# The variance of 20 observations is 5. If each observation is multiplied by 2.

Question:

The variance of 20 observations is 5. If each observation is multiplied by 2. Find the variance of the resulting observations

Solution:

Let the observations are $\mathrm{X}_{1}, \mathrm{X}_{2}, \mathrm{X}_{3}, \mathrm{X}_{4}, \ldots, \mathrm{X}_{20}$

and Let mean $=\overline{\mathrm{x}}$

Given: Variance = 5 and n = 20

We know that,

Variance, $\sigma^{2}=\frac{1}{\mathrm{n}} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$

Putting the given values, we get

$5=\frac{1}{20} \sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$

$\Rightarrow 5 \times 20=\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$

$\Rightarrow 100=\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$

or $\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=100 \ldots$ (i)

It is given that each observation is multiplied by 2, we get new observations

Let the new observation be $y_{1}, y_{2}, y_{3}, \ldots, y_{20}$

where $y_{i}=2\left(x_{i}\right) \ldots$ (ii)

or $\mathrm{x}_{\mathrm{i}}=\frac{1}{2} \mathrm{y}_{\mathrm{i}}$ …(iii)

Now, we find the variance of new observations

i. e. New Variance $=\frac{1}{n} \sum\left(y_{i}-\bar{y}\right)^{2}$

Now, we calculate the value of $\bar{y}$

We know that,

Mean $=\frac{\text { Sum of observations }}{\text { Total number of observations }}$

$\Rightarrow \overline{\mathrm{y}}=\frac{\sum \mathrm{y}_{\mathrm{i}}}{\mathrm{n}}$

$\Rightarrow \overline{\mathrm{y}}=\frac{\sum\left(2 \mathrm{x}_{\mathrm{i}}\right)}{20}$ [from eq. (ii)]

$\Rightarrow \overline{\mathrm{y}}=2\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{20}\right)$

$\Rightarrow \overline{\mathrm{y}}=2 \overline{\mathrm{x}}$

$\Rightarrow \overline{\mathrm{X}}=\frac{1}{2} \overline{\mathrm{V}}$ …(iv)

Putting the value of eq. (iii) and (iv) in eq. (i), we get

$\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=100$

$\sum\left(\frac{1}{2} \mathrm{y}_{\mathrm{i}}-\frac{1}{2} \overline{\mathrm{y}}\right)^{2}=100$

$\Rightarrow \sum\left(\frac{1}{2}\right)^{2}\left(\mathrm{y}_{\mathrm{i}}-\overline{\mathrm{y}}\right)^{2}=100$

$\Rightarrow\left(\frac{1}{2}\right)^{2} \sum\left(\mathrm{y}_{\mathrm{i}}-\overline{\mathrm{y}}\right)^{2}=100$

$\Rightarrow \sum\left(\mathrm{y}_{\mathrm{i}}-\overline{\mathrm{y}}\right)^{2}=100 \times 4$

$\Rightarrow \sum\left(\mathrm{y}_{\mathrm{i}}-\overline{\mathrm{y}}\right)^{2}=400$

So,

New Variance $=\frac{1}{n} \sum\left(y_{i}-\bar{y}\right)^{2}$

$=\frac{1}{20} \times 400$

$=20$