STA 4321/5325 – Spring 2007 – Exam 5
A supplier faces a daily demand ( Y , in proportion of her daily supply) with the following density function. f ( y )

2 ( 1
 y )
0
0
 y o.w
.

1
Her profit is given by U =10 Y -2. Find the density function of her daily profit.
U f (

10 Y y )


2
2 ( 1


Y y )

U


10

2

1
 u
2

10
2
 f
U
( u )

8
 u
50

2
 u

8

8
 u
5 dy du

1
10
What is her average daily profit?
E ( U )
  8

2 u
50 u du

1
50
 8

2
( 8 u
 u
2
) du

1
50

 4 u
2  u
3
3


8

2

1
50

 256

512
3
1
50
( 240

168 )

72
50

1 .
44
What is the probability she will lose money on a day?
P ( U

0 )
  0

2
8
 u du
50

1
50

 8 u
 u
2
2


0

2

1
50
 
0

0
 

16

2
 

18
50

0 .
36



16


8
3




The daily amounts of Coke and Diet Coke sold by a vendor are independent and follow exponential distributions with a mean of 1 (units are 100s of gallons). Use the method of conditioning to obtain the distribution of the ratio of Coke to Diet Coke sold. f ( x
1
)
 e
 x
1 x
1

0 f ( x
2
)
 e
 x
2 x
2

0
U

X
1
X
2
Set X
2
 x
2

U

X
1 x
2

X
1

Ux
2 f ( u

| x
2
)
 e
 ux
2 dux
2 du
 f ( u , x
2
)
 x
2 e
 ux
2 e
 x
2 x
2 e
 ux
2
 x
2 e
 x
2
( u

1 )
 x
2
, u

0 f
U
( u )
 
0
 x
2 e
 x
2
( u

1 ) dx
2
 
0
 x
2
2

1 e
 x
2
/( u

1 )

1 dx
2
 
( 2 )

( u

1 )

1
2


( u
1

1 )
2 u

0
Let X
1
,…,X n
be a set of independent observations from a normal (

2
) distribution. Find the mean and variance of the sample variance S
2
. State any results you use.
( n

1 ) S
2

2
~
V

( n

1 ) S
 2
2
 2 n

1

E

( n


1 ) S
2
2



2 ( n

1 )



 n

1

( n

1 )
2
 4
V n


1
E
2

2 ( n

1 )

V
 n

1

E

2
 n
4

1
  2

A game is played where a player generates 4 independent Uniform(0,100) random variables and wins the maximum of the 4 numbers.
Give the distribution of a player’s winnings.
X ~ U ( 0 , 100 )
 f ( x )

1
100
0
 x

100 F ( x )
 x
100
0
 x

100
G
4
( x )

P ( X
( 4 )
 x )
 x
100
4
 g
4
( x )


4
 x
100
3
1
100

4
100
4 x
3
What is her expected winnings?
E ( X
( 4 )
)
 
0
100 4 x
100
4 x
3 dx

4
100
4 x
5
5
100

80
0
Let X
1
,…,X n
be independent random variables, each with probability density function: f ( x )

2 x 0

0 o.w.
x

1
Show that X

 i n

1
X i n
converges in probability to a constant as n
 
and find the constant.
E ( X )
 
0
1 x ( 2 x ) dx

2 x
3
1
3
0

2
3
 
E ( X
2
)
 
0
1 x
2
( 2 x ) dx

2 x
4
4
1
0

1
2

V ( X )

1
2


X
P

2
3
2
3
2

1
18
 

Imperfections in rolls of fabric follow a Poisson process with a mean of 64 per 10000 feet. Give the approximate probability a roll will have at least 75 imperfections.
P ( Y

75 )

P

 Z

Y
 


75

64
8

1 .
38



.
5

.
4162

.
0838
Delivery times for shipments from a central warehouse are exponentially distributed with a mean of 2.4 days (note that times are measured continuously, not just in number of days). A random sample of n =100 shipments are selected, and their shipping times are observed. What is the approximate probability that the average shipping time is less than
2.0 days?
Y ~ Exp ( 2 .
4 )

E ( Y )

2 .
4 V ( Y )

2 .
4 2 
5 .
76
Y

~ N 2 .
4 ,
5 .
76
100

.
0576

P

Y

2 .
0


P

 Z

2

2 .
4
.
0576


0 .
4
.
24
 
1 .
67



.
5

.
4525

.
0475