Homework 6 Solutions
1. Ross 5.5 p. 247
The probability of the gasoline supply being exhausted for a tank capacity c is obtained by
integrating f (x) from c to 1. Here, we want the probability of the supply exhausted to be 0.01;
hence, we must choose c such that
1
1
c
c
0.01   f ( x)dx   5(1  x) 4 dx  (1  c) 5 .
Solving for c gives us c  1  (0.01)1 / 5  0.6019 , so the tank capacity needs to be about 601.9
gallons in order for the probability of the supply exhausted in a given week to be 0.01.
2. Ross 5.6 p. 248


1
1
1
1

1 2 x / 2
x e dx  2 y 2 e  y dy  2(3)  4

40
0
0
(changing variables using y  x / 2 )
a) E ( X )   xf ( x)dx 
b) E ( X )   xf ( x)dx  c  x(1  x 2 )dx  0
(since the integrand is an odd function)


dx

 5 ln( x) 5  
x
5
c) E ( X )   xf ( x)dx  5
5
3. Ross 5.11 p. 248
The statement tells us that X (a random variable representing the location of the point) has a
uniform distribution on (0, L).
In order for the ratio of the shorter to the longer segment to be less than 1/4, the shorter segment
cannot be more than 1/5 of the total bar length L; in other words, X must either be less than L / 5
or greater than 4L / 5 .
Since X is uniformly distributed, this means the probability that the ratio of the shorter to the
longer line segment is less than 1/4 is just ( L / 5) / L  (1  4 L / 5) / L  2 / 5 .
4. Ross 5.12 p. 248
Let X be the random variable of the distance from the breakdown to the nearest service station.
For the first case, if the bus breaks down between 0 and 25, then it is closest to city A’s station; if
it breaks down between 25 and 75, it is closest to the station in the center of the route; and if it
breaks down between 75 and 100, it is closest to city B’s station. Thus, the expected distance
from a breakdown to the nearest service station in the first case is just
25
50
75
100
x
50  x
x  50
100  x
E( X )  
dx  
dx  
dx  
dx  12.5 .
100
100
100
100
0
25
50
75
On the other hand, if the 3 service stations are located 25, 50, and 75 miles from city A, then a
bus that breaks down between 0 and 37.5 is closest to the 1st station; a bus that breaks down
between 37.5 and 62.5 is closest to the 2nd station; and a bus that breaks down between 62.5 and
100 is closest to station 3. Thus, the expected distance from a breakdown to the nearest service
station in the second case is
25
37.5
50
62.5
25  x
x  25
50  x
x  50
E( X )  
dx  
dx  
dx  
dx
100
100
100
100
0
25
37.5
50
75  x
62.5 100 dx 
75

x  75
dx  9.375
100
75
100

So, the 2nd method yields a smaller expected distance from the breakdown to the nearest service
station; hence, the station set-up under the 2nd case is indeed more efficient.
5. Ross 5.20 p. 249
Let X be the number of people in favor of the proposition in a random sample of 100. X has a
binomial distribution with n  100 and p  0.65 . Since np(1  p)  22.75 is greater than 10, we
can use the normal approximation to the binomial distribution. Then X is approximately normal
with parameters   100(0.65)  65 and  2  100(0.65)(0.35)  22.75 .
a) Then the probability that at least 50 favor the proposition is

49.5  65 
P( X  49.5)  P Z 
  P( Z  3.250)  0.9994
22.75 

b) The probability that between 60 and 70 inclusive are in favor of the proposal is
 59.5  65
70.5  65 
P(59.5  X  70.5)  P
Z
  P(1.153  Z  1.153)  0.7498
22.75 
 22.75
c) The probability that fewer than 75 are in favor is

74.5  65 
P( X  74.5)  P Z 
  P( Z  1.992)  0.9767
22.75 

6. Ross 5.21 p. 249
Let X be a random variable representing the height of a 25-year-old man. Then X has a normal
distribution with parameters   71 and  2  6.25 .
The probability that a randomly chosen 25-year-old man is over 6’2” tall is just
74  71 

P( X  74)  P Z 
  P( Z  1.2)  0.1151 .
2.5 

So, approximately 11.51% of 25-year-old men are over 6’2” tall.
The conditional probability that a 25-year-old man will be over 6’5” tall given that he is over 6’
tall is just
P( X  77) P( Z  6 / 2.5) P( Z  2.4) 0.0082
P( X  77 | X  72) 



 0.0238 .
P( X  72) P( Z  1 / 2.5) P( Z  0.4) 0.3446
So, about 2.4% of men in the 6’ club are over 6’5” tall.
7. Ross 5.29 p. 250
Let s be the initial price of the stock, and let X denote the number of times (in the 1000 time
periods) the stock increases in value; then the price of the stock at the end of the 1000 periods is
X
X
su d
1000 X
 sd
1000
u
  .
d 
For the price to be up at least 30%, we need to have
X
u
d 1000    1.3
d 
X
u
1000
   1.3d
d 
u
X log    log( 1.3)  1000 log( d )
d 
log( 1.3)  1000 log( d )
X 
log( u / d )
Substituting in u  1.012 and d  0.990 gives us X  469.2 ; i.e., the stock would have to rise
in at least 470 time periods.
Now, X has a binomial distribution with parameters n  1000 and p  0.52 . Since
np(1  p)  1000(0.52)(0.48)  249.6 , we can use the normal approximation to the binomial
distribution. So

469.5  1000(0.52) 
P( X  469.5)  P Z 
 P( Z  3.196)  0.9993 .

1000(0.52)(0.48) 

8. Ross 5.30 p. 250
The probability that a point with a reading of 5 is in the black region is

P(5 | black )
P(in black ) 

P(5 | black )  P(5 | white )(1   )

3 2
3 2
e ( 56 )
2
/ 18
e  ( 56 )

2
/ 18
1
2 2
e  ( 5 4 )
2
/8
Thus, we need to find the value of  that yields P(in black )  1 / 2 (i.e., the probability of
making an error is the same regardless of which region one concludes the point is in).


3

3
3
e 1 / 18
e 1 / 18 
e 1 / 18 
1   1 / 8
e
2

e 1 / 18 

1
2
1   1 / 8
e
4
6
 1 / 18 1 1 / 8  1 / 8
e
 e
 e
6
4
4
1 / 18
1 / 8
e
 1 1 / 8
e
 e
 

4  4
 6
   0.5832
9. Ross 5.32 p. 250
a) Let X be the time required to repair a machine; then X ~ Exp(1/2). Hence, the probability
that a repair time exceeds 2 hours is just
 x / 2

e
P ( X  2)  
dx   e  x / 2  e 1  0.3679
2
2
2
b) The conditional probability that a repair takes at least 10 hours, given that its duration
exceeds 9 hours is exactly the probability that a repair takes at least 1 hour (thanks to the
memory-less property of exponential random variables):
P( X  10 | X  9)  P( X  1)   e  x / 2
10. Ross 5.34 p. 250

1
 e 1 / 2  0.6065
a) Let X be the total number of miles (in thousands) that a car can be driven before needing
to be junked. In this part, X ~ Exp(1/20). The probability that Jones gets at least 20,000
additional miles out of a used car that already has at least 10,000 miles under it is just
(thanks again to the memory-less property of exponential random variables) the
probability that one gets at least 20,000 miles out of a car before junking it:
P( X  20)   e  x / 20

20
 e 1  0.3679
b) In this part, instead assume that X ~ Uniform(0, 40). The probability that Jones gets at
least 20,000 additional miles out of a used car that already has at least 10,000 miles under
it is now
P( X  30) 1 / 4 1
P( X  30 | X  10) 


P( X  10) 3 / 4 3