The median of a continuous random variable having distribution function F is that value m such that F(m) = $\frac{1}{2}$. That is, a random variable is just as likely to be larger than its median as it is to be smaller. Find the median of X if X is

(a) uniformly distributed over (a, b);

(b) normal with parameters μ,σ^$2$;

(c) exponential with rate λ.

Given the value that, F(M)=$\frac{1}{2}$

Find the median of X if X is uniformly distributed over (a, b);

$F\left(m\right)=\frac{1}{2}\text{(given)}$

$P[x<m]=P[x>m]\left(\text{given}\right)$

a) Uniform distribution

$F_x\left(x\right)=\{\begin{array}{ll}0& x<a\\ \frac{x-a}{b-a}& a\le x\le b\\ 1& x>b\end{array}$

$F_x\left(m\right)=\frac{m-a}{b-a}=1/2$

$2(m-a)=b-a$

$m=\frac{a+b}{2}$

uniformly distributed over (a, b) is$m=\frac{a+b}{2}$

Find the median of X if X is normal with parameters μ,σ^$2$;

$F\left(m\right)=P[x\le m]=1/2$

$P[z\le \frac{m-\mu }{{\sigma }^2}]=0.5$

So we have to calculate that value of z for which the probability will be equal to ' $0.5$'. From the normal distribution table, we can see that at z=$0$, the probability

$z=\frac{m-\mu }{\sigma }=0$

$m=\mu$

normal with parameters μ,σ^$2$; is$m=\mu$

Find the median of X if X is exponential with rate λ.

Here,

$F_x\left(x\right)={\int }_0^{x}d{e}^{-dx}dx$

$={\left[\frac{\lambda e^{-\lambda x}}{-\lambda }\right]}_0^x$

$F_x\left(x\right)=-{\left[e^{-\lambda x}\right]}_0^x$

$F_x\left(x\right)=1-e^{-\lambda x}$ ( cummulative distribution function)

$F_x\left(m\right)=1-e^{-\lambda m}=1/2$(given)

$e^{-\lambda m}=1/2$

$\lambda m=\ln 2$

$m=\frac{\ln 2}{\lambda }$

If x is exponential with rate λ. is $m=\frac{\ln 2}{\lambda }$