Probability density function, question?

Question

pdf is

f(x) = 1/5e^-x/5 x>=0

0 x<0

so for the antidif, the lower bound/limit would be 8, but what would the upper one be? Infinity? But how do u sub that into x?

Answer 1

The cumulative probability density function is:

F(x)=∫[t=0 to x] f(t) dt=

x

∫[(1/5)e^(-t/5)]dt=

0

x

∫[-e^(-t/5)] d(-t/5)=-e^(-t/5) [t=0 to x]=

0

-e^(-x/5)+e^(-0/5)=-e^(-x/5)+1

F(x)=1-e^(-x/5), 0≤x&lt;∞

F(x)=0, x&lt;0

If you need more detailed solution of

x

∫[(1/5)e^(-t/5)]dt, don&#039;t hesitate to ask.

0

Answer 2

ANSWER: F(x) [cumulative density function] = e^-(x/5) - 1  for  0 ≤ x ≤ ∞