For the distribution of ages in the example in Section 1.3.1:

(a) Compute$\mathbf{⟨}{\mathbf{j}}^{\mathbf{2}}\mathbf{⟩}$ and${\mathbf{⟨}\mathbf{j}\mathbf{⟩}}^{\mathbf{2}}$ .

(b) Determine ∆j for each j, and use Equation 1.11 to compute the standard deviation.

(c) Use your results in (a) and (b) to check Equation 1.12.

The given equation 1.12 is,

Here, j represents the ages of the individuals.

Let${\mathrm{N}}_{\left(\mathrm{j}\right)}$be the number of individuals with age j.

Hence,

role="math" localid="1655374067007" $\begin{array}{c}{\mathrm{N}}_{\left(14\right)}=1\\ {\mathrm{N}}_{\left(15\right)}=1\\ {\mathrm{N}}_{\left(16\right)}=3\\ {\mathrm{N}}_{\left(22\right)}=2\\ {\mathrm{N}}_{\left(24\right)}=2\\ {\mathrm{N}}_{\left(25\right)}=5\end{array}$

Average of the values is given by,

Since we just calculated the expectation value, hence squaring it for the required answer:

$\begin{array}{l}<j{>}^2={21}^2\\ <j{>}^2=441\end{array}$

Now solving for $<j^2>$.

$\begin{array}{l}<j^2>=\frac{{\displaystyle \sum _j^{}{j}^2{N}_j}}{{\displaystyle \sum _j^{}{N}_j}}\\ <j^2>=\frac{{14}^2\times 1+{15}^2\times 1+{16}^2\times 3+{22}^2\times 2+{24}^2\times 2+{25}^2\times 5}{1+1+3+2+2+5}\\ <j^2>=\frac{196+225+768+968+1152+3125}{14}\\ <j^2>=\frac{3217}{7}\\ <j^2>=459.57\end{array}$

The standard deviation is given by the equation,

$\begin{array}{l}\mathrm{\sigma }=\sqrt{<{\mathrm{j}}^2>-<\mathrm{j}{>}^2}\\ \mathrm{\sigma }=\sqrt{459.57-441}\\ \mathrm{\sigma }=4.309\end{array}$

Difference between each j and the average value $<j>$is given by Δj.

Hence, $\Delta j=j-⟨j⟩$

$\boxed{\begin{array}{cccc}j& \Delta j=j-21& {\left(\Delta j\right)}^2& N_j\\ 14& -7& 49& 1\\ 15& -6& 36& 1\\ 16& -5& 25& 3\\ 22& 1& 1& 2\\ 24& 3& 9& 2\\ 25& 4& 16& 5\end{array}}$

Given equation is, ${\sigma }^2=⟨{\left(\Delta j\right)}^2⟩$

Since the answer for σ is same, therefore the equation 1.12 is satisfied.