NIMCET Previous Year Questions (PYQs) – Statistics Measures Of Central Tendency – Page 1 of 3

Given: 5 observations, mean = 5 ⇒ total sum = $5\times 5=25$. Three values are 1, 2, 6. Let the other two be $a,b$.

From mean: $a+b=25-(1+2+6)=16$.

Variance (about mean): $12.4$ ⇒ $\sum (x_i-5{)}^2=5\times 12.4=62$.
Known part: $(1-5{)}^2+(2-5{)}^2+(6-5{)}^2=16+9+1=26$.
Hence $(a-5{)}^2+(b-5{)}^2=62-26=36$.

Let $u=a-5,v=b-5$. Then $u+v=(a+b)-10=6$ and $u^2+v^2=36$.
$(u+v{)}^2=u^2+v^2+2uv\Rightarrow 36=36+2uv\Rightarrow uv=0$.
So one of $u,v$ is 0 ⇒ one of $a,b$ is 5, the other is $16-5=11$.

Mean deviation about mean:
${\displaystyle \text{MD}=\frac{1}{5}(|1-5|+|2-5|+|6-5|+|5-5|+|11-5|)}$ $=\frac{1}{5}(4+3+1+0+6)=\frac{14}{5}=2.8.$

Answer: 2.8