What is the Arithmetic Mean of the first 16 natural numbers with weights being the number itself?
\[\left( a \right)\dfrac{17}{2}\]
\[\left( b \right)\dfrac{33}{2}\]
\[\left( c \right)11\]
\[\left( d \right)\dfrac{187}{2}\]

Hint: To solve this question, we will use the formula of the Arithmetic Mean given as \[AM=\dfrac{\text{Total Weights}}{\text{Number of Observations}}.\] The total weights is given by \[\sum\limits_{n=1}^{16}{{{n}^{2}}}\] whose formula is \[\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}\] and the number of observations is given by \[\sum{n}\] whose formula is \[\sum{n}=\dfrac{n\left( n+1 \right)}{2}.\]

Complete step by step answer:
The formula of Arithmetic Mean is given by \[\text{Arithmetic Mean}=\dfrac{\text{Total Weights}}{\text{Number of Observations}}.\] Here we are using the numbers from 1, 2, 3, ….16. Given that the weights of the numbers are itself. The weight of 1 is 1. Therefore, the total weight of 1 will be \[1=1\times 1.\]
Now, similarly, the weight of 2 is 2. Therefore, the total weight of 2 will be \[2=2\times 2.\]
Similarly, the weight of 3 is 3. Therefore, the total weight of 3 will be \[3=3\times 3.\]
So, the total weights of numbers from 1 to 16 is given by
\[1\times 1+2\times 2+3\times 3+4\times 4+.....+16\times 16\]
This is nothing by \[\sum\limits_{n=1}^{16}{{{n}^{2}}}.\]
The sum of the square of number n is given by the formula
\[\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}}\]
When n = 1 till 16,
\[\sum\limits_{n=1}^{16}{{{\left( n \right)}^{2}}}=\dfrac{16\left( 16+1 \right)\left( 16\times 2+1 \right)}{6}\]
\[\Rightarrow \sum\limits_{n=1}^{16}{{{\left( n \right)}^{2}}}=\dfrac{16\left( 17 \right)\left( 33 \right)}{6}\]
Therefore, the total weight is given by the value,
\[\Rightarrow \text{Total Weight}=\dfrac{16\left( 17 \right)\left( 33 \right)}{6}.......\left( i \right)\]
Now, we have to calculate the total number of observations. The total number of observations is given by \[1+2+3+4+.....+16\] that is \[\sum{n}.\]
And the formula of \[\sum\limits_{n=1}^{16}{n}=\dfrac{n\left( n+1 \right)}{2}.\]
Substituting n = 16, we get,
\[\sum\limits_{n=1}^{16}{n}=\dfrac{16\left( 17 \right)}{2}\]
\[\Rightarrow \text{Number of observations}=\dfrac{16\left( 17 \right)}{2}.......\left( ii \right)\]
Hence we have Arithmetic Mean as given by the equation (i) and (ii) is
\[\text{Arithmetic Mean}=\dfrac{\text{Total Weights}}{\text{Number of Observations}}\]
\[\Rightarrow \text{Arithmetic Mean}=\dfrac{16\left( 17 \right)\left( 33 \right)}{6\times 16\left( 17 \right)}\times 2\]
\[\Rightarrow \text{Arithmetic Mean}=11\]
Therefore, we have the Arithmetic Mean of the first 16 natural numbers with the weights being the number itself is 11.

So, the correct answer is “Option C”.

Note: The Natural numbers are the numbers starting from 1, 2, 3,…. Whole numbers are numbers starting from 0. This should be taken care of in this question. Although adding 0 to the numerator and denominator would not matter as 0 + a = a for any real number a. Therefore, the Arithmetic Mean remains unchanged in that case too.