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

##### 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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