Q.6.47 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

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

Step 1 : Uniform distribution

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

Step 2 : Explanation :

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,

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

Step 3 : Explanation :

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 $0.793$.