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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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