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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} $

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1 Answer

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  • 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.
answered Jun 27, 2014 by meena.p
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