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 observations.

Asked by Abhisek | 11 months ago |  117

1 Answer

Solution :-

Let the observations be x1, x2, x3, x4, x5, and x6.

It is given that mean is 8 and standard deviation is 4

\( Mean,\overline X=\dfrac{x_1+ x_2+x_3+ x_4+ x_5+x_6}{6}\)

If each observation is multiplied by 3 and the resulting observations are yi , then

y= 3xi

\( Mean,\overline y=\dfrac{y_1+ y_2+y_3+ y_4+ y_5+y_6}{6}\)

 \( =\dfrac{3(x_1+ x_2+x_3+ x_4+ x_5+x_6)}{6}\)

\( 3\times 8 =24\)

We know that,

Standard deviation,σ =

\( \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{6} (X_i - \overline X})^2\)

By squaring both sides,

\( (4)^2={\dfrac{1}{6} \displaystyle\sum_{i=1}^{6} (X_i-\overline {X})^2}\)
\( \displaystyle\sum_{i=1}^{6} (X_i-\overline {X})^2=96\)

From (1) and (2), it can be observed that

\( \overline{y}=3\overline{x}\)

Substituting the values of x1 and \( \overline{x}\) in (2), we obtain

\( \displaystyle\sum_{i=1}^{6}(\dfrac{1}{3}y_i-\dfrac{1}{a}\overline{y})^2 =96\)

\(\displaystyle\sum_{i=1}^{6}(y_i-\overline{y})^2=864\)

Therefore, variance of new observations =

\( \dfrac{1}{6}\times 864\) = 144

Hence, the standard deviation of new observations is

\( \sqrt{144}=12\)

Answered by Pragya Singh | 11 months ago

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