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If $\mathrm{\bar{x}}$ is the mean and mean deviation from mean is $\mathrm{\mathrm{MD}(\bar{x})}$ , then the number of observations lying between $\mathrm{\bar{x}-\mathrm{MD}(\bar{x})\: \: and \: \: \bar{x}+\mathrm{MD}(\bar{x})}$ for the data $\mathrm{34,66,30,38,44,50,40,60,42,51}$ , are

Option: 1
$6$

Option: 2
$5$

Option: 3
$4$

Option: 4
$3$

Answers (1)

best_answer

Given data can be arranged in ascending order as

$30,34,38,40,42,44,50,51,60,66$

$\mathrm{ \therefore \bar{x}=\frac{30+34+38+40+42+44+50+51+60+66}{10} }$

$\mathrm{ =\frac{455}{10}=45.5 }$

Now,

$\mathrm{ \left|x_i-\bar{x}\right|=15.5,11.5,7.5,5.5,3.5,1.5,4.5,5.5,14.5,20.5 }$

Mean deviation from the mean $\mathrm{ =\operatorname{MD}(\bar{x})=\frac{\mathcal{2}\left|\left(x_i-\bar{x}\right)\right|}{10} }$

$\mathrm{ =\frac{1}{10}(15.5+11.5+7.5+5.5+3.5+1.5+4.5 +5.5+14.5+20.5) }$

$\mathrm{ =\frac{1}{10}(90.0)=9 }$

Now, $\mathrm{ \bar{x}-\operatorname{MD}(\bar{x})=45.5-9=36.5 }$

and $\mathrm{ \bar{x}+\mathrm{MD}(\bar{x})=45.5+9=54.5 }$

Given observations which lie between $\mathrm{ 36.5 \: and \: 54.5 }$ , are $\mathrm{ 38,40,42,44,50,51, }$ which are six in number.

Hence option 1 is correct.

Posted by

Ritika Jonwal

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