Answer
a) $\lt{j}^{2}\gt$ =459.571
$\hspace{.4cm}{\lt{j}\gt}^{2}\hspace{.1cm}$= 441
b)
j $\hspace{2cm}\Delta{j}$
14 $\hspace{1.7cm}$ -7
15 $\hspace{1.7cm}$ -6
16 $\hspace{1.7cm}$ -5
22 $\hspace{1.8cm}$ 1
24 $\hspace{1.8cm}$ 3
25 $\hspace{1.8cm}$ 4
$\sigma=4.309$
c)Equation 1.12 is verified
Work Step by Step
a) $\lt{j}^{2}\gt$ =$\sum_{j=0}^{\infty}{j}^{2}P(j)$ , $P(j)=\frac{N(j)}{N}$ (Probability of getting age j = number of people of age j/total number of people)
N=$\sum_{j=0}^{\infty}N(j)$=1+1+3+2+2+5=14
$\lt{j}^{2}\gt$=$14^{2}(\frac{1}{14}) + 15^{2}(\frac{1}{14}) + 16^{2}(\frac{3}{14}) + 22^{2}(\frac{2}{14}) + 24^{2}(\frac{2}{14}) + 25^{2}(\frac{5}{14})$
$\hspace{1.3cm}$=$\large\frac{(196 + 225 + 768 + 968 + 1152 + 3125)}{14} =\frac{6434}{14} =\normalsize 459.571$
${\lt{j}\gt}^2$ =${(\sum_{j=0}^{\infty}{j}P(j))}^{2}$
${\lt{j}\gt}^2$=${(14(\frac{1}{14}) + 15(\frac{1}{14}) + 16(\frac{3}{14}) + 22(\frac{2}{14}) + 24(\frac{2}{14}) + 25(\frac{5}{14}))}^{2}$
$\hspace{1.3cm}$=$\large\frac{{(14 + 15 + 16 + 44 + 48 + 125)}^{2}}{196}=\frac{{(294)}^{2}}{196}=\frac{86436}{196}=\normalsize441$
b) $\Delta{j}=j-\lt{j}\gt$
$\lt{j}\gt=\sqrt{441}=21$ (found in part a))
$\Delta{j}\vert_{j=14}=14-21=-7$
$\Delta{j}\vert_{j=15}=15-21=-6$
$\Delta{j}\vert_{j=16}=16-21=-5$
$\Delta{j}\vert_{j=22}=22-21=1$
$\Delta{j}\vert_{j=24}=24-21=3$
$\Delta{j}\vert_{j=25}=25-21=4$
$\sigma=\sqrt{\lt{{(\Delta{j})}^2}\gt}$=$\sum_{j=0}^{\infty}{(\Delta{j})}^{2}P(j)$ , $P(j)=\frac{N(j)}{N}$
$\sigma$=$\small\sqrt{(-7)^{2}(\frac{1}{14}) + (-6)^{2}(\frac{1}{14}) + (-5)^{2}(\frac{3}{14}) + (1)^{2}(\frac{2}{14}) + (3)^{2}(\frac{2}{14}) + (4)^{2}(\frac{5}{14})}$
$\hspace{.3cm}$=$\large\sqrt{\frac{49+ 36+75+2+18+80}{14}}={\normalsize\sqrt{18.57}}=\normalsize4.309$
c) $\sigma=\sqrt{\lt{j}^{2}\gt-{\lt{j}\gt}^2}$
From a)
$\lt{j}^{2}\gt=459.571$
${\lt{j}\gt}^2=441$
$\Rightarrow\sigma=\sqrt{459.571-441}=\sqrt{18.571}=4.309$,
which agrees with the result obtained in part b)
