Applied Statistics and Probability for Engineers

5-3 TWO CONTINUOUS RANDOM VARIABLES 161

The first integral is

The second integral is

Therefore,

Alternatively, the probability can be calculated from the marginal probability distribution of Y

as follows. For

 610 ^3 e0.002y^11 e0.001y 2 for y

0

 610 ^6 e0.002y a

e0.001x

0.001

`

y

0

b 610 ^6 e0.002y a

1 e0.001y

0.001

b

fY 1 y 2 

y

0

610 ^6 e0.001x0.002y dx 610 ^6 e0.002y (^) 
y
0
e0.001x^ dx
y
0
P 1 Y
20002 0.04750.00250.05.

610 ^6
0.002
a
e^6
0.003
b0.0025

610 ^6 

2000

°

e0.002y

0.002

`

x

¢ e0.001x dx

610 ^6

0.002

(^) 
2000
e0.003x dx

610 ^6
0.002
e^4 a
1 e^2
0.001
b0.0475

610 ^6 

2000

0

°

e0.002y

0.002

`

2000

¢ e0.001x dx

610 ^6

0.002

e^4 

2000

0

e0.001x dx

Figure 5-10 Region of integration

for the probability that is

darkly shaded and it is partitioned

into two regions withx 2000 and

x 2000.

Y 2000

y

0 x

0

2000

2000

c 05 .qxd 5/13/02 1:49 PM Page 161 RK UL 6 RK UL 6:Desktop Folder:TEMP WORK:MONTGOMERY:REVISES UPLO D CH114 FIN L:Quark Files: