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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Determine the mean and standard deviation for the following distribution:

$\begin{array}{1 1}(A)\;4.15,3.049\\(B)\;25,1600\\(C)\;790,87\\(D)\;5.975,2.85\end{array} $

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1 Answer

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Toolbox:
  • The formula to solve this problem:
  • Mean $(\bar {x} )=A+\large\frac{\sum f_id_i}{\sum f_i}$
  • Standard deviation (SD) $=\sqrt {\large\frac{\sum f_id_i^2}{\sum f_i}- \bigg(\large\frac{\sum f_id_ui}{\sum f_i} \bigg)^2}$
Step 2:
Mean $(\bar {x} )=A+\large\frac{\sum f_id_i}{\sum f_i}$
$\qquad= 9+\large\frac{-121}{40}$
$\qquad= 9- 3.025$
$\qquad= 5.975$
Step 3:
Standard deviation (SD) $=\sqrt {\large\frac{\sum f_id_i^2}{\sum f_i}- \bigg(\large\frac{\sum f_id_ui}{\sum f_i} \bigg)^2}$
$\qquad= \sqrt { \large\frac{691}{40} -\bigg(\large\frac{-121}{40}\bigg)^2}$
$\qquad= \large\frac{1}{40} $$\sqrt {27640 -14641}$
$\qquad=\large\frac{1}{40}$$ \sqrt {12999}$
$\qquad= \large\frac{114.013}{40}$
$\qquad= 2.85$
Hence D is the correct answer.
answered Jul 4, 2014 by meena.p
 

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