Statistics

Given $12+f_1+f_2=19$

Median = 6

$\Rightarrow \quad f_1+f_2=7$

Median $=\left(\frac{N+1}{2}\right)^{\text {th }}$ term $=\left(\frac{19+1}{2}\right)^{\text {th }}$ term $=10^{\text {th }}$ term

$ \begin{aligned} & \Rightarrow \quad 6+f_1=10 \Rightarrow f_1=4 \text { and } f_2=3 \\ & \therefore \quad 7 f_1+9 f_2=7 \times 4+9 \times 3 \\ & =55 \Rightarrow P \rightarrow 5 \end{aligned} $

Mean $(\bar{X})=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{133}{19}=7$

$ \begin{aligned} & \text { Mean deviation about mean }(\alpha)=\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i} \\ & =\frac{5(3)+4(2)+1(1)+3(1)+2(2)+3(4)+1(5)}{19}=\frac{48}{19} \\ & \Rightarrow 19 \alpha=48 \quad Q \rightarrow 3 \end{aligned} $

Mean deviation about median $(\beta)$

$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-M\right|}{\Sigma f_i} \\ & =\frac{5(2)+4(1)+1(0)+3(2)+2(3)+3(5)+1(6)}{19} \\ & =\frac{47}{19} \\ & 19 \beta=47 \quad R \rightarrow 2 \end{aligned} $

$\begin{aligned} & \text { Variance }\left(\sigma^2\right)=\frac{\Sigma f_i f_i^2}{\Sigma f_i}-(\bar{X})^2 \\ & =\frac{5 \times 16+4 \times 25+1 \times 36+3 \times 64+2 \times 81+3 \times 121+1 \times 144}{19}-(7)^2 \\ & =\frac{1077}{19}-49=\frac{146}{19} \\ & \Rightarrow 19 \sigma^2=146 \\ & S \rightarrow 1 \\ & P \rightarrow 5, Q \rightarrow 3, R \rightarrow 2, S \rightarrow 1\end{aligned}$

$ \begin{array}{|c|c|c|c|c|} \hline x_1 & f_i & f_i x_i & f_i\left|x_i-\bar{x}\right| & f_i\left|x_i-N\right| \\ \hline 3 & 5 & 15 & 15 & 10 \\ \hline 4 & 4 & 16 & 8 & 4 \\ \hline 5 & 4 & 20 & 4 & 0 \\ \hline 8 & 2 & 16 & 4 & 6 \\ \hline 10 & 2 & 20 & 8 & 10 \\ \hline 11 & 3 & 33 & 15 & 18 \\ \hline & \Sigma f_i=20 & \Sigma f_i x_i=120 & \text { sum }=54 & \text { sum }=48 \\ \hline \end{array} $

(P) $\quad$ Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{120}{20}=6$

(Q)

$ \begin{aligned} \text { Median } & =\frac{\left(10^{\text {th }}+11^{\text {th }}\right) \text { observation }}{2} \\\\ & =\frac{5+5}{2}=5 \end{aligned} $

(both observation are same)

(R) Mean deviation

$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{54}{20} \\\\ & =2.7 \end{aligned} $

(S) Mean deviation about median

$ \begin{aligned} & =\frac{\Sigma f_i\left|x_i-\mathrm{M}\right|}{\Sigma f_i}=\frac{48}{20} \\\\ & =2.4 \end{aligned} $