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Home  >>  CBSE XI  >>  Math  >>  Statistics
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Mean and standard deviation of 100 items are 50 and 4 , respectively . find the sum of all items and sum of the squares of the items.

$\begin{array}{1 1}(A)\;1122\\(B)\;251600\\(C)\;79087\\(D)\;15345\end{array} $

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1 Answer

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Mean $\bar{x} =\large\frac{\sum x_i}{n}$
Given $\bar{x}=50$
$\qquad n=100$
$\qquad 50 =\large\frac{\sum x_i}{100}$
$\qquad =50 \times 100$
$\qquad =5000$
Sum of all the item= 5000
Step 2:
Standard deviation $\sigma =\sqrt { \large\frac{ \sum x_i^2}{n} - \bigg( \large\frac{\sum x_i}{n}\bigg)^2}$
$\sigma^2= \large\frac{\sum x_i ^2}{n} - \bigg( \large\frac{\sum x_i^2}{n}\bigg)^2$
Given $\sigma= 4, n=100,mean= 50$
$4^2=\large\frac{\sum x_i^2}{100}$$-50^2$
$16+2500=\large\frac{\sum x_i^2}{100}$
$=> 2516 \times 100 = \sum x_i^2$
$\sum x_i^2 =251600$
Hence B is the correct answer.
answered Jul 4, 2014 by meena.p
 

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