Statistics Part 2

Concept:Use mean and variance to find the two missing observations. Then arrange all data, find median, and compute mean deviation about median.Explanation:Let the missing observations be $x$ and $y$.Given mean $=9$:$$\frac{2+3+5+10+11+13+15+21+x+y}{10}=9$$$$80+x+y=90 \Rightarrow x+y=10$$Given variance $=34.2$:$$\frac{\sum x_i^2}{10}-9^2=34.2$$Squares of given: $4+9+25+100+121+169+225+441=1094$.So $$\frac{1094+x^2+y^2}{10}=115.2 \Rightarrow 1094+x^2+y^2=1152 \Rightarrow x^2+y^2=58$$Solve $x+y=10$ and $x^2+y^2=58$.Use $(x-y)^2=2(x^2+y^2)-(x+y)^2 = 2(58)-100=16 \Rightarrow x-y=4$.Adding: $2x=14 \Rightarrow x=7$, then $y=3$.All 10 observations in ascending order: $2,3,3,5,7,10,11,13,15,21$.For $n=10$, median is average of 5th and 6th terms: $(7+10)/2=8.5$.Mean deviation about median:$$\frac{\sum |x_i-8.5|}{10} = \frac{|2-8.5|+|3-8.5|+|3-8.5|+|5-8.5|+|7-8.5|+|10-8.5|+|11-8.5|+|13-8.5|+|15-8.5|+|21-8.5|}{10}$$$= \frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10} = \frac{50}{10} = 5$.Answer:Mean deviation about median $=5$. Hence option C is correct.