If the probability density of a random variable is given by:
f(x)= Kx^2 for 0

Question

If the probability density of a random variable is given by:
f(x)= Kx^2 for 0<x<1
       0   elsewhere

find the value k and the probability that the random variable takes on a value between

(a) 1/4 and 3/4 

(b) greater than 2/3

anonymous · Answer

So that I'm understanding this correctly. I'm finding the value of "k"
in order to verify that the function f(x)=kx^2 is a probability density function
based on these conditions:

f(x)>=0 and f(x) dx = 1 when x is between -infinity and infinity.

anonymous · Answer

Therefore, 

$$f(x)=\int\limits_{0}^{1}kx^2dx=k \frac{ x^3 }{ 3 }_{0}^{1}=k/3=1$$
so k=3

anonymous · Answer

From here I'm plugging in the limits between 1/4 (.25) and 3/4 (.75).
 $$\int\limits_{.25}^{.75}kx^2dx=k \frac{ x^3 }{ 3 }_{.25}^{.75}=k(\frac{ .4218-.0156 }{ 3 })=.4062/3=.1354k$$

anonymous · Answer

I have a filling I did that last step wrong.

hartnn · Answer

thats correct, 0.1354*3=0.4062 is the correct probability for a)

hartnn · Answer

you know how to do part b), right ?

anonymous · Answer

^ alright good. For (b) do I convert f(x) into F(x), which is the distribution function then plug in the limit .67 (or 2/3) for F(x) and then minus that from 1 like so?
Find:
$$P(x>.67)$$
$$F(x)=\int\limits_{0}^{x}kx^2dx=kx^3/3$$
$$F(.67)=3*.67^3/3=.902289/3=.3008$$
$$1-.3008=.70$$

hartnn · Answer

that is correct, but there is also another way.
$\huge \int \limits_{(2/3)}^1 kx^2dx=....$
and u get same result 0.703

anonymous · Answer

yes that makes sense as well. Thanks.

hartnn · Answer

welcome ^_^