Question

The mean and variance of seven observations are $8$ and $16$, respectively. If $5$ of the observations are $2,4, 10, 12, 14$, then the product of the remaining two observations is :

• A. 40
• B. 49
• C. 48
• D. 45

Answer 1

Answer: (C)

Let $7$ observations be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$
$\bar{x} = 8 \Rightarrow \sum^7_{i=1} x_i = 56 $ .......(1)
Also $\sigma^2 = 16$
$\Rightarrow 16 = \frac{1}{7} \left(\sum^{7}_{i=1} x_{i}^{2}\right) - \left(\bar{x}\right)^{2} $
$ \Rightarrow 16 = \frac{1}{7} \left(\sum^{7}_{i=1} x_{i}^{2}\right) - 64 $
$ \Rightarrow \left(\sum^{7}_{i=1} x^{2}_{i}\right) = 560 $
$ x_{1} = 2 , x_{2} = 4 , x_{3} = 10, x_{4} = 12 , x_{5} =14$
$ \Rightarrow x_{6} + x_{7} = 14 $ (from (1))
& $ x_{6}^{2} + x_{7}^{2} = 100$ (from (2))
$ \therefore x_{6}^{2} + x_{7}^{2} = \left(x_{6} +x_{7}\right)^{2} - 2x_{6}.x_{7} \Rightarrow x_{6}.x_{7} = 48 $