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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find mean and variance for the following frequency distribution :

$\begin{array}{1 1}(A)\;27,132\\(B)\;43,984\\(C)\;48,87\\(D)\;5,125\end{array} $

Can you answer this question?
 
 

1 Answer

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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:
$h=10$
$A=As \;n\;is \;odd$
$\qquad= \bigg( \large\frac{n+1}{2}\bigg)^{th} $ observation
$\qquad= \bigg( \large\frac{5+1}{2}\bigg)^{th} $ observation
$\qquad= 3$rd observation
$A= 25$
Step 2:
Mean $= A+\large\frac{f_id_i}{\sum f_i}$$ \times h$
$\qquad= 25+ \large\frac{10}{50} $$ \times 10$
$\qquad= 25 +\large\frac{100}{50}$
$\qquad= 25+2=27$
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]$$ \times h^2$
$\qquad= \bigg[\large\frac{68}{50} - \bigg( \large\frac{10}{50} \bigg)^2\bigg] $$ \times (10)^2$
$\qquad= \large\frac{68 \times 50 \times - 100 }{50 \times 50 } $$ \times 100$
$\qquad= \large\frac{3400-100}{50} $$ \times 2$
$\qquad= \large\frac{3300}{50 } $$ \times 2$
$\qquad= \large\frac{6600}{50}$
$\qquad= 132$
Hence A is the correct answer.
answered Jun 30, 2014 by meena.p
 
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