91影视

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 $<j^2>$.

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

The standard deviation is given by the equation,

$\begin{array}{l}\mathrm{蟽}=\sqrt{<{\mathrm{j}}^2>鈭�<\mathrm{j}{>}^2}\\ \mathrm{蟽}=\sqrt{459.57鈭�441}\\ \mathrm{蟽}=4.309\end{array}$

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

Hence, $螖j=j鈭�鉄�j鉄�$

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

Given equation is, ${蟽}^2=鉄�{\left(螖j\right)}^2鉄�$

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