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Home  >>  CBSE XI  >>  Math  >>  Statistics
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The mean deviation of the data $3,10,10,4,7,10,5$ from mean is

$\begin{array}{1 1}(A)\; 2\\(B)\;2.57\\(C)\;3\\(D)\;3.75\end{array} $

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1 Answer

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Toolbox:
  • The formulae to solve the problem are Mean $=\large\frac{\sum x_i}{n} $
  • Mean deviation about mean = $ \large\frac{\sum|x_i -\bar{x}|}{n}$
Mean of the given series $3,10,10,4,7,10,5$
$\bar{x}=\large\frac{\text{ sum of the terms}}{\text{Number of terms}}$
$\qquad= \large\frac{\sum x_i}{n}$
$\qquad= \large\frac{3+10+10+4+7+10+5}{7}$
$\qquad= \large\frac{49}{7}$$=7$
$x_i =3; \qquad |x_i -\bar{x} |= |3-7 |=4$
$x_i =10; \qquad |x_i -\bar{x} |= |10-7 |=3$
$x_i =10; \qquad |x_i -\bar{x} |= |10-7 |=3$
$x_i =4; \qquad |x_i -\bar{x} |= |4-7 |=3$
$x_i =7; \qquad |x_i -\bar{x} |= |7-7 |=0$
$x_i =10; \qquad |x_i -\bar{x} |= |10-7 |=4$
$x_i =5; \qquad |x_i -\bar{x} |= |5-7 |=2$
Total $= |x_i -\bar {x} |=8$
Mean deviation about mean = $ \large\frac{\sum|x_i -\bar{x}|}{n}$
$\qquad= \large\frac{18}{7}$$=2.57$
Hence B is the correct answer.
answered Jul 8, 2014 by meena.p
 

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