Midterm Exam
Solution and discussion
1.A random variable
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A variable is used in order to describe the
quantities of interest that are determined by
the outcomes of an experiment.
A random variable is usually not a fixed
parameter, but a variable composed of all
possible experimental results. So the
distribution of values of a random variable
becomes the focus of experimentation.
2. p.m.f. vs. p.d.f.
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The p.m.f is used to describe the
relative probability for at most a
countable number of a discrete random
variable.
In contrast, the p.d.f. is used for
describing the relative interval
probability of a continuous random
variable.
3. A competitive campaign

A proper sampling frame should be
exhaustively 1-to-1 mapping to the entire
population.
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The members of Internet club are just belonging
to a segment of total voters who may be more
innovative, or more high-income, or possessing a
special interest.
The campaign manager must be careful to clarify
the size of this special segment relative to the
total voters.
4. Exclusive vs. independent
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Suppose two events A, and B
If A and B are mutually exclusive, then
A∩B=φ, and P{A∩B}=0, i.e., P{A/B}=0
If A and B are independent, then
P{A/B}=P{A}, or P{A∩B}=P{A} ×P{B}
5. The existence of covariance
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Var(X1+X2)=Var(X1)+Var(X2)+2Cov(X1,X2)
Let X=X1=X2, then
Var(X1)+Var(X2)+2Cov(X1,X2)=2Var(X)+2
Var(X)=4Var(X)
1. The stem-and-leaf plot
8
3
7
2,7
6
2,5
5
1,1,3,4,6,8,9
4
1,3,7
3
2,4
2
1
3,5,6
7,9
2. The box-and-whisker plot

∵N=22,
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p=0.5, np=11, the median=(the 11th+the 12th)/2=51
p=0.25, np=5.5, the Q1=the 6th number=32
p=0.75, np=16.5, the Q3=the 17th number=59
∵Interquartile range=Q3-Q1=27, ∴the lower fence=-8.5, the
upper fence=99.5

Min=17>the lower fence, max=83<the upper fence, ∴ no significant
outliers
Q1
Min=17
32
Q2
Q3
Max=83
51
59
3. Marginal p.d.f.

f x ( X )   xe
 x y
0

x
dy  xe
f y (Y )   xe
 x y
0


0
dx  e
y


0
x
e dy  xe (e )
y
y

0
 xe x
xe x dx,
x
x
let x  u , e dx  dv, dx  du , v  e ,


0
x 
x
xe dx   xe
 f y (Y )  e
0

   e dx  0  ( e
x
0
x 
0
y
∵f(X,Y)< fx(X)× fy(Y), ∴ X is not independent from Y
此題因原joint p.d.f.有誤, 只要列式正確則一律給分
) 1
4. The insurance risk

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Using the Bayes’ rule
P{accident}=P{g}×P{accident/g}
+P{a}×P{accident/a}+P{b}×P{accident/b}
=0.2×0.05+0.5×0.15+0.3×0.3 =0.175
P{no accidents}=1-0.175=0.825
P{g/no accident}=(P{g}×P{no accidents/g})
/P{no accidents} =(0.2×0.95)/0.825=0.23
5. Operations of expectation
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E[X]=2, E[X2]=8,
E[(2+4X)2]=E[4+16X+16X2]
=4+16×2+16×8=164
E[X2+(X+1)2]= E[X2]+E[X2+2X+1]
=8+8+4+1=21
6. The circuit problem
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(a) The successful probability=1-the failure
probability=[1-(1-p)2][1-(1-p)2] =[1-(1-p)2]2=4p24p3+p4
(b) If the central bridge is closed, then the situation
is the same as (a). But, we have to consider the
additional situation that occurs when the central
bridge is open.

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The failed upper/bottom circuit flow: 1-p2;
The successful chance in the situation of open bridge:
(1-p)[1-(1-p2)(1-p2)]
∴p(4p2-4p3+p4)+(1-p)[1-(1-p2)(1-p2)]=p2(2+2p-5p2+2p3)
7. The birthday problem
• The probability of no two having the same
birthday:
• (365/365)(364/365)(363/365)…[(365n+1)/365]=(365)(364)…(365-n+1)/(365)n
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If n>23, the probability is less than 0.5, i.e.,
P{at least one pair having the same
birthday}>0.5;
Moreover, the number of students is usually
more than 23. So, the needed gifts may cost
too much. Don’t accept this marketing proposal.