Introduction to Quantum Mechanics 2nd Edition

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
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)