Download Answer Key for Homework 3

“An Aggie does not lie, cheat, or steal or tolerate those who do.”
Answer Key for Homework 3 (Collected: February 18, 2005, Friday in class)
1.
Exercise 3.10
Let A be the number of pumps in use at six-pump station and B be the number of pumps in use at
four-pump station. A can take the values 0 to 6 (integers) and B can take the values 0 to 4 (integers).
(a) T=A+B=0,1,2,3,4,5,6,7,8,9,10
(b) X=A-B=-4,-3,-2,-1,0,1,2,3,4,5,6
(c) U=max(A,B)=0,1,2,3,4,5,6
(d) Z=0,1,2
2.
Roll two dice only once. Give the possible values for each of the following
(a) T= total of their face numbers = 2,3,4,5,6,7,8,9,10,11,12
(b) X: the difference between the face numbers = -5,-4,-3,-2,-1,0,1,2,3,4,5
(c) U: the minimum value on the face numbers = 1,2,3,4,5,6
(d) Z=number of dice with the face number of 2 = 0,1,2
3.
Exercise 3.16(a)(c)(d)
(a)
x
0
1
2
3
4
otherwise
p(x)
0.2401 0.4116 0.2646 0.0756 0.0081 0
P(X=0)=1(0.3)0(0.7)4=0.2401
P(X=1)=4(0.3) 1(0.7)3=0.4116
P(X=2)=6(0.3)2(0.7)2=0.2646
P(X=3)=4(0.3) 3 (0.7)1=0.0756
P((X=4)=1(0.3)4(0.7)0=0.0081
(c) P(X=1) is the largest, the answer is X=1.
(d) P(X2)=0.3483
4.
When manufacturing DVDs for Sony, batches of DVDs are randomly selected and the number of
defects (X) is found for each batch.
x
0
1
2
3
4
p(x)
0.502 0.365 0.098 0.034 0.001
(a) Is p(x) a valid pmf of X?
Yes. All probabilities are between 0 and 1 but they add up to 1.
(b) Calculate the expected number of defects for each batch?
E(X)=
 x  p(x) =0.667
(c) Calculate the variance in the number of defects?
E(X2)=
x
2
p( x) =1.079
2
Var(X)= E(X )- [E(X)] 2=1.079-0.6672=0.634
5.
Exercise 3.31
(a) E(X)=
 x p( x ) =16.38
i
i
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
E(X2)=
x
2
i
p( xi ) =272.298
2
Var(X)= E(X )- [E(X)] 2=3.9936
(b) E(25X-8.5)=25E(X)-8.5=401
(c) Var(25X-8.5)=252Var(X)=2496
(d) E(h(X))=E(X-0.01X2)=E(X)-0.01E(X2)=13.657
6.
Exercise 3.52
(This is not in the 5th edition)
 25 
0.02(0.98) 24 =0.3079
1 
(a) P(X=1)= 
(b) P(X1)=1-P(X=0)=1- (0.98)
25
=0.3965
(c) P(X2)=1-P(X=0)-P(X=1)=1- (0.98)
25
 25 
0.02(0.98) 24 =0.0887
1
 
- 
(d) =np=25(0.02)=0.5 and 2=np(1-p)=25(0.02)(0.98)=0.49
P  2    X    2     P(0.5  2(0.7)  X  0.5  2(0.7))  P(0.9  X  1.9)
24
25  25 
=P(X=0)+P(X=1)= (0.98) +  0.02(0.98) =0.9114
1 
(e) E[3(25-X)+4.5X]/25=[75+1.5E(X)]/25=[75+1.5(0.5)]/25=3.03 hours
7.
Exercise 3.66 (Exercise 3.64 in the 5th edition)
 20  30 
 

x 15  x 

P(X=x)=
, x=0,1,2,3,……,15 where X be the number of students from the first section
 50 
 
15 
 20  30 

 
15  y  y 

P(Y=y)=
, y=0,1,2,3,……,15 where Y be the number of students from the second section
 50 
 
15 
 20  30 
  
 5 10  =0.207
(a) P(X=5)= P(Y=10)=
 50 
 
15 
(b) P(Y10)=P(X5)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=0.3798
(c) P(Y10)+ P(Y5)= P(X10)+ P(X5)=0.3938
50  15
30  30 
(15) 1   =2.5714 then std dev=1.6036
50  1
50  50 
(d) E(Y)=15(30)/50=9
Var(Y)=
(e) E(15-Y)=15-E(Y)=15-9=6
Var(15-Y)=Var(Y)=2.5714 then the standard deviation is 1.6036
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
8.
Exercise 3.83 (Exercise 3.81 in the 5th edition)
4 assistance requests per hour
e 8 810
=0.0993
10!
e 2 2 0
(b) =0.5(4)=2 then P(X=0)=
=0.1353
0!
(a) =2(4)=8 then P(X=10)=
(c) E(X)=2
9.
Exercise 3.84 (Exercise 3.82 in the 5th edition)
X: number fails ~ Binomial(n=200, p=0.01)
(a) E(X)=200(0.01)=2
Var(X)=200(0.01)(0.99)=1.98 then the standard deviation is 1.407
(b) Approximate distribution is Poisson with =200(0.01)=2
P(X4)=1-P(X3)=1-0.857=0.143 by using the Poisson table
or P(X3)=
e 2 2 0 e 2 21 e 2 2 2 e 2 2 3
+
+
+
=0.857
0!
2!
3!
1!
(c) P(at least 4 boards properly)=P(exactly 4 works properly)+P(all 5 properly)
= 5(0.1353)4(1-0.1353)+ (0.1353)5 =0.0015
P(any board work properly)=P(All diodes work)=P(X=0)=
e 2 2 0
=0.1353
0!