item is defective in the first 4 items}= C40C122/C162=0.55 ht tp :// ih or, more directly, 1 gx x/ Pr{Acceptance|First item of the sample is defective} =(15/19)(14/18)(13/17)(12/16)(11/15)=0.2583 om e. us t.h (d) More PastPaper: http://ihome.ust.hk/~cs_gxx Pr{Acceptance|1 item is defective in the first 4 items } =(12/16)(11/15)=0.55 Question 3: [16 Marks] (a) Let P be the event that a purchase is made by the random customer and let H,U,O be the even that the random customer is of high school age, university age and older, respectively. Then, by total probability theorem, Pr{ P} =Pr{P|H}Pr{H}+Pr{P|U}Pr{U}+Pr{P|O}Pr{O} =(0.2)(0.3)+(0.6)(0.5)+(0.8)(0.2)=0.52 (b) Pr(H|P)=Pr(P|H)Pr(H)/Pr(P)=(0.2)(0.3)/0.52=0.115 (c) Pr(Hc|Pc)=Pr(Hc ∩ Pc)/Pr(Pc) and Pr(Hc ∩ Pc) =1-Pr(H ∪ P) =1-Pr(H)-Pr(P)+Pr(H∩P) =1- Pr(H)-Pr(P)+Pr(P|H)Pr(H)=1-0.3-0.52+0.2(0.3)=0.24 That is, Pr(Hc|Pc)=(0.24)/(1-0.52)=0.5 Question 4: [14 Marks] (a) The cost of Jerry’ three pizzas is 3×$70 = $210, the probability distribution of his profit s is # demanded 1 2 3 Profit ($) 120-210 = -90 240-210 = 30 360-210 = 150 Probability 0.4 0.3 0.3 ht tp :// ih om e. us t.h Likewise, if he prepares one pizza in advance, the probability distribution will be # demanded 1 2 3 Profit ($) 120-70 = 50 50-40 = 10 50-2×40 = -30 Probability 0.4 0.3 0.3 Expected profit = $14 k/ ~c s_ gx x/ (b) 0.4×(-$90) + 0.3 × $30 + 0.3×$150 = $18. (c) If Jerry stores two pizzas in advance, it will cost him $40 in the future if three pizzas are demanded, therefore, his probability distribution is # demanded 1 2 3 Profit ($) 120-140 = -20 240-140 = 100 100-40 = 60 Probability 0.4 0.3 0.3 Expected profit = $40 2 More PastPaper: http://ihome.ust.hk/~cs_gxx Jerry should stock two pizzas. Question 5: [16 Marks] (a) Let X be the amount of time needed to travel to Japan from Hong Kong. Then, X~N(5,0.252) and Pr(4.5<X<5)=Pr((4.5-5)/0.25<Z<(5-5)/0.25)=Pr(-2<Z<0)=0.477 (b) Let 5+c and 5-c be the upper bound and the lower bound of the required range, respectively. We need to solve P(-c<X-5<c)=.8 which is equivalent to Pr((-c/0.25)<Z<c/.25)=.8 That is, Pr(0<Z< c/0.25)=0.4 That gives: c/0.25=1.28 and hence, c=1.28(0.25)=0.32. Therefore, the required range is [5-0.32,5+0.32]=[4.68,5.32]. Question 6: [20 Marks] 1500 − 1700 ) = P( Z > −0.5 ) = 0.6915 400 (a) choose Brand A 1500 − 1600 P(Y > 1500) = P( Z > ) = P( Z > −0.4) = 0.6554 250 (b) P(Z < z) = 0.1 ⇒ z = −1.28; x = 1700 − 1.28 × 400 = 1188 (c) 5 2000 − 1700 P( X < 2000) = P(Z < ) = P(Z < 0.75) = 0.7734 ; (0.7734 ) 1 (0.2266 ) 4 = 0.0102 1 400 2000 − 1600 P( X < 2000) = 0.7734; P(Y < 2000) = P( Z < ) = P ( Z < 1.6) = 0.9452 (d) 250 0.7 × 0.7734 + 0.3 × 0.9452 = 0.825 0.7 × 0.7734 (e) P( A) = = 0.656 0.825 ht tp :// ih om e. us t.h k/ ~c s_ gx x/ P( X > 1500) = P( Z > 3 ...
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