Let X denote the time to failure (in years) of a certain engineering component.
Supposethe pdf of X is f(x)=3/(x+3)^2 for x>0 -verify that f(x) is apdf and determine its
cdf. -calculate the probability thatthe time to failure is between2 and 5 years. -obtain the
expected time to failure.
Here it is given that,
3
, x>0
f ( x) 
( x  3)2

If f ( x) is a p.d.f. then
 f ( x)dx  1 .
0

Here,
3
 ( x  3)
2
dx  1
0
Therefore f ( x) is a p.d.f.
The c.d.f. of X is given by
x
x
3
x
dx =
,x>0
2
( x  3)
3 x
0
0
The probability that the time to failure is between 2 and 5 years is given by
5
5
3
P[ 2 < X < 5] =  f ( x)dx = 
dx = 9/40 = 0.225
( x  3)2
2
2
The Expected time to failure is given by,


3x
dx  
E(X) =  xf ( x)dx = 
( x  3) 2
0
0
F ( x)   f ( x)dx = 