2.3
(a)
0.4
f ( x)
0.3
0.2
0.1
0.0
0
1
2
4
3
5
7
6
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
2.6
(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.
obtain
Similarly, for f (1) we
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
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 2year 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 

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.
2.26
(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
 2850 lb.
1
That is, if
15
 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 

  0.001
P Z  n
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
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.