The median of 31, 16, 19, 25, 14, 13, 12, 4, 28, 45 is.

A. 14 B. 20 C.17.5 D. None of these.

Hint: Since all these numbers appear only once, we don’t need to make a frequency table. We will arrange them in ascending order and then find the average of the \[{\dfrac{n}{2}^{th}}\] and ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$ tem because n is even in this case.

Complete step-by-step answer:
Arranging in ascending order:
4, 12, 13, 14, 16, 19, 25, 28, 31, 45
clearly n = 10. Which is even
so to find the median, we need the average of the ${\dfrac{n}{2}^{th}} = {5^{th}}$ and ${\left( {\dfrac{n}{2} + 1} \right)^{th}} = {6^{th}}$ term.
${5^{th}}\;{\text{term}} = 16$
${6^{th}}{\text{term}}\; = 19$

$\boxed{{\text{Average}} = \dfrac{{16 + 19}}{2} = \dfrac{{35}}{2} = 17.5}$

Therefore, option C is right.

Note: In case of odd n, the ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term in itself is the median.
In case of repetition of data, try to make a table of data and frequency to help with easier calculation.