Question

Match the value of standard deviation with respective method and their data.

• A. $\sqrt{2.5}$
• B. $\sqrt{2}$
• C. $\sqrt{0.5}$

Answer 1

(i) By Direct method
Given data: 1, 2, 3, 4, 5

$\sigma =\sqrt{\frac{\sum x_i^2}{n}-(\frac{\sum x_i}{n}{)}^2}$

$\sum x_i=1+2+3+4+5=15$
$\sum x_i^2=1+4+9+16+25=55$

$\sigma =\sqrt{\frac{55}{5}-(\frac{15}{5}{)}^2}$
$\sigma =\sqrt{11-9}=\sqrt{2}$

(ii) By Mean method
Given data: 3, 4, 6, 7

$\sigma =\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{n}}$

$\overline{x}=\frac{3+4+6+7}{4}=5$
$\sum (x_i-\overline{x}{)}^2=(3-5{)}^2+(4-5{)}^2+(6-5{)}^2+(7-5{)}^2$$=4+1+1+4=10$

$\Rightarrow \sigma =\sqrt{\frac{10}{4}}=\sqrt{2.5}$

(iii) By Assumed mean method
Given data: 3, 4, 5

$\sigma =\sqrt{\frac{\sum d_i^2}{n}-(\frac{\sum d_i}{n}{)}^2}$
Here, A = 4 (Middle value of the given data)
$\sum d_i=\sum (x_i-A)=(3-4)+(4-4)+(5-4)=-2-1+0+1+2=0$
$\sum d_i^2=\sum (x_i-A{)}^2=(3-4{)}^2+(4-4{)}^2+(5-4{)}^2=1+0+1=2$

$\sigma =\sqrt{\frac{2}{4}-(\frac{0}{4}{)}^2}=\sqrt{0.5}$