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2.
Answer Key for Homework 3 (Due date: February 20, 2004 in class)
Make sure to write your name and section number on each page clearly. Do not write your SSN or
ID.
Staple everything together. Place it on the table before class starts. There is a penalty of 5 pt. if
you do not staple them together.
If you come to class after the lecture has started, do not walk through the classroom to turn your
homework in. Wait until the class ends and give it to me before I leave the classroom.
There is no partial credit for the numerical mistakes that you may make. Include 4 digits after the
period to reduce the rounding error. Each question has the same weight with the total of 100 pt.
for the homework.
Remember that the 6th edition can be found on the reserve if you did not purchase the book.
Exercise 3.22 (additional to the parts, (e) write the pmf of X)
(a) P(X=2)=F(2) - F(1) = 0.39 - 0.19 = 0.20
(b) P(X > 3) = 1 - P(X3) = 1 - F(3) = 1 - 0.67 = 0.33
(c) P(2X5)=F(5)-F(1)=0.97-0.19=0.78
(d) P(2<X<5)=F(4)-F(2)=0.92 -0.39=0.53
(e) The pmf of X is
x
0
1
2
3
4
5
P(X=x)
0.06
0.13
0.20
0.28
0.25
0.05
6
0.03
Use the pmf of Y and related information given for Exercise 3.29 then only answer the following
(a) Calculate the expected number of moving violations for which the individual was cited during
the last 3 years.
E(Y)=0(0.6)+1(0.25)+2(0.1)+3(0.05)=0.6
(b) Suppose an individual with Y violations incurs surcharge of 100Y 2, calculate the expected
amount of the surcharge.
E(100Y2)=100E(Y2)=100[02 (0.6)+12 (0.25)+22 (0.1)+32 (0.05)]=100(1.1)=110
(c) Calculate the variance of number of moving violations for which the individual was cited during
the last 3 years.
Var(Y)= E(Y2)-[E(Y)] 2=1.1-0.62=0.74
(d) Calculate the probability that Y being within 1 standard deviation of its mean value. (it means
P  1   Y    1     ?)
P  1    Y    1     P(0.6  0.74  Y  0.6  0.74 )  P(0.26  Y  1.46)
= P(Y=0)+P(Y=1)=0.85
Suppose that the surcharge was really U= 100Y+50.
(e) Calculate the pmf of U?
u
50
150
p(u)
0.60
0.25
250
0.10
350
0.05
otherwise
0
(f) Calculate the expected value of the surcharge.
E(U)=100E(Y)+50=110
Or E(U)=50(0.6)+150(0.25)+250(0.10)+350(0.05)=110
(g) Calculate the variance of the surcharge.
Var(U)=1002Var(Y)=7400
Or Var(U)= E(Y2)-[E(Y)] 2=[502(0.6)+1502 (0.25)+2502 (0.10)+3502 (0.05)]-1102=7400
3.
Exercise 3.50
X: number of customers want the oversize version ~Binomial(n=10,p=0.60)
(a) P(X≥6)=1-P(X≤5)=1-0.367=0.633
(b) =E(X)=np=10(0.6)=6 and 2=Var(Y)=np(1-p)=10(0.6)(0.4)=2.4 then =1.5492
P( - ≤ X ≤  +)= P(4.45 ≤ X ≤ 7.55)=P(X ≤ 7.55)-P(X <4.45)=0.833-0.166=0.667
(c) P(3 ≤ X ≤7)=P(X≤7)-P(X≤2)=0.833-0.012=0.821
4.
Exercise 3.64 (This is not in the 5th edition)
(a) X:number of 3-mega-pixel resolution ~ Hypergeometric(M=6,N=15,n=5)
x
0
1
2
3
4
5
p(x) 126/3003
756/3003 1260/3003 720/3003 135/3003
6/3003
 6  9 
 

x  5  x 

P(X=x)=

15 
   3003
5 
otherwise
0
 6  9 
 

 x  5  x  , x=0,1,2,3,4,5
3003
(b) P(X=2)=1260/3003=0.4195
P(X≤2)=P(X=0)+P(X=1)+P(X=2)=2142/3003=0.7133
P(X≥2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)=2121/3003=0.7063
M
=5(6)/15=2
N
N n
M  M  15  5
6
6
 n   1   
 5  1   =30/35=0.8571
2=Var(X)=
N 1
N 
N  15  1
15  15 
(c) E(X)= n 
then
=0.9258
5.
Exercise 3.67 (Exercise 3.65 in the 5th edition)
10  10 
 

x 15  x 

(a) X~hypergeometric with N=10+10=20, M=10, n=15 then P ( X  x) 
,x=5,..,10
 20 
 
15 
(b) P(all
specimens
of
one
of
the
two
types
selected)=P(all
are
granite)=P(all
10 10  10 10 
     
10  5    5 10  =0.0325
basaltic)=P(X=10)+P(X=5)=
 20 
 20 
 
 
15 
15 
10 10  10 10 
     
 7  8    8  7 
(c) P ( | X -  |  1 ) = P(6.51  X  8.49)=P(X=7)+P(X=8)=
 20 
 20 
 
 
15 
15 
M
=0.3483+0.3483=0.6966
where =E(X) = n 
=15(10/20)=7.5
N
N n
M  M  20  15
10  10 
 n   1   
 15  1   =0.9868 then =0.9934
2=Var(X) =
N 1
N 
N  20  1
20  20 
are
6.
Exercise 3.73
X: total number of males until each family has 2 girls ~ negative binomial (r=6, p=0.5)
E(X)=6 which is 2+2+2 where 2 is the expected males per brother
7.
Exercise 3.80 (This is not in the 5th edition)
X: defective number of computers ~ Binomial(n=10000,p=0.001) can be approximated Poisson(=np=10)
(a) E(X)=np10 and Var(X)=np(1-p)=9.99
(b) P(X> 10)=1-P(X≤10)=1-0.583=0.417
(c) P(X=0)=0
8.
Exercise 3.87 (Exercise 3.85 in the 5th edition)
X: number of trees in t acres ~ Poisson(=80t)
(a) =80(1/4)=20 then P(X≤16)=0.221
(b) t=85000 acres then E(X)= =80(85000)=6.800,000
(c) Area of the circle= r =0.031416
X~Poisson(=80(20.106)=1608.48)
2
square
miles
or
20.106
9.
Exercise 3.106 (Exercise 3.104 in the 5th edition)
X: number of grasshoppers per square yard ~ Poisson(=2)
Y: number of grasshoppers in a circular region with radius r ~ Poisson(=2r2)
P(Y1)=0.99
2
e 2r (2r 2 ) 0
 1  e  r =0.99 where r=0.8561
Then P(at least one)=1-P(none)= 10!
2
acres.
Thus