Question

Consider the statistics of two sets of observations as follows:

	Size	Mean	Variance
Observation I	$10$	$2$	$2$
Observation II	$n$	$3$	$1$

If the variance of the combined set of these two observations is$\frac{17}{9}$, then the value of $n$ is equal to

Answer 1

Finding the value of n:

We have given

$\begin{array}{rcl}{n}_1& =& 10,\\ n_2& =& n,\\ {{\sigma }_1}^2& =& 2,\\ {{\sigma }_2}^2& =& 1,\\ {\overline{X}}_1& =& 2,\\ {\overline{X}}_2& =& 3,\\ {\sigma }^2& =& \frac{17}{9}\end{array}$

Putting all values in the formula.${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{1}}{{\mathbf{\sigma }}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{+}{\mathbf{n}}_{\mathbf{2}}{{\mathbf{\sigma }}_{\mathbf{2}}}^{\mathbf{2}}}{{\mathbf{n}}_{\mathbf{1}}\mathbf{+}{\mathbf{n}}_{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{2}}}{{\mathbf{(}{\mathbf{n}}_{\mathbf{1}}\mathbf{+}{\mathbf{n}}_{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}{\mathbf{(}\overline{{\mathbf{X}}_{\mathbf{1}}}\mathbf{-}{\overline{\mathbf{X}}}_{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}$

$$\begin{gathered} \Rightarrow \left(\frac{17}{9}\right)=\frac{10\left(2\right)+n{\left(1\right)}^2}{10+n}+\frac{10n}{{(10+n)}^2}{(2-3)}^2 \\ \Rightarrow \left(\frac{17}{9}\right)=\frac{(20+n)(10+n)+10n}{{(10+n)}^2} \\ \Rightarrow 17n^2+340n+1700=9[200+n^2+30n]+90n \\ \Rightarrow 8n^2-20n-100=0 \\ \Rightarrow 4\{2n^2-5n-25\}=0 \\ \Rightarrow (2n+5)(n-5)=0 \\ \Rightarrow n=\frac{-5}{2},5 \\ \Rightarrow n=5 \end{gathered}$$

Take the positive value only.

Hence, the required answer is $5$.