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Home  >>  CBSE XI  >>  Math  >>  Statistics
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The mean and standard deviation of six observations are $8$ and $4$ respectively. If each observation is multiplied by $3$, find the new mean and new standard deviation of the resulting observation.

$\begin{array}{1 1}(A)\;144,12\\(B)\;5,8\\(C)\;3,8\\(D)\;5,7\end{array} $

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1 Answer

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Given Mean $\bar {x} =8$
Variance $\sigma ^2= 4$
Number of observation =6
Step 1:
Let us assume the observations as $x_1,x_2,x_3,x_4 \;and\;x_6$
Mean $\bar{x} =\large\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}$
$8= \large\frac{x_1+x_2+x_3+x_4+x_5+x_6}{6}$
$48= x_1+x_2+x_3+x_4+x_5+x_6$
Step 2:
If each observation is multiplied by 3 the new mean will be
$3x_1+3x_2+3x_3+3x_4+3x_5+3x_6=48 \times 3$
$\sum \limits_{i=1}^{6} x_i =144$
$\therefore$ New Mean $\bar{x}=\large\frac{144}{6}$$=24$
Hence the new mean =24
Step 3:
Given standard deviation =4
Variance $=4^2=16$
$\therefore \large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i}{n}\bigg)^2$$=16$
$\therefore \large\frac{\sum x_i^2}{6} $$-8^2=16$
=> $\large\frac{\sum x_i^2}{6}$$=16+64$
=> $\sum x_i^2 =480$
Step 4:
New variance can be calculated as
New $ \sum x_i^2=(3x_1)^2+(3x_2)^2+(3x_3)^2+(3x_4)^2+3(x_5)^2+3(x_6)^2$
$\qquad= 9(x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2)$
$\qquad= 9 \times 480$
$\qquad= 4320$
New variance $=\large\frac{\sum x_1^2}{n} $$-(\bar{x})^2$
$\qquad= \large\frac{4320}{6} $$-(24)^2$
$\qquad= 720-576$
$\qquad =144$
New standard deviation $= \sqrt {144}$
$\qquad= 12$
Hence A is the correct answer.
answered Jul 2, 2014 by meena.p
 

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