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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Find mean and variance and standard deviation using short cut method.

$\begin{array}{1 1}(A)\;27,132,87\\(B)\;93,105.58,10.27\\(C)\;48,87,46\\(D)\;5,125,45\end{array} $

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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$
  • Standard deviation $ \sigma =\sqrt {variance}$
Step 1:
$h=5$
$A=As \;n\;is \;odd$
$\qquad= \bigg( \large\frac{n+1}{2}\bigg)^{th} $ observation
$\qquad= \bigg( \large\frac{9+1}{2}\bigg)^{th} $ observation
$\qquad= 5$th observation
$\qquad= 92.5$
Step 2:
Mean $= A+\large\frac{f_id_i}{\sum f_i}$$ \times h$
$\qquad= 92.5 \times \large\frac{6}{60} $$ \times 5$
$\qquad= 92.5 +0.5$
$\qquad= 93$
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]$$ \times h^2$
$\qquad= \bigg[\large\frac{254}{60} -\bigg( \large\frac{65}{60}\bigg) \bigg]$$ \times 5^2$
$\qquad= \large\frac{254 \times 60 -36}{60 \times 60} $$ \times 25$
$\qquad= \large\frac{15204 -36 }{60 \times 12} $$ \times 5$
$\qquad =\large\frac{15204}{12 \times 12} $$=105.58$
Step 4:
Standard deviation $\sigma =\sqrt {variance}$
$\qquad= \sqrt {105.58}$
$\qquad= 10.27$
Mean =93
Variance= 105.58
Standard deviation =10.27
Hence B is the correct answer.
answered Jun 30, 2014 by meena.p
 
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