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

For a random variable $X$ with probability density function $f(x)=\frac{3}{2}(2-x)^{1/2}, \, 1<x<2,$ find its median.

Find $m$ such that $P(X\le m)=0.5$ for the distribution with pdf $f(x)=\frac{3}{2}(2-x)^{1/2},\,1<x<2.$

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

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 $f(x)=\frac{3}{2}\sqrt{2-x} \text{ for }1\le x\le2.$ Determine its median $m$.

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

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

Determine the median $m$ of the random variable with pdf $f(x)=\frac{3}{2}\sqrt{2-x} \text{ for } 1<x<2.$

Given the probability density function $f(x)=4x^3 \quad \text{for } 0 < x < 1$, find the value $m$ satisfying $\int_0^m f(x)\,dx = \frac{1}{2}$.

A continuous random variable has probability density function $f(x)=4x^3$ for $0<x<1$. Compute its median.

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