The mean and variance of 10 observations are 9 and 34.2 , respectively.

Question

The mean and variance of 10 observations are 9 and 34.2 , respectively. If 8 of these observations are $2,3,5,10,11,13,15,21$, then the mean deviation about the median of all the 10 observations is

Accepted Answer

Let the missing observations be $x$ and $y$.

$(\vec{x})=9:\frac{2+3+5+10+11+13+15+21+x+y}{10}=9$

$80+x+y=90\Rightarrow x+y=10$

$({\sigma }^{2})=34.2:{\sigma }^{2}=\frac{\sum {x}_{2}^{i}}{n}-(\vec{x}{)}^{2}$

$\frac{{2}^{2}+{3}^{2}+{5}^{2}+{10}^{2}+{11}^{2}+{13}^{2}+{15}^{2}+{21}^{2}+{x}^{2}+{y}^{2}}{10}-{9}^{2}=34.2$

$\frac{4+9+25+100+121+169+225+441+{x}^{2}+{y}^{2}}{10}-81=34.2$

$\frac{1094+{x}^{2}+{y}^{2}}{10}=115.2$

$1094+{x}^{2}+{y}^{2}=1152\Rightarrow {x}^{2}+{y}^{2}=58$

Solving $x+y=10$ and ${x}^{2}+{y}^{2}=58$

To solve the system $x+y=10$ and ${x}^{2}+{y}^{2}=58$ by finding $x-y$ use the formula $x-y=\sqrt{2({x}^{2}+{y}^{2})-(x+y{)}^{2}}$ :

$x-y=\sqrt{2({x}^{2}+{y}^{2})-(x+y{)}^{2}}$

$x-y=\sqrt{2(58)-(10{)}^{2}}=\sqrt{116-100}=\sqrt{16}$

$x-y=4$

Now add these equations $x+y=10\&x-y=4$

$2x=14\implies x=7$

Put $x=7$ in $x+y=10$

$7+y=10\implies y=3$

missing numbers are 3 and 7 .

Find the Median

The 10 observations in ascending order are: $2,3,3,5,7,10,11,13,15,21$.

Since $n=10$, the median is the average of the 5th and 6th terms :

Median $=\frac{7+10}{2}=$ 8.5

Calculate Mean Deviation about Median

$\begin{gathered} ()=\frac{\sum |{x}_{i}-|}{n} \ =\frac{|2-8.5|+|3-8.5|+|3-8.5|+|5-8.5|+|7-8.5|+|10-8.5|+|11-8.5|+|13-8.5|+|15-8.5|+|21-8.5|}{10} \ =\frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10}=\frac{50}{10}=5 \end{gathered}$

Solutions · 1