Measures of Dispersion — Maths STD 11 Science — Question

$\bar{x}=\frac{\sum f_i x_1}{ N }=\frac{5820}{100}=58.2$

$\operatorname{Var}( X )=\sigma_x^2=\frac{\sum f _{ i } x_{ i }^2}{ N }-(\bar{x})^2$

$\begin{aligned} & =\frac{404100}{100}-(58.2)^2 \\ & =4041-3387.24\end{aligned}$

$=653.76$

S.D. $=\sigma_x=\sqrt{\operatorname{Var}(X)}=\sqrt{653.76}=25.56$

Alternate Method:

Let $u =\frac{ x - A }{ h }=\frac{ x -55}{10}$

Calculation of variance of u:

$\overline{ u }=\frac{\sum f _{ f } u _i}{ N }=\frac{32}{100}=0.32$

$\begin{aligned} \bar{x} & =\overline{ u } \times h + A \\ & =0.32 \times 10+55\end{aligned}$

$=58.2$

$\operatorname{Var}( u )=\sigma_{ u }{ }^2=\frac{\sum f _i u _i{ }^2}{ N }-(\overline{ u })^2$

$\begin{aligned} & =\frac{664}{100}-(0.32)^2 \\ & =6.64-0.1024 \\ & =6.5376\end{aligned}$

$\begin{aligned} \operatorname{Var}(X) & =h^2 \operatorname{Var}(u) \\ & =(10)^2 \times 6.5376 \\ & =100 \times 6.5376 \\ & =653.76\end{aligned}$

S.D. $=\sigma_x=\sqrt{\operatorname{Var}(X)}=\sqrt{653.76}=25.56$