Question

5 students of a class have an average height $150\text{cm}$ and variance $18{\text{cm}}^2$. A new student, whose height is $156\text{cm}$, joined them. The variance $\left(\text{in}{\text{cm}}^2\right)$ of the height of these six students is:

• A. $20$
• B. $16$
• C. $18$
• D. $22$

Answer 1

The correct option is A $20$

Let $x_1,x_2,x_3,x_4,x_5$ be the height of the five students.
So $\sum _{i=1}^5{x}_i=5\times 150=750$
and $\frac{\sum _{i=1}^5{x}_i^2}{5}-{150}^2=18$
$\Rightarrow \sum _{i=1}^5{x}_i^2=5\times (18+{150}^2)$

Let $x_6$ be the height of the new student.
So $x_6=156\text{cm}$
Now
$\sum _{i=1}^6{x}_i=750+156=906$ and $\sum _{i=1}^6{x}_i^2=5\times (18+{150}^2)+{156}^2$

Then, new variance$=\frac{\sum _{i=1}^6{x}_i^2}{6}-{\left(\frac{\sum _{i=1}^6{x}_i}{6}\right)}^2$
$=\frac{5\times (18+{150}^2)+{156}^2}{6}-{\left(\frac{906}{6}\right)}^2$
$=\frac{5\times (18+{150}^2)+{156}^2}{6}-{\left(151\right)}^2$
$=20$

Alternate solution :
From given data ,
Let $\overline{x_1}=150,\overline{x_2}=156$
${\sigma }_1^2=18,{\sigma }_2^2=0\text{(considering only one observation}$
$n_1=5,n_2=1$

$\overline{x}=\frac{5\left(150\right)+156}{6}=151$
Now using combined variance formulae,
${\sigma }^2=\frac{1}{n_1+n_2}\left[n_1\right({\sigma }_1^2+d_1^2)+n_2({\sigma }_2^2+d_2^2\left)\right]$
where $d_1=\overline{x}-\overline{x_1}=151-150=1$
$d_2=\overline{x}-\overline{x_2}=151-156=-5$
${\sigma }^2=\frac{1}{6}\left[5\right(18+1)+1(0^2+25\left)\right]$
$\Rightarrow {\sigma }^2=\frac{120}{6}=20$