# Find mean and variance for the following frequency distribution :

$\begin{array}{1 1}(A)\;107,2276\\(B)\;43,984\\(C)\;48,87\\(D)\;5,125\end{array}$

Toolbox:
• The formula required to solve this problem are : Mean $A+ \large\frac{\sum f_i d_i}{\sum f_i} $$\times h • Variance = \bigg[\large\frac{\sum f_i d_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_i d_i}{\sum f_i} \bigg)^2 \bigg]$$ \times h^2$
Step 1:
$h=30$ [class difference is 30]
A= As nth is odd
$\qquad= \bigg( \large\frac{(n+1)}{2}\bigg)^{th}$ observation
$\qquad= \bigg( \large\frac{(7+1)}{2}\bigg)^{th}$ observation
$\qquad= 4th$ observation
$A= 105$
Step 2:
Mean $= A+\large\frac{f_id_i}{\sum f_i}$$\times h \qquad= 105 +\large\frac{2}{30}$$\times 30$
$\qquad= 105 +2 =107$
Step 3:
Variance $= \bigg[\large\frac{\sum f_i d_i^2}{\sum f_i} - \bigg( \large\frac{\sum f_i d_i}{\sum f_i} \bigg)^2 \bigg]$$\times h^2 \qquad= \bigg[\large\frac{76}{30} - \bigg( \large\frac{2}{30}\bigg) ^2 \bigg]$$ \times (30)^2$
$\qquad= \large\frac{2276}{900}$$\times 900=2276$
Hence A is the correct answer.