Cont. Random Var has PDF f(x)=1/2 (0<=x<=1),=(3-x)/4 (1<=x<=3),=0 (otherwise)

Cont. Random Var has PDF f(x)=1/2 (0<=x<=1),=(3-x)/4 (1<=x<=3),=0 (otherwise)


the questions that i do not get are b and d
my teacher told me for q.b that 1/2 is half of the data we have therefore 1/2*2=1 but what i dont get is that the question is asking for the mediam of X so does that mean the random continuous variables are all median??
For question d.
by completing question c i know that u=13/12, 3u=13/4, 13/4-1=9/4
P(X<9/4)
then i 1/2*9/4*(3-9/4)/4 which isnt right as the mark scheme states i should have for this stage 1/2*3/4*(3-9/4)/4
what i do not understand is where did the 3/4 come from??
please explain thank you very much

ka923 said:

View attachment 10513
the questions that i do not get are b and d
my teacher told me for q.b that 1/2 is half of the data we have therefore 1/2*2=1 but what i dont get is that the question is asking for the mediam of X so does that mean the random continuous variables are all median?

Click to expand...

There is the function:

\(\displaystyle f(x)=\begin{cases}\dfrac{1}{2} & 0\le x< 1 \\ \dfrac{3-x}{4} & 1\le x<3 \\ 0 &\text{otherwise}\end{cases}\)


Is it true that \(\displaystyle \int_0^1 {\frac{1}{2}dx} = \int_1^3 {\frac{{3 - x}}{4}dx = \frac{1}{2}}~? \)

How is the mean calculated?

should it not be adding the two numbers together and dividing it by 2?
assuming that the other function is also 1/2
my first guess would've been 1/2+1/2=1 1/2
but i know that i am wrong for sure

(b) With a probability density function, the median is the x value where you can draw a vertical line to split the area under the curve in half. Since the area under a pdf is 1, then the area in each half must be 0.5. On your graph, draw a vertical line up from 1 (the given median). Does this give the area of 0.5 on both sides?
If so, then you have confirmed it is the median.

(d) to find P(X<2.25), you need to find the area under your graph to the left of 2.25. There are several ways to do this:
(1) integrate the first part of the hybrid function from 0 to 1 + integrate the second part of the hybrid function from 1 to 2.25
or
(2) forget about integrating and just find the area (as the sum of the area of a rectangle and a triangle)
or
(3) easiest way: You know the area under the pdf is 1, so just subtract the area of the little triangle on the right from 1.
(The area of the triangle is 0.5 x 0.75 x (3-2.25)/4). Maybe that's where the 3/4 (or 0.75) comes in.

ka923 said:

should it not be adding the two numbers together and dividing it by 2?
assuming that the other function is also 1/2 - but it's not so you cant do that.
my first guess would've been 1/2+1/2=1 1/2
but i know that i am wrong for sure

Click to expand...

To calculate the mean you will need to use integration.

Mean = \(\displaystyle \int x*f(x) dx\) for the domain 0 to 3.

But because your case is a hybrid function, you will need to split it up from 0 to 1 (using f(x) =1/2) and then from 1 to 3 using the other part of the hybrid function.