The diameters of circles(in mm) drawn in a design are given below . Calculate the std. deviation and mean diameter of the circles.

$\begin{array}{1 1}(A)\;27,132\\(B)\;43.5,5.55\\(C)\;48,87\\(D)\;5,125\end{array}$

Toolbox:
• The formula required to solve this problem are : Mean $A+ \large\frac{\sum f_i d_i}{\sum f_i} $$\times h • Standard deviation \sigma= \sqrt {\large\frac{\sum f_i d_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_i d_i}{\sum f_i} \bigg)^2 }$$ \times h$
As the given data is discontinuous, first make the data continuous by making the classes as $32.5 -36.5, 36.5-40.5,40.5-44.5,44.5-48.5,48.5-52.5$
Step 1:
$h=4$
$A=As \;n\;is \;odd$
$\qquad= \bigg( \large\frac{n+1}{2}\bigg)^{th}$ observation
$\qquad= \bigg( \large\frac{5+1}{2}\bigg)^{th}$ observation
$\qquad= 3$rd observation
$\qquad= 42.5$
Step 2:
Mean $A+ \large\frac{\sum f_i d_i}{\sum f_i} $$\times h \qquad= 42.5 +\large\frac{25}{100}$$ \times 4$
$\qquad= 42.5 +1$
$\qquad= 43.5$
Step 3:
Standard deviation $\sigma= \sqrt {\large\frac{\sum f_i d_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_i d_i}{\sum f_i} \bigg)^2 }$$\times h \qquad= \sqrt {\large\frac{199}{100} -\bigg( \large\frac{25}{100}\bigg)^2 } \times h \qquad= \sqrt{\large\frac{199 \times 100 -625}{100 \times 100}}$$ \times 4$
$\qquad=\large\frac{4}{100} $$\sqrt {19900-625} \qquad= \large\frac{1}{25}$$ \sqrt {19275}$
$\qquad=\large\frac{138.83}{25}$$=5.55$
Hence B is the correct answer.