Applied Statistics and Probability for Engineers

(Chris Devlin) #1
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

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