IB Mathematics Analysis and Approaches 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 $$f(x)=4x^3\text{for}0<x<1$$ 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 $$f(x)=\frac{3}{2}\sqrt{2-x}\text{ for }1\le x\le 2.$$ 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 $$f(x)=4x^3\text{for }0<x<1.$$

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 $f(x)=4x^3\text{for }0<x<1$, 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 $X$.