Question

The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.

• A. 4 and 8
• B. 8 and 6
• C. 2 and 6
• D. 2 and 8

Answer 1

The correct option is B 8 and 6

Let the missing two observations be ‘a’ and ‘b’
Arithmetic mean = 8

$\begin{array}{cc}{x}_i& x_i^2\\ 2& 4\\ 4& 16\\ 10& 100\\ 12& 144\\ 14& 196\\ a& a^2\\ b& b^2\\ \sum x_i=42+a+b& \sum x_i^2=460+a^2+b^2\end{array}$

Now, $\frac{42+a+b}{7}=8\Rightarrow 42+a+b=56\Rightarrow a+b=14\dots \left(1\right)$

Also, variance =
$16=\frac{\sum x_i^2}{n}-(\frac{x_i}{n}{)}^2=\frac{460+a^2+b^2}{7}-8^2=\frac{460+a^2+b^2}{7}-64$

$\Rightarrow \frac{460+a^2+b^2}{7}=80a^2+b^2=100(a+b{)}^2-2ab=100$

$2ab=196-100=96$
$ab=48$
$b=\frac{48}{a}$
Substituting $b=\frac{48}{a}$ in (1),
$a+\frac{48}{a}=14$
$a^2-14a+48=0$
$(a-6)(a-8)=0$
$a=6,8$
When $a=6,b=\frac{48}{6}=8$
When $a=8,b=\frac{48}{8}=6$
Hence, the two observations are 6 and 8 or 8 and 6.