## If (f(x) = e^-x find ..

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### If (f(x) = e^-x find ..

Hi

I found 2 questions with answers in my handout.

1 - If f(x) = e^-x for x>0 find
a) p(x = 3 )
b) p(x<4 )
I found the answer of this question for
a
= 0 but why??
for b
the integration was from x= 0 to 4
why from 0 to 4 and we given in question x>0 so that mean x not become 0

------------------------

2 - If f(x) = x/8 for 3<x<5
find
a) p(x<4)
b) the mean of x
Why for a the integration from 3 to 4 and we given that 3<x<5?
b) How we can find mean?
r-soy

Posts: 872
Joined: Thu Oct 08, 2009 2:37 pm

### Re: If (f(x) = e^-x find ..

r-soy wrote:Hi

I found 2 questions with answers in my handout

1 - If f(x) = e^-x for x>0 find
a) p(x = 3 )
b) p(x<4 )

As stated, the problem makes no sense. Are we to assume that "f(x)" is a probability density function and that "p(x)" is the corresponding cumulative probability function? It would have helped if you had told us that!

Assuming that, then, by definition of "probability density function" and "cumulative probability functions", $p(x< a)= \int_0^a f(x)dx$. Did you not know that? And, from that, $p(a\le x\le b)= \int_a^b f(x) dx$. In particular, $p(x= a)= p(a\le x\le a)= \int_a^a f(x)dx= 0$ for any f(x).

I found the answer of this question for
a
= 0 but why??
For b
the integration was from x= 0 to 4.
Why from 0 to 4 and we given in question x>0 so that mean x not become 0

Well, what lower bound would you use if x can be arbitrarily close to 0? The "area" of a line is 0 so the "area" of a region for "x> 0" is exactly the same as the area of a region for "$x\ge 0$". Here, I am thinking of the integral as giving the "area under the curve".

-------------

2 - If f(x) = x/8 for 3<x<5
find
a) p(x<4)
b) the mean of x
Why for a the integration from 3 to 4 and we given that 3<x<5?

The integral "to 4" because you were asked for p(x< 4). It is "from 3" because f(x)= 0 below x= 3.
(You could think of this as $\int_{-\infty}^4 f(x)dx$ but because f(x)= 0 for x< 3, that is the same as $\int_3^4 f(x) dx$.)

As for the mean, do you not know the definition of "mean" of a probability distribution? The mean value for probability density function f(x) is defined as $\int x f(x)dx$. Here, since f(x) is defined to be x/8 for x between 3 and 5 (and 0 for all other x), that is $\int_3^5 x(x/8)dx= \frac{1}{8}\int_3^5 x^2 dx$.
b) How we can find mean?
HallsofIvy
Super User

Posts: 2334
Joined: Tue Sep 11, 2007 9:14 am

Thanks so much.
r-soy