IS 314 - OPLOSSINGS : TUTORIAAL VIER
3-97
X has a hypergeometric distribution N=100, n=4, K=20
    20(82160)  0.4191
a.) P( X  1) 
  3921225
20
1
80
3
100
4
b.) P ( X  6)  0 , the sample size is only 4
    4845(1)  0.001236
c.) P( X  4) 
  3921225
20 80
4
0
100
4
3-99
N=10, n=3 and K=4
P(x)
x
0
1
2
3
1
1
3-99(vervolg).
 4  6 

0

 3

f (0)      0.1667, f (1) 
10 

 3

 
 4  6 

 2

1



   0.3, f (3) 
f ( 2) 
10 

 3

 
 4  6 

1

 

  2   0.5,
10 

 3

 
 4  6 

 2

 

  1   0.0333
10 

 3

 
Dus volg:

0,

1 / 6,


F ( x)  2 / 3,

29 / 30,


1,

3104



0  x 1


1  x  2

2  x  3



3 x

x0
Let X denote the count of the numbers in the state's sample that match those in the
player's
sample. Then, X has a hypergeometric distribution with N = 49, n = 6, and K = 6.
     49!   1  7.15 10
a) P( X  6) 
   6!43!  13,983,816
    6  43  1.85 10
b) P( X  5) 
   
    9.686 10
c) P( X  4) 
 
6
6
1
43
0
8
49
6
6
5
43
1
49
6
6
4
5
49
6
43
2
4
49
6
d) Let Y denote the number of weeks needed to match all six numbers. Then, Y
1
has a geometric distribution with p =
and E(Y) = 1/p = 13,983,816
13,983,816
weeks. This is more than 2389 centuries!
2
e4 40
 e4  0.0183
0!
b) P( X  2)  P( X  0)  P( X  1)  P( X  2)
e4 41 e4 42
 e4 

1!
2!
 0.2381
e4 44
c) P( X  4) 
 0.1954
4!
e4 48
d) P( X  8) 
 0.0298
8!
3-107. a) P( X  0) 
3-116. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson
random variable with  = 10.
P( X  0)  e10  4.54  105
b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a
Poisson random variable with  = 1.
P(Y  1)  1  P(Y  0)  1  e1  0.6321
c) The assumptions of a Poisson process require that the probability of a count is
constant for all intervals. If the probability of a count depends on traffic load
and the load varies, then the assumptions of a Poisson process are not valid.
Separate Poisson random variables might be appropriate for the heavy and
light load sections of the highway.
Section 4-2

4-1.

a) P(1  X )   e  x dx  (e  x )  e 1  0.3679
1
1
2.5
b) P(1  X  2.5)   e  x dx  (e  x )
2.5
1
 e1  e 2.5  0.2858
1
3
3
c) P( X  3)   e x dx  0
3
4
d) P( X  4)   e x dx  (e x )  1  e 4  0.9817
4
0
0


e) P(3  X )   e x dx  (e x )  e3  0.0498
3
3

4-2.

a) P( x  X )   e  x dx  (e  x )  e  x  0.10 .
x
x
Then, x = ln(0.10) = 2.3
x
b) P( X  x)   e  x dx  (e  x )  1  e  x  0.10 .
x
0
0
Then, x = ln(0.9) = 0.1054
4-5
a) P(0  X )  0.5 , by symmetry.
1
b) P(0.5  X )  1.5 x 2 dx  0.5x3
1
0.5
 0.5  0.0625  0.4375
0.5
0.5
c) P(0.5  X  0.5) 
1.5x dx  0.5x
2
 0.5
3 0.5
 0.5
 0.125
d) P(X < 2) = 0
e) P(X < 0 or X > 0.5) = 1
1
f) P( x  X )  1.5 x 2dx  0.5 x3  0.5  0.5 x3  0.05
1
x
x
Then, x = 0.9655
4-11. a) P( X  2.8)  P( X  2.8) because X is a continuous random variable.
Thus, P(X<2.8) = F(2.8) = 0.2(2.8) = 0.56.
b) P( X  1.5)  1  P( X  1.5)  1  0.2(1.5)  0.7
c) P( X  2)  FX (2)  0
d) P( X  6)  1  FX (6)  0
4
x
4-15. Now, f ( x)  e  x for 0 < x and FX ( x)   e  x dx   e  x
x
0
 1 e x
0
0, x  0

for 0 < x. Then, FX ( x)  
x
1  e , x  0
4-20.
f ( x)  2e2 x , x  0
4
4
4-23.
x2
E ( X )   0.25 xdx  0.25
2
2 0
0
4
( x  2)3
2 2 4
V ( X )   0.25( x  2) dx  0.25
  
3
3 3 3
0
0
4
2
4
4
4-24.
x3
E ( X )   0.125 x dx  0.125
 2.6667
3
0
0
2
4
x4 4
E  X    0.125 x dx  0.125 |0
4
0
2
3
 3.55559
Var  X   E  X 2    2  0.88889
120
4-29. a.) E ( X ) 
x
100
120
600
120
dx  600 ln x 100  109.39
2
x
V ( X )   ( x  109.39) 2
100
120
600
dx  600  1 
x2
100
 600( x  218.78 ln x  109.39 2 x 1 )
2 (109.39)
x
120
100

(109.39) 2
x2
dx
 33.19
b.) Average cost per part = $0.50*109.39 = $54.70
5

4-31. a) E ( X )   x10e 10( x 5) dx .
5
Thus E ( X )   xe
10( x  5 ) 
5

 e
5
EX
10( x  5 )
e 10( x 5)
dx  5 
10

 5.1
5

2
   x 10 exp  10  x  5 dx
2
5

Further


 x 2 [ exp  10  x  5  ]5   2 x  exp  10  x  5   dx
5
 5.1 
 25 1  2 
 from   5.1
 10 
 26.02
Thus 2  E  X 2    2  26.02  5.12  0.01

b) P( X  5.1)   10e ( x 5) dx   e 10( x 5)

5.1
 e 10(5.15)  0.3679
5.1
6