Question

The random variable Xhas a probability density function,.f{x) where

f(x)= $\begin{Bmatrix} a-5+x; 5<x<=6 \\ 0 ; otherwise \end{Bmatrix}$ {

(a) Show that a=1/2

(b) Calculate P( X > 5. 7) .

(c) Find E (X), Var (X) and Var(3+X).

Answer 1

Solution:

f(x)= $\begin{Bmatrix} a-5+x; 5<x<=6 \\ 0 ; otherwise \end{Bmatrix}$

(a) Since, f(x) is pdf, $\int_5^6f(x)dx=1$

$\int_5^6[a-5+x]dx=1 \\ \Rightarrow [ax-5x+\dfrac{x^2}2]_5^6=1 \\ \Rightarrow [6a-30+18]-[5a-25+12.5]=1 \\ \Rightarrow a-5+5.5=1 \\ \Rightarrow a=0.5$

(b):

Now, f(x)= $\begin{Bmatrix} x-4.5; 5<x<=6 \\ 0 ; otherwise \end{Bmatrix}$

$P(X>5.7)=\int_{5.7}^6(x-4.5)dx \\=(\dfrac{x^2}2-4x)_{5.7}^6 \\=(18-24)-(16.245-22.8) \\=0.555$

(c):

$E[X]=\int_5^6x(x-4.5)dx \\=\int_5^6(x^2-4.5x)dx \\=(\dfrac{x^3}3-4.5\times\dfrac{x^2}2)_5^6 \\=(\dfrac{x^3}3-2.25x^2)_5^6 \\=(\dfrac{216}3-2.25(36))-(\dfrac{125}3-2.25(25)) \\=5\dfrac{7}{12}$

$E[X^2]=\int_5^6x^2(x-4.5)dx \\=\int_5^6(x^3-4.5x^2)dx \\=(\dfrac{x^4}4-4.5\times\dfrac{x^3}3)_5^6 \\=(\dfrac{x^4}4-1.5x^3)_5^6 \\=(\dfrac{1296}4-1.5(216))-(\dfrac{625}4-1.5(125)) \\=31.25$

Now, $Var[X]=E[X^2]-(E[X])^2=31.25-(5\dfrac7{12})^2=\dfrac{11}{144}$

And $Var[3+X]=0+Var[X]=\dfrac{11}{144}$