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Home  >>  CBSE XI  >>  Math  >>  Statistics
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If for a distribution $ \sum (x-5) =3,\sum (x-5)^2 =43$ and the total number of items is $18$. Find the mean and standard deviation .

$\begin{array}{1 1}(A)\;11,22\\(B)\;25,1600\\(C)\;5.16,1.54\\(D)\;15,3.45\end{array} $

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1 Answer

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Toolbox:
  • Formulae used to solve this problem are :
  • Mean $=\large\frac{\sum x}{n}$
  • $SD(\sigma)= \sqrt { \large\frac{ \sum x^2}{n} -\bigg( \large\frac{ \sum x}{n}\bigg)^2}$
Step 1:
Let us calculated the value of $\sum x $ and $ \sum x^2$
Given $ \sum (x-5) =3 \qquad n=18$
$=> \sum x - \sum 5=3$
$=> \sum x - 18 \times 5=3$
$=> \sum x = 3+90$
$=> \sum x=93$
$\sum (x-5)^2=43$
$\sum (x^2 +25-10x) =43$
$\sum x^2+\sum 25 -10 \sum x =43$
$\sum x^2+25 \times 18 -10 \times 93 =43$
$\sum x^2 =43 +930 -450$
$\qquad=523$
Step 2:
Mean $=\large\frac{\sum x}{n}=\frac{93}{18} $$=5.16$
Step 3:
$SD(\sigma)= \sqrt { \large\frac{ \sum x^2}{n} -\bigg( \large\frac{ \sum x}{n}\bigg)^2}$
$\qquad= \sqrt {\large\frac{523}{18} - \bigg( \large\frac{93}{18}\bigg)^2}$
$\qquad= \sqrt { \large\frac{523 \times 18 -93 \times 93}{18 \times 18} }$
$\qquad= \large\frac{1}{18} $$ \sqrt{ 9414-8649}$
$\qquad= \large\frac{27.66}{18} $
$\qquad= 1.54$
Hence C is the correct answer.
answered Jul 4, 2014 by meena.p
 

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