1. 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 .

2. 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 .
$\displaystyle F(x) = \int\limits_0^x {\frac{{\exp ( - t/2000)}}{{2000}}dt}$

3. 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 ?

4. 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.

5. if x>0, then it would integrate from 0 to infinity ?

6. 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.