**Question:**

Consider the statistics of two sets of observations as follows :

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

**Solution:**

$\sigma^{2}=\frac{\mathrm{n}_{1} \sigma_{1}^{2}+\mathrm{n}_{2} \sigma_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}$

$\mathrm{n}_{1}=10, \mathrm{n}_{2}=\mathrm{n}, \sigma_{1}^{2}=2, \sigma_{2}^{2}=1$

$\bar{x}_{1}=2, \bar{x}_{2}=3, \sigma^{2}=\frac{17}{9}$

$\frac{17}{9}=\frac{10 \times 2+n}{n+10}+\frac{10 n}{(n+10)^{2}}(3-2)^{2}$

$\Rightarrow \quad \frac{17}{9}=\frac{(\mathrm{n}+20)(\mathrm{n}+10)+10 \mathrm{n}}{(\mathrm{n}+10)^{2}}$

$\Rightarrow \quad 17 n^{2}+1700+340 n=90 n+9\left(n^{2}+30 n+200\right)$

$\Rightarrow \quad 8 n^{2}-20 n-100=0$

$2 n^{2}-5 n-25=0$

Hence $n=5$

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.