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3. The mean and standard deviation of six observations are $\small 8$ and $\small 4$ , respectively. If each observation is multiplied by $\small 3$ , find the new mean and new standard deviation of the resulting observations.

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Given,

Mean = 8 and Standard deviation = 4

Let the observations be $x_1, x_2, x_3, x_4, x_5\ and\ x_6$

Mean, $\overline x = \frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} = 8$

Now, Let $y_i$ be the the resulting observations if each observation is multiplied by 3:

$\\ \overline y_i = 3\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{3}$

New mean, $\overline y = \frac{y_1+ y_2+ y_3+ y_4+ y_5+ y_6}{6}$

$= 3\left [\frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} \right] =3\times 8$

= 24

We know that,

Standard Deviation = $\sigma = \sqrt{Variance}$

$\dpi{100} =\sqrt{ \frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2}$

$\dpi{100}\\ \implies 4^2=\frac{1}{6}\sum_{i=1}^6(x_i - \overline x)^2 \\ \implies \sum_{i=1}^6(x_i - \overline x)^2 = 6\times16 = 96$ -(i)

Now, Substituting the values of $x_i\ and\ \overline x$ in (i):

$\dpi{100} \\ \implies \sum_{i=1}^6(\frac{y_i}{3} - \frac{\overline y}{3})^2 = 96 \\ \implies \sum_{i=1}^6(y_i - \overline y)^2 = 96\times9 =864$

Hence, the variance of the new observations = $\frac{1}{6}\times864 = 144$

Therefore, Standard Deviation = $\sigma = \sqrt{Variance}$ = $\sqrt{144}$ = 12

Posted by

HARSH KANKARIA

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