Simple mean

Amswer is $\dfrac{1 + 3 + 9 + 16 + 25}{5} = \color{#69047E}{\boxed{11}}$ .

Its simple

The set of natural numbers or counting numbers ( denoted by N ) are :

N = { 1, 2, 3, 4, 5, 6, 7, ...........................}

Let the set of squares of natural numbers or counting numbers ( denoted by $N^{2}$ ) are :

$N^{2}$ = { $1^{2}$ , $2^{2}$ , $3^{2}$ , $4^{2}$ , $5^{2}$ , ............................}

= { 1, 4, 9, 16, 25, 36, 49 .................................}

Now,

The first five squares of natural numbers are :

1, 4, 9, 16, 25

Hence,

Average or mean = $\frac{1 + 4 + 9 + 16 + 25}{5}$

= $\frac{55}{5}$

= $\boxed{11}$

$\begin{aligned} Mean & = \dfrac{1^2 +2^2 + 3^2 + 4^2 + 5^2}{5} \\ & = \dfrac{1+ 4 + 9 + 16 + 25}{5} \\ & = \dfrac{55}{5} \\ & = \color{#D61F06}{\boxed{11}} \end{aligned}$

The mean of the squares of the first $p$ natural numbers is:

$\dfrac{\displaystyle \sum_{n=1}^p n^2}{p}\\ =\dfrac{1}{p} \left(\dfrac{p(p+1)(2p+1)}{6}\right)\\ =\dfrac{(p+1)(2p+1)}{6}$

Given that $p=5$ , the mean is

$\dfrac{(5+1)(2(5)+1)}{6}\\ =\dfrac{6(10+1)}{6}\\ =\boxed{11}$