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Home  >>  CBSE XI  >>  Math  >>  Statistics
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The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:Number of observations =$25$,mean =$18.2$ seconds, standard deviation $=3.25$ seconds. Further another set of $15$ observations $x_1,x_2,......x_5$ also in seconds is now available and we have $\sum\limits_{i=1} ^{15}x_i =279$ and $\sum \limits_{i=1} ^{15} x_i^2 =5524$Calculate the standard deviation based on all $40$ observations

$\begin{array}{1 1}(A)\;1.99\\(B)\;3.873\\(C)\;8\\(D)\;1.25\end{array} $

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1 Answer

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Toolbox:
  • Formula used to solve this problem are :
  • Combined SD $=\sqrt {\large\frac{n_1\sigma_1 ^2 + n_2 \sigma_2 ^2}{n_1+n_2} + \frac{n_1n_2 (\bar{x_1} - \bar {x_2})^2}{(n_1+n_2)}}$
Step 1:
First consider the second set of 15 observation.
Mean of 15 observation $= \large\frac{\sum \limits_{i=1} ^{15} x_i}{n}$
Given $\sum \limits_{i=1} ^{15} x_i =279$
$mean =\large\frac{279}{15}$
$\qquad= 18.6 \;seconds$
Step 2:
$SD= \sqrt { \large\frac{\sum x^2}{n}-\bigg( \large\frac{\sum x_i}{n}\bigg)^2}$
Given $\sum \limits_{-i1} ^{15} x_i^2=5524$
$\sum \limits_{i=1}^n x_i =279$
$\therefore SD= \sqrt { \large\frac{5524}{15} -\bigg( \large\frac{279}{15}\bigg)^2}$
$\qquad= \sqrt { \large\frac{5524 \times 15 -(279)^2}{15 \times 15}}$
$\qquad= \large\frac{1}{15}$$ \sqrt { 82860 -77841}$
$\qquad= \large\frac{1}{15} $$\sqrt {5019} =\large\frac{70.8449}{15}$
$\qquad= 4.722 second$
Step 3:
Combined SD $=\sqrt {\large\frac{n_1\sigma_1 ^2 + n_2 \sigma_2 ^2}{n_1+n_2} + \frac{n_1n_2 (\bar{x_1} - \bar {x_2})^2}{(n_1+n_2)}}$
$\qquad= \sqrt {\large\frac{25 \times 3.25 ^2 +15 \times 4.722^2}{25+15} +\frac{25 \times 15 (18.2 -18.6)^2}{(25+15)^2}}$
$\qquad= \sqrt {\large\frac{25 \times 10.5625 + 15 \times 22.297 }{40} +\frac{375 \times 0.16}{40^2}}$
$\qquad= \sqrt { \large\frac{264.06 +334.455}{40} +\frac{60}{40^2}}$
$\qquad= \sqrt { \large\frac{598.515 \times 40 +60}{40^2}}$
$\qquad= \large\frac{1}{40}$$ \sqrt{24000.6}$
$\qquad= \large\frac{1}{40} $$ \times 154.92$
$\qquad= 3.873\;seconds$
$\therefore $ standard deviation based on all 40 observations is $3.873\;seconds$
Hence B is the correct answer.
answered Jul 3, 2014 by meena.p
edited Jul 3, 2014 by meena.p
 

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