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(ISMT111)Mid03sol.pdf
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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.

(d)
. xi = 1.27 10 +1.20 5 = 18.7


15
i=1
15
2 22

. xi = 290.01+ [4.78 4 +1.2 5] = 388.60
i=1
Combined sample standard deviation =

Question 2: [18 Marks]
(a)
Let X be the number of defective items in the sample. Then, X~Hypergeometric(N=20,A=5,n=6) and E(X)=n(A/N)=6(5/20)=1.5

(b)
Pr(X1)=Pr(X=0)+Pr(X=1)


0C15 1C15
=(C5 6/C206)+(C5 5/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
0C15
Pr{X-1=0|First item of the sample is defective}= C4 5/C195=0.2583
or, more directly,
Pr{Acceptance|First item of the sample is defective}
=(15/19)(14/18)(13/17)(12/16)(11/15)=0.2583

(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
0C122/C16
Pr{X-1=0|1 item is defective in the first 4 items}= C4 2=0.55
or, more directly,
1
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(HP)
=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 Jerrys three pizzas is 3$70 = $210, the probability distribution of his profit is

(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 ($) 120-210 = -90 240-210 = 30 360-210 = 150
Probability 0.4 0.3 0.3

# demanded 1 2 3
Profit ($) 120-140 = -20 240-140 = 100 100-40 = 60
Probability 0.4 0.3 0.3

Expected profit = $40 Likewise, if he prepares one pizza in advance, the probability distribution will be
# demanded 1 2 3
Profit ($) 120-70 = 50 50-40 = 1