Question Type 4: Finding the median for a given PDF Bootcamps

Given the probability density $f(x)=4x^3 \text{ for }0<x<1,$ find the value $m$ satisfying $\int_0^m f(x)\,dx =\tfrac12.$

Find the median $m$ of the random variable with probability density function $f(x)=4x^3 \text{ for }0<x<1.$

Let $X$ have density $f(x)=4x^3 \text{ for }0<x<1.$ Find the median of $X$.

For a continuous random variable $X$ with density $f(x)=4x^3 \text{ for }0<x<1,$ determine the value $m$ such that $P(X\le m)=0.5$.

A continuous random variable has density $f(x)=4x^3 \text{ for }0<x<1.$ Compute its median.

Show that the median of the distribution with pdf $f(x)=4x^3 \text{ for }0<x<1$ is $m=(1/2)^{1/4}$.

Find $m$ such that $P(X\le m)=0.5$ for the distribution with pdf $f(x)=\tfrac32(2-x)^{1/2},\,1<x<2.$

Determine the median $m$ of the random variable with pdf $f(x)=\tfrac32\sqrt{2-x} \text{ for }1<x<2.$

A continuous random variable has density $f(x)=\tfrac32\sqrt{2-x} \text{ for }1\le x\le2.$ Determine its median $m$.

Given the pdf $f(x)=\tfrac32(2-x)^{1/2}\text{ for }1\le x<2,$ show that the median is $m=2-2^{-2/3}$.

For a random variable $X$ with pdf $f(x)=\tfrac32(2-x)^{1/2},\,1<x<2,$ find its median.