1
Chapter 2
Solutions to Exercises
Solutions to Exercises in Chapter 2
4
2.1
(a)
x1  x2  x3  x4   xi
i 1
4
(b) x3  x4   xi
i 3
4
(c)
x1 y1  x2 y2  x3 y3  x4 y4   xi yi
i 1
4
(d) x1 y2  x2 y3  x3 y4  x4 y5   xi yi 1
i 1
3
(e)
x2 y22  x3 y32   xi yi2
i 2
3
(f)
( x1  y1 )  ( x2  y2 )  ( x3  y3 )   ( xi  yi )
i 1
4
2.2
(a)
 (a  bxi )  (a  bx1 )  (a  bx2 )  (a  bx3 )  (a  bx4 )
i 1
 4a  b( x1  x2  x3  x4 )
3
(b)
 i 2  12  22  32  1  4  9  14
i 1
3
(c)
 ( x 2  2 x  2)  (02  2  0  2) (12  2  1  2)  (22  2  2  2)  (32  2  3  2)
x 0
= 2 + 5 + 10 + 17 = 34
4
(d)
 f ( x  2) 
x 2
 f (4)  f (5)  f (6)
2
(e)
f (2  2)  f (3  2)  f (4  2)
 f ( x, y ) 
f (0, y )  f (1, y )  f (2, y )
x 0
4
(f)
2
4
4
x 2
x 2
 ( x  2 y)  ( x  2 1)  ( x  2  2)   (2 x  6)
x 2 y 1
 (2  2  6)  (2  3  6)  (2  4  6)  10  12  14  36
2
Chapter 2
2.3
Solutions to Exercises
(a)
0.4
f ( x)
0.3
0.2
0.1
0.0
0
1
2
3
4
5
6
7
FX
(b) The probability that, on a given Monday, either 2, or 3, or 4 students will be absent is
4
 f ( x)  f (2)  f (3)  f (4)  .310  .340  .220  0.87 .
x2
(c) The probability that, on a given Monday, more than 3 students are absent is
7
 f ( x)  f (4)  f (5)  f (6)  f (7)  .220  .080  .019  .001  0.32.
x4
7
(d) E ( X )   x. f ( x)
x 0
 0  .005  1 .025  2  .310  3  .340  4  .220  5  .080  6  .019  7  .001  3.066
3.066 is the average number of students absent on Mondays after considering infinitely
many Mondays.
(e) 2  Var(x)  E( x2 )  [ E( x)]2
7
E ( x 2 )   x 2 f ( x)  02  .005  12  .025  22  .310  32  .340  42  .220
x 0
52  .080  62  .019  72  .001
= 10.5776
2  10.5776  (3.066)2  1.1776
(f)
  2  1.08519.
E (Y )  E (7 x  3)  7 E ( x)  3  7  3.066  3  24.462
Var(Y )  Var(7 x  3)  72 Var( x)  49 1.776  57.7046
3
Chapter 2
2.4
Solutions to Exercises
(a)
f ( x)
1
f ( x)
x
0
1
2
(b) Total area of the triangle is half the base multiplied by the height. That is
area = 0.5  2  1  1
(c) When x = 1, f ( x)  f (1)  0.5. Then, P ( X  1) is given by the area to the right of 1
which is
P( X  1)  0.5  1  0.5  0.25
(d) Now f  0.5 = 0.75, and thus
1
P( X  0.5)  1  P( X  0.5)  1   1.5  0.75  0.4375
2
(e) For a continuous random variable the probability of observing a single point is zero.
Thus, P( X  1.5)  0 .
2.5
(a) The probability density function for x is
f ( x) 
1
,
ba
a xb
Since our case has a = 10, b = 20,
f ( x) 
1
1
 ,
20  10 10
f ( x)
1
10
x
10
20
10  x  20
4
Chapter 2
Solutions to Exercises
(b) The total area beneath the pdf for 10  x  20 is the area of a rectangle. Namely,
Area = (20  10) 
(c)
2.6
1
1
10
P(17  X  19)  (19  17) 
1
 0.2
10
(a) For 4-year schools
P(men and 4-year) 
4,076,416
 0.27
15,064,859
P(women and 4-year) 
4,755,790
 0.32
15,064,859
(b) Let
X=0
if a male is selected
X=1
if a female is selected
Y=1
if 4-year student is chosen
Y=2
if 2-year student is chosen
Y=3
if a less than 2-year student is chosen
P( X  0)  P( X  0, Y  1)  P( X  0, Y  2)  P( X  0, Y  3)
 0.27  0.16  0.01  0.44
P(Y  2)  P( X  0, Y  2)  P( X  1, Y  2)  0.16  0.22  0.38
(c) For f ( x), we note from part (b) that f (0)  0.44. Similarly, for f (1) we obtain
f (1)  P( X  1)  0.32  0.22  0.02  0.56
Thus, the marginal probability function f ( x) is given by
0.44 for x  0
f ( x)  
0.56 for x  1
For g ( y ) we have
0.27  0.32  0.59 for y  1

g ( y )  0.16  0.22  0.38 for y  2
0.01  0.02  0.03 for y  3

(d) The conditional probability function for Y given that X = 1 can be obtained using the
result
5
Chapter 2
Solutions to Exercises
P(Y  y X  x) 
P(Y  y, X  x) f ( x, y)

P( X  x)
f ( x)
Thus,
0.32 0.56  0.57 for y  1

f ( y x  1)  0.22 0.56  0.39 for y  2
0.02 0.56  0.04 for y  3

(e) If a randomly chosen student is male, the probability that he attends a 2-year college is
P(Y  2 X  0) 
P(Y  2, X  0) .16

 .36
P( X  0)
.44
(f) If a randomly chosen student is a male, the probability that he attends either a 2-year
college or a 4-year college is
P(Y  1 or Y  2 X  0)  P(Y  1 X  0)  P(Y  2 X  0)

P(Y  1, X  0) P(Y  2, X  0)

P( X  0)
P( X  0)

.27 .16

 .61  .36  .97
.44 .44
(g) For gender and type of college institution to be statistically independent, we need
f ( x, y )  f ( x ) g ( y )
for all x and y.
For x = 0 and y = 1, we have
f (0,1)  0.27  f (0) g (1)  0.44  0.59  0.26
While these two values are close, they are not identical. Hence, gender and type of
institution are not independent.
2.7
The mutual fund has an annual rate of return, X ~ N (0.1,0.042 ) .
(a)
0  0.1 

P( X  0)  P  Z 
  P( Z  2.5)  1  .9938  .0062
0.04 

(b)
0.15  0.1 

P( X  0.15)  P  Z 
  P( Z  1.25)  1  .8844  .1056
0.04 

(c)
Now, X ~ N (0.12,0.052 )
0  0.12 

P( X  0)  P  Z 
  P( Z  2.4)  1  .9918  .0082
0.05 

6
Chapter 2
Solutions to Exercises
0.15  0.12 

P( X  .15)  P  Z 
  P( Z  .6)  1  .7257  .2743
0.05 

From the modification, the probability that a 1-year return will be negative is increased
from 6.2% to 8.2%, and the probability that a 1-year return will exceed 15% is increased
from 10.56% to 27.43%. Since the chance of negative return has increased only slightly,
and the chance of a return above 15% has increased considerably, I would advise the
fund managers to make the portfolio change.
(d) Computer software files xr2-7.xls, xr2-7.szm, xr2-7.wf1 and xr2-7.sas contain the
instructions for computing these probabilities.
cov( X ,Y )  E ( X  E ( X ))( X  E (Y ))
2.8
 E  XY  X .E (Y )  E ( X ).Y  E ( X ) E (Y ) 
 E ( XY )  E  X .E (Y )   E  E ( X ).Y   E  E ( X ).E (Y ) 
 E ( XY )  E ( X ).E (Y )  E ( X ).E (Y )  E ( X ).E (Y )
 E ( XY )   X Y
2.9
2.10
(a)
P(Z  3)  P(Z  0)  P(0  Z  3)  .5  .4987  0.9987
(b)
P(Z  1)  P(Z  0)  P(0  Z  1)  .5  .3413  0.8413
(c)
P(Z  1)  P(Z  0)  P(0  Z  1)  .5  .3413  0.1587
(d)
P(Z  3)  P(Z  0)  P(0  Z  3)  .5  .4987  .0013
(e)
P(1  Z  1)  2  P(0  Z  1)  .3413  2  .6826
(f)
P(3  Z  1)  P(0  Z  3)  P(0  Z  1)  .4987  .3413  0.84
(g)
Computer software files xr2-9.xls, xr2-9.szm, xr2-9.wf1 and xr2-9.sas contain the
instructions for computing these probabilities.
Using the result E[ g x ]   g ( x) f  x  , we have
x
(a)
E[c]   c f ( x)  c  f ( x)  c
x
x
(b) E[cX ]   c x f ( x)  c  x f ( x)  c E[ X ]
x
(c)
x
E[a  cX ]   a  cx f ( x)   a f ( x )   cx f ( x)  a  cE[ X ]
x
x
x
7
Chapter 2
2.11
Solutions to Exercises
var(a+cX) = E[(a+cX)  E(a+cX)]2 = E[a+cX  a  cE(X)]2
= E[c(X  E(X))]2 = c2E[X  E(X)]2 = c2var(X)
2.12
(a)
2 x 0  x  1
f  x  
otherwise
0
f(x)
2
1
2.13
x
(b)
P(0  X  21 )  21 ( 21 )(1)  41
(c)
9  1 1
P( 41  X  43 )  P( X  43 )  P( X  41 )  21 ( 43 )( 43 )( 23 )  21 ( 41 )( 21 )  16
16 2
As X1 and X2 are independent continuous random variables, f  x1 , x2   f  x1  f  x2  .
0  Xi  1
2 xi
f  xi   
0
otherwise
for i = 1,2 (from Exercise 2.12)
Thus, the joint probability density function is
4 x1 x 2
f  x1 , x 2   
0
2.14
0  X 1  1, 0  X 2  1
otherwise
The joint probability density function for the random variables, gender (G) and political
affiliation (P), is
G
P
0
1
2
0
.20
.30
.06
1
.27
.10
.07
8
Chapter 2
Solutions to Exercises
The marginal probability density functions are
0.44 if g  0
f g  
0.56 if g  1
0.47 if

f  p   0.40 if
0.13 if

and
p0
p 1
p2
(a) For the two random variables, G and P, to be independent,
f  g , p  f  g  f  p 
When g = 0 and p = 0, we have f (0,0) = 0.27. However, f ( g  0) = 0.44 and f ( p  0)
= 0.47. Thus, f ( g  0). f ( p  0) = 0.44  0.47 = 0.2068  0.27 = f (0,0) . Hence the
random variables P and G are not independent. Similarly, f (11
, ) = 0.3, f ( g  1) = 0.56
and f ( p  1) = 0.4. Again 0.3  0.56  0.4, providing further evidence that the two
random variables G and P are not independent.
(b)
f  g, p 
f  p | G  1 
f  g  1
. Thus we can set up the following table.
f  g , p
p
2.15
f  x  p x 1  p
f  p G  1
f  g  1
0
0.20
0.56
0.3571
1
0.30
0.56
0.5357
2
0.06
0.56
0.1071
1 x
for x = 0,1.
(a) The mean of the discrete random variable, X, is
E  X    x f  x   0  f  0   1 f 1  p1 1  p 
11
p
x
The variance of X is
 2  var  X   E  X  E  X     x  p f  x     x  p p x 1  p
2
2
2
x
   p p 1  p
2
0
1 0
x
 1  p p 1  p
2
1
11
 p 2 1  p  1  p p  p1  p p  1  p  p1  p
2
1 x
9
Chapter 2
Solutions to Exercises
(b) E[ B]  E[ X 1  X 2 ... X n ]  E[ X 1 ]  E[ X 2 ]  ...  E[ X n ]  p  p  p  np
Given X 1 , X 2 ,..., X n are independent, we have
var  B  var  X 1  X 2 ... X n   var  X 1   var  X 2 ... var  X n 
 p1  p  p1  p... p1  p  np1  p
(c)
np
B 1
E[Y ]  E    E[ B] 
p
n
n n
np1  p p1  p
 B 1
var Y   var    2 var  B 

 n n
n
n2
2.16
X
2
4
6
1
Y
3
9
1/8
1/4
1/8
1/2
1/24
1/4
1/24
1/3
1/12
0
1/12
1/6
1/4
1/2
1/4
(a) The marginal probability density function of Y is h(y) where
h(1) = 1/2
h(3) = 1/3
h(9) = 1/6
(b) The conditional probability density function for y is given by the equation
f ( x, y)
. Applying this rule, we can obtain f ( y |2) , which is given by
f ( y | x) 
f ( x)
y
f ( y |2)
1
f (2,1) f (2)  (1 / 8) (1 / 4)
= 1/2
3
f (2,3) f (2)  (1 / 24) (1 / 4)
= 1/6
9
f (2,9) f (2)  (1 / 12) (1 / 4)
= 1/3
Therefore, the conditional probability density function, f ( y |2) , is given by
f (1|2) = 1/2
f (3|2) = 1/6
(c) cov X , Y   E[ X  E ( X )][Y  E (Y )]
where E  X   2  41  4  21  6  41  4
E Y   1  21  3  13  9  16  3
f (9|2) = 1/3
10
Chapter 2
Solutions to Exercises
From first principles
cov X , Y      x  4 y  3 f  x , y 
x
y
1 
 2  41  3 81  2  43  3 24
1  64 93 1
 6  43  3 24
   12
0
Alternatively, using results in Section 2.5 it is possible to show that
cov X , Y   E[ X  E  X ][Y  E Y ]  E  XY   E  X  E Y 
This is an important result that will be used throughout the text. In terms of the current
example we can show that E(XY) = 12 and hence that
cov X , Y   12  4  3  0
(d) This is an example where X and Y are not independent, despite the fact that their
covariance is zero. If X and Y are independent, then f  x, y  g xh y . To prove that
this result does not hold let us consider two outcomes. First, if x = 2 and y = 3,
f 2,3 
1
24
 g2h3 
1
 
1
1
4
3
12
Also, if x = 4 and y = 9
f 4,9  0  g4h9 
2.17
1

2
1
1

6
12
Let X1 and X2 denote the number of points showing on the first and second rolls, respectively,
of a die. It is clear that X1 and X2 are independent random variables. Let
U = X1 + X2 and V = X1  X2.
(a) The mean of V is
E V   E  X 1  X 2   E  X 1   E  X 2 
=0
because X1 and X2 have identical probability density functions and hence
identical means.
For the variance of V, first note that
varV   var X 1  X 2 
 var X 1   var X 2  since X1, X2 are independent.
 
Now, var  X 1   E  X 1  E  X 1   E X 12   E  X 1 
2
2
where
E  X 1    x1 f  x1  
x1
 
E X 12   x12 f  x1  
x1
1
6
1  2
1
6
1
2
 ...  6 
21
6

 2 2  ...  6 2 
91
6
11
Chapter 2
Solutions to Exercises
var  X 1   E
Thus,
    E  X 
X 12
2
1
2
 21
     2.9167
 6
6
91
Similarly, var  X 2   2.9167 .
Hence, the mean and variance of V are 0 and 5.8333, respectively.
covU ,V   E U  E U V  E V   E UV   E U  E V 
(b)
E U   E ( X 1  X 2 ) 
21
6

21
6
7

 

 
E UV   E[( X 1  X 2 )( X 1  X 2 )]  E X 12  X 22  E X 12  E X 22

91
6

91
0
6
Thus, covU ,V   E UV   E U  E V   0  7  0  0
(c) To examine whether or not U and V are independent we will begin by deriving the joint
density function for U and V from that for X1 and X2. The following table contains the
values for U(V) that correspond to each possible outcome for X1 and X2. The numbers
2,3,...,12 are the possible values that U can take while the numbers in brackets (5,
4,...,4,5) represent the possible values of V.
Value of U(V) for each possible (X1, X2) outcome
2nd roll of die (X2)
1
1st
roll
of die
(X1)
1
2
3
4
5
6
2 (0)
3 (1)
4 (2)
5 (3)
6 (4)
7 (5)
2
3
4
5
6
3 (1)
4 (0)
5 (1)
6 (2)
7 (3)
8 (4)
4 (2)
5 (1)
6 (0)
7 (1)
8 (2)
9 (3)
5 (3)
6 (2)
7 (1)
8 (0)
9 (1)
10 (2)
6 (4)
7 (3)
8 (2)
9 (1)
10 (0)
11 (1)
7 (5)
8 (4)
9 (3)
10 (2)
11 (1)
12 (0)
Note that each cell has a probability of 1/36. The probability density functions for U and
V, respectively, are:u
h(u)
v
h(v)
2
1/36
-5
1/36
3
2/36
-4
2/36
4
3/36
-3
3/36
5
4/36
-2
4/36
6
5/36
-1
5/36
7
6/36
0
6/36
8
5/36
1
5/36
9
4/36
2
4/36
10
3/36
3
3/36
11
2/36
4
2/36
12
1/36
5
1/36
If U and V are independent random variables, then they must satisfy the following
equation f u, v   hu gv  . To show this equation does not hold, consider u = 3, v = 1.
12
Chapter 2
Solutions to Exercises
In this case, f 3,1 = 1/36, h(3) = 2/36, and g(1) = 5/36. Obviously, 1/36  2/36  5/36.
Thus the random variables U and V are not independent.
2.18
Given that X1 and X2 are independent discrete random variables with joint probability density
function f  x1 , x 2  , we can write
f  x1 , x 2   f  x1  f  x 2 
E  X 1 X 2     x1 x 2 f  x1 , x 2     x1 x 2 f  x1  f  x 2 
x1 x2
x1 x2
  x1 f  x1   x 2 f  x 2   E  X 1  E  X 2 
x1
2.19
x2
var( X )  E[( X  E ( X )) 2 ]  E[ X 2  2 XE ( X )   E ( X ) ]
2
 E[ X 2 ]  2 E ( X ) E[ X ]  [ E ( X )]2  E[ X 2 ]  [ E ( X )]2
2.20
Using equation (2.5.2), we have
E[c1 X  c2Y ]    c1 x  c2 y  f  x , y     c1 x f  x , y     c2 y f  x , y 
x
y
x
y
x
y
 c1  x  f  x , y  c2  y  f  x , y 
x
y
y
x
 c1  x f  x   c2  y f  y   c1 E[ X ]  c2 E[Y ]
x
2.21
2.22
y
  
 X  1
E[ Z ]  E     E [ X ]     0
  
   
2
 X  1
var Z   var     2 var X   2  1
   

Thus, Z ~ N(0,1).
1
 1
E[ X ]  E   X1  X 2 ... X n    E[ X1 ]  E[ X 2 ]... E[ X n ]
n
 n
n
1
   ... 

n
n
1
2
1
 1
var X  var   X1  X 2 ... X n   2  var X1   var X 2 ... var X n   2 n 2 
n
 n
n
n
 
13
Chapter 2
2.23 (a)
Solutions to Exercises
f 0,0  0.6 0.4 0.3 052
.  2 0  0.208
0
1
0
1
f 0,1  (0.6) 0 0.4 0.3 0.52 2 0  0120
.
1
1
0
f 1,0  0.6 0.4 0.3 0.52 2 0  0.312
1
0
0
1
f 11
,   0.6 0.4 0.3 0.52 2 1  0.360
1
(b)
(c)
0
1
0
By summing the relevant values we obtain
x
f  x
y
f  y
0
1
0.328
0.672
0
1
0.52
0.48
This function is given by
f  y | x  0 
(d)
f  0, y 
f  x  0
f 0, y 
f  y | x  0
y
f  x  0
0
0.208
0.328
0.6341
1
0120
.
0.328
0.3659
E[ X ]   x f  x   0.672
x
E[ X 2 ]   x 2 f  x   0.672
x
var X   E[ X 2 ]  [ E  X ]2  0.672  0.672  0.2204
2
(e)
E[Y ]   y f  y   0.48
y
E [Y ]   y 2 f  y   0.48
2
y
var Y   0.48  0.48  0.2496
2
(f)
E[ XY ]    xy f  x , y   0.36
x
y
(Only the term where both x and y are equal to one needs to be considered in the above
summation. Other terms are zero.)
cov X , Y   E [ XY ]  E [ X ]E [Y ]  0.36  0.6720.48  0.03744

cov X , Y 
var  X  var Y 

0.03744
0.22040.2496
 01596
.
14
Chapter 2
(g)
Solutions to Exercises
E[ X  Y ]  E[ X ]  E[Y ]  0.672  0.48  1152
.
var  X + Y   var  X   var Y   2 cov X , Y 
 0.2204  0.2496  20.03744  0.5449
2.24
We have
3
 3 
E Y   E  13  Yi   13  E Yi   13  3   
i 1
 i 1 
 3 
var Y   var  13  Yi   19 var Y1  Y2  Y3 
 i 1 
 19  var Y1   var Y2  + var Y3   2cov(Y1 , Y2 )  2cov(Y1 , Y3 )  2cov(Y2 , Y3 ) 
 19  32   92 1.52   23 2
2.25
Let X represent the life length of the computer. The fraction of computers lasting for a given
time is equal to the probability of one computer, selected at random, lasting for that given
time.
(a)
1  2.9 

P X  1  P Z 
. 
  P Z  1357

196
. 
 0.5  P0  Z  1357
.   0.5  0.413  0.087
(b)
4  2.9 

P X  4  P Z 
  P Z  0.786

14
. 
 0.5  P0  Z  0.786  0.5  0.284  0.216
(c)
2  2.9 

P X  2  P Z 
  P Z  0.643

14
. 
 0.5  P0  Z  0.643  0.5  0.240  0.740
(d)
4  2.9 
 2.5  2.9
P2.5  X  4  P
Z
  P 0.286  Z  0.786
 14
.
14
. 
 P 0.286  Z  0  P0  Z  0.786  0112
.
 0.284  0.396
(e) We want X0 such that P(X < X0) = 0.05. Now, P(0 < Z < 1.645) = 0.45 (from tables).
Hence, P(Z < 1.645) = 0.05. Thus, an appropriate X0 is defined by
1645
.

X 0  2.9
. 14
.   0.6.
or X 0  2.9  1645
14
.
15
Chapter 2
2.26
Solutions to Exercises
(a) If p is the probability that one of the economists selected at random weighs less than 189
lb, then the probability that all fifteen of them weigh less than 189 lb is p15. Let X be the
weight of one male economist selected at random. Then X ~ N[178, (17)2] and
189  178 

p  P X  189  P Z 
  P Z  0.647  0.741


17
15
15
Hence, (p) = (0.741) = 0.011
(b) Let X1, X2,...,X15 be the weights of the fifteen men. Now the raft will be overloaded if
15
 Xi
i 1
1
That is, if
15
 2850 lb.
 Xi 

17 2  . Hence,
2850
lb or if X  190 . Now, X ~ N 178,

15 
15


190  178 
  P Z  2.734  0.0031
P X  190  P Z 
17 15 



(c) We need n such that
2

17 

2850 

.
P X 
  0.001 where X ~ N  178,

n 
n 

That is, we need n such that
2850


 178 

n
  0.001
P Z 
17 n 



Now, from the tables,
P Z  3.09  0.001.
Thus, the value of n which gives a probability of overloading of exactly 0.001 is given by
solving the following equation
2850
 178
2850
52.53
n
 178 
 3.09 or
n
n
17 n
or
178n  52.53 n  2850  0 .
This is a quadratic equation in


n . Applying the formula for solving a quadratic
equation, b  b 2  4ac 2a , we have
n
52.53  52.532  41782850
2178

52.53  1425.47
 386
.
356
16
Chapter 2
Solutions to Exercises
We have ignored the negative root which obviously is an impossible value of n .
Hence, n = 14.9. Now we cannot have fractional economists, so the maximum number of
economists must be 14 or 15. Increasing n to 15 would increase the probability above
0.001, whereas reducing n to 14 would give a probability less than 0.001. Thus, the
maximum number of economists is 14.