Question

Suppose X follows the uniform distribution from the range (-a,a) where a is greater than zero. Find the value of a such that P(X<1)=(1/3)=P(X>1)

Answer 1

A continuous rv $X$ is said to have a uniform distribution on the interval $[A,B]$ if the pdf of $X$ is

$$f(x;A,B) = \begin{cases} {1\over B-A} &A\leq x\leq B\\ 0 &otherwise \end{cases}$$

Suppose $X$ follows the uniform distribution from the range (-a,a) where a is greater than zero.

$$f(x;-a,a) = \begin{cases} {1\over 2a} & -a\leq x\leq a\\ 0 &otherwise \end{cases},\ a>0$$

$$P(X<-1)=\displaystyle\int_{-\infin}^{-1}f(x)dx=\displaystyle\int_{-a}^{-1}{1\over 2a}dx=$$ $$={1\over 2a}[x]\begin{matrix} -1 \\ -a \end{matrix}=-{1\over 2a}+{1\over 2},a\geq1$$

$$P(X<1)=\displaystyle\int_{-\infin}^{1}f(x)dx=\displaystyle\int_{-a}^{1}{1\over 2a}dx=$$ $$={1\over 2a}[x]\begin{matrix} 1 \\ -a \end{matrix}={1\over 2a}+{1\over 2},a\geq1$$

Given $P(X<-1)=\dfrac{1}{3}$

$$P(X<-1)=-{1\over 2a}+{1\over 2}={1\over 3}$$ $${1\over 2a}={1\over 6}$$ $$a=3$$

$$P(X>1)=1-P(X<1)=1-({1\over 2(3)}+{1\over 2})={1\over 2}-{1\over 6}={1\over 3}$$

$a=3.$