# Continuous Probability Distributions

• Nov 19th 2009, 07:51 AM
hazel
Continuous Probability Distributions
3.27 walpole

The time to failure in hours of an important piece of electronic equipment used in a manufactured DVD player has the density function

f(x) = 1/2000 exp (-x/2000), x>=0
0, x<0

find F(x).

my queries :
- so meaning integrate from 0 to x or 0 to infinity ?
- and could i use du=-x/2000 dx to solve ? i saw some of my mates using dt. which i don't know how does dt fit in . (Thinking)

• Nov 19th 2009, 08:45 AM
Plato
Quote:

Originally Posted by hazel
3.27 walpole

The time to failure in hours of an important piece of electronic equipment used in a manufactured DVD player has the density function

f(x) = 1/2000 exp (-x/2000), x>=0
0, x<0

find F(x).

my queries :
- so meaning integrate from 0 to x or 0 to infinity ?
- and could i use du=-x/2000 dx to solve ? i saw some of my mates using dt. which i don't know how does dt fit in .

$F(x) = \int\limits_0^x {\frac{{\exp ( - t/2000)}}{{2000}}dt}$
• Nov 20th 2009, 07:18 PM
hazel
ok, o to x

but why do you use t ?
it doesn't matter if i use u or t (du or dt ) right ?

and also, if x>0, then it would integrate from 0 to x.
but since x>= 0, so integrate from 0 to x.
am i right to conclude this ?
• Nov 20th 2009, 09:01 PM
mr fantastic
Quote:

Originally Posted by hazel
ok, o to x

but why do you use t ?
it doesn't matter if i use u or t (du or dt ) right ?

and also, if x>0, then it would integrate from 0 to x.
but since x>= 0, so integrate from 0 to x.
am i right to conclude this ?

t is a dummy variable of integration.
Yes, you integrate from 0 to x.
• Nov 20th 2009, 10:29 PM
hazel
if x>0, then it would integrate from 0 to infinity ?
• Nov 20th 2009, 10:49 PM
mr fantastic
Quote:

Originally Posted by hazel
if x>0, then it would integrate from 0 to infinity ?

If you want to get F(x) then I've already said that you integrate from 0 to x.