The variance of 15 observations is 6.

Let the observations are $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}, \ldots, \mathrm{x}_{15}$

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

Given: Variance $=6$ and $n=15$

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

$6=\frac{1}{15} \sum\left(x_{i}-\bar{x}\right)^{2}$

$\Rightarrow 6 \times 15=\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$

$\Rightarrow 90=\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}$

or $\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=90$ .........(i)

It is given that each observation is increased by 8, we get new observations

Let the new observation be $\mathrm{y}_{1}, \mathrm{y}_{2}, \mathrm{y}_{3}, \ldots, \mathrm{y}_{15}$

where $y_{i}=x_{i}+8 \ldots$ (ii) or $x_{i}=y_{i}-8 \ldots$ (iii)

Now, we find the variance of new observations

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

Now, we calculate the value of $\overline{\mathrm{Y}}$

We know that,

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

$\Rightarrow \overline{\mathrm{y}}=\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{y}_{\mathrm{i}}}{\mathrm{n}}$

$\Rightarrow \bar{y}=\frac{\sum_{1=1}^{15} x_{1}+8}{15}$ [from eq. (ii)]

$\Rightarrow \overline{\mathrm{y}}=\left(\frac{1}{15}\right)\left\{\sum_{\mathrm{i}=1}^{15}\left(\mathrm{x}_{\mathrm{i}}+8\right)\right\}$

$\Rightarrow \overline{\mathrm{y}}=\frac{1}{15}\left[\sum_{\mathrm{i}=1}^{15} \mathrm{x}_{\mathrm{i}}+8 \sum_{\mathrm{i}=1}^{15} 1\right]$

$\Rightarrow \bar{y}=\frac{1}{15} \sum_{i=1}^{15} \mathrm{x}_{\mathrm{i}}+8 \times \frac{15}{15}$

$\Rightarrow \overline{\mathrm{y}}=\overline{\mathrm{x}}+8$

$\Rightarrow \overline{\mathrm{x}}=\overline{\mathrm{y}}-8$  …(iv)

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

$\sum\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=90$

$\sum\left(\mathrm{y}_{\mathrm{i}}-8-(\overline{\mathrm{y}}-8)\right)^{2}=90$

$\Rightarrow \sum\left(y_{i}-8-\bar{y}+8\right)^{2}=90$

$\Rightarrow \sum\left(y_{i}-\bar{y}\right)^{2}=90$ so

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

$=\frac{1}{15} \times 90$

$=6$