Consider the statistics of two sets of observations as follows : $\begin{array}{llcc} & ext { Size

Question

Consider the statistics of two sets of observations as follows : $\begin{array}{llcc} & 	ext { Size } & 	ext { Mean } & 	ext { Variance } \ 	ext { Observation I } & 10 & 2 & 2 \ 	ext { Observation II } & n & 3 & 1\end{array}$ \If the variance of the combined set of these two observations is $\frac{17}{9}$, then the value of $n$ is equal to

Professor PC Thomas Classes · Answer

(5) : We have, $\bar{x}_{1}=2, n_{1}=10, \sigma_{1}^{2}=2$;\ $\bar{x}_{2}=3, n_{2}=n, \sigma_{2}^{2}=1$,\ $\bar{x}=\frac{10 	imes 2+n 	imes 3}{10+n}=\frac{20+3 n}{10+n}$\ So, $d_{1}^{2}=\left(\bar{x}-\bar{x}_{1}ight)^{2}=\frac{n^{2}}{(10+n)^{2}}$\ and $d_{2}^{2}=\left(\bar{x}-\bar{x}_{2}ight)^{2}=\left(\frac{-10}{10+n}ight)^{2}=\frac{100}{(10+n)^{2}}$\ Now, $\sigma^{2}=\frac{n_{1}\left(\sigma_{1}^{2}+d_{1}^{2}ight)+n_{2}\left(\sigma_{2}^{2}+d_{2}^{2}ight)}{n_{1}+n_{2}}$\ $\Rightarrow \frac{17}{9}=\frac{10\left(2+\frac{n^{2}}{(10+n)^{2}}ight)+n\left(1+\frac{100}{(10+n)^{2}}ight)}{10+n}$\ $\Rightarrow 17(10+n)=\left[20+n+\frac{10 n^{2}+100 n}{(10+n)^{2}}ight] 	imes 9$\ $\Rightarrow 17(10+n)^{2}=[(20+n)(10+n)+10 n] 	imes 9$\ $\Rightarrow 17 n^{2}+1700+340 n=9 n^{2}+360 n+1800$\ $\Rightarrow 8 n^{2}-20 n-100=0 \Rightarrow 2 n^{2}-5 n-25=0$\ $\Rightarrow 2 n^{2}-10 n+5 n-25=0$\ $\Rightarrow(n-5)(2 n+5)=0$\ $\Rightarrow n=5 \quad(\because n$ can't be negative)\