Question

What is the median of the first 12 prime numbers?
A) $13$
B) $14$
C) $15$
D) $17$

Answer 1

Hint: In this question, we have been asked the median of first 12 prime numbers. Start by noting down the first 12 prime numbers. Since it is given that there are 12 prime numbers, we will use the formula of even numbers of medians putting $n = 12$ to find the two middle most terms. Once we have found the two middle most terms, we will find their average. The average will be the required median.

Complete step-by-step solution:
We have to find the median of the first 12 prime numbers. Before moving towards the solution, let us know what prime numbers are.
Prime numbers are those numbers that are divisible only by 1 and the number itself. Other numbers which have more than 2 factors are called composite numbers. It should be noted that the number- 1 is neither prime, nor composite.
Now, let us note down the first 12 prime numbers.
$2,3,5,7,11,13,17,19,23,29,31,37$.
Since there are 12 (even) numbers, we will have to find the average of the two middle numbers.
The two middle terms can be founded by using the formula- ${\left( {\dfrac{n}{2}} \right)^{th}}$ and ${\left( {\dfrac{n}{2} + 1} \right)^{th}}$ term
Putting $n = 12$,
$ \Rightarrow {\left( {\dfrac{{12}}{2}} \right)^{th}}$ and ${\left( {\dfrac{{12}}{2} + 1} \right)^{th}}$
$ \Rightarrow {6^{th}}$ and ${7^{th}}$ term
$ \Rightarrow {6^{th}}$ term $ = 13$, and ${7^{th}}$ term $ = 17$
Now, we have both our numbers and we have to find their average. That average will be our mean.
$ \Rightarrow \dfrac{{13 + 17}}{2}$
$ \Rightarrow \dfrac{{30}}{2} = 15$

Hence, the median of the first 12 prime numbers is option C.

Note: The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average. The median is sometimes used as opposed to the mean when there are outliers in the sequence that might skew the average of the values. We had 12 prime numbers in this question. If we had 11 numbers, we would have used the formula- ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term or the middle most term would have been our median.