91Ó°ÊÓ

The random variable X has a discrete distribution for which the p.f. is as shown below.

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{cx}&{{\rm{for }}x = 1,2,3,4,5,6}\\0&{{\rm{otherwise}}}\end{aligned}} \right.\)

We know:

Then, the probability function of X is:

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{\frac{x}{{21}}}&{{\rm{for }}x = 1,2,3,4,5,6}\\0&{{\rm{otherwise}}}\end{aligned}} \right.\)

Using the probability function defined in the previous step, find the probability of each value that the random variable X can take as follows:

\(\begin{aligned}{}f\left( 1 \right) &= \frac{1}{{21}}\\f\left( 2 \right) &= \frac{2}{{21}}\\f\left( 3 \right) &= \frac{3}{{21}}\\f\left( 4 \right) &= \frac{4}{{21}}\\f\left( 5 \right) &= \frac{5}{{21}}\\f\left( 6 \right) &= \frac{6}{{21}}\end{aligned}\)

From the above values, it can be inferred that, \(\Pr \left( {X \le 5} \right) = \frac{{15}}{{21}} \ge \frac{1}{2}{\rm{ and }}\Pr \left( {X \ge 5} \right) = \frac{{11}}{{21}} \ge \frac{1}{2}\)

Hence, \(x = 5\) satisfies both the conditions of a median. It should be noted that no other value of X satisfies both the conditions of a median.

Thus, the median of X is 5.