Question

The pdf of a continuous random variable R is given by f(r)={kr for 0<x<4 both inclusive

{0,elsewhere

a)determine c hence compute p(10<r<2)both inclusive, E(r) and var(r)

Answer 1

Solution:

(a): We know that $\int_a^bf(x)dx=1$

$\int_0^4(kr)dr=1$

$\Rightarrow [kr^2/2]_0^4=1 \\ \Rightarrow \dfrac12\times [16k]=1 \\ \Rightarrow 8k=1 \\ \Rightarrow k=\dfrac18$

(b):

So, $f(x)=\dfrac r8$

Now, $P(-10\le r\le2)=P(-10\le r<0)+P(0\le r\le 2)$

$=0+\int_0^2(\dfrac r8)dr \\=\dfrac{r^2}{16}|_0^2 \\=\dfrac14$

Next, $E(r)=\int_0^4r(\dfrac r8)dr=\int_0^4(\dfrac {r^2}8)dr$

$=\dfrac{r^3}{24}|_0^4 \\=\dfrac{64}{24}-0 \\=\dfrac 83$

$E(r^2)=\int_0^4r^2(\dfrac r8)dr=\int_0^4(\dfrac {r^3}8)dr \\ =\dfrac{r^4}{32}|_0^4 \\=\dfrac{16\times 16}{32}-0 \\=8$

Now, $Var(r)=E(r^2)-[E(r)]^2=8-(\dfrac83)^2=\dfrac89$