Consider a sample of size $5$from a uniform distribution over $(0,1)$. Compute the probability that the median is in the interval $\left(\frac{1}{4},\frac{3}{4}\right)$.

In statistics, a uniform distribution is a probability distribution in which all outcomes have the same probability.

A sample of size $5$drawn from a uniform distribution with a standard deviation of $\left(0,1\right)$.

The size of the samples, $n=5$is known.

The sample's median value is $j=3$. As a result, use the density function of a third-order statistic.

The uniform distribution's density function is,

$$\begin{gathered} f\left(x\right)=\frac{1}{b-a} \\ =\frac{1}{1-0} \\ =1 \end{gathered}$$

The distribution function is,

localid="1647268220737" $$\begin{gathered} f\left(x\right)=P(X\le x) \\ ={\int }_0^{x}1dx \\ ={\left(x\right)}_0^x \\ =x \end{gathered}$$

The density function of third-order statistics may be written as follows:

$$\begin{gathered} f_X\left(x\right)=\frac{n!}{(n-j)!(j-1)}{\left[F\left(x\right)\right]}^{j-1}{\left[1-F\left(x\right)\right]}^{n-j}f\left(x\right) \\ =\frac{5!}{(5-3)!(3-1)}{\left[x\right]}^{2-1}{[1-x]}^{5-3}\left(1\right) \\ =\frac{5!}{2!2!}{x}^2{(1-x)}^2 \end{gathered}$$

Assess the probability that the median is inside the interval $\left(\frac{1}{4},\frac{3}{4}\right)$.

That is, find $P\left(\frac{1}{4}<X_3<\frac{3}{4}\right)$

$$\begin{gathered} P\left(\frac{1}{4}<X_3<\frac{3}{4}\right)=30{\int }_{\frac{1}{4}}^{\frac{3}{4}}{x}^2{(}_{-}^{1}dx \\ =30{\left|\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^4}{2}\right|}_{\frac{1}{4}}^{\frac{3}{4}} \\ =30\left|\frac{9}{4^3}+\frac{3^5}{5\times 4^5}-\frac{3^4}{2\times 4^4}-\frac{1}{3\times 4^3}-\frac{1}{5\times 4^5}+\frac{1}{2\times 4^4}\right| \\ =30\left|\frac{26}{3\times 4^3}+\frac{242}{5\times 4^5}-\frac{80}{2\times 4^4}\right| \\ =30\left|0.1354+0.0473-0.1562\right| \\ =0.793 \end{gathered}$$

As a result, the probability that the median in the interval$\left(\frac{1}{4},\frac{3}{4}\right)$is$0.793$.