1. Use the given mean and variance of 7 observations

Let the 7 observations be $x_1,x_2,\dots,x_7$.

Given:

• Mean $=8$
• Variance $=16$
• Number of observations $n=7$

So, $\sum x_i = 7\cdot 8 = 56$

Using the formula $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2$ we get $16 = \frac{\sum x_i^2}{7} - 8^2$ $16 = \frac{\sum x_i^2}{7} - 64$ $\frac{\sum x_i^2}{7} = 80$ $\sum x_i^2 = 560$

2. Remove the two known observations 6 and 8

For the remaining 5 observations, let their sum be $S$ and sum of squares be $Q$.

Then $S = 56 - (6+8) = 56-14 = 42$

and $Q = 560 - (6^2+8^2) = 560 - (36+64) = 560-100 = 460$

3. Find mean of the remaining 5 observations

$\bar{x} = \frac{42}{5}$

4. Find variance of the remaining 5 observations

Again use $\text{Variance} = \frac{Q}{5} - \left(\frac{S}{5}\right)^2$

So, $\text{Variance} = \frac{460}{5} - \left(\frac{42}{5}\right)^2$ $= 92 - \frac{1764}{25}$ $= \frac{2300-1764}{25}$ $= \frac{536}{25}$

5. Match with the options

$\frac{536}{25}$ corresponds to Option C.

Therefore, the variance of the remaining 5 observations is $\boxed{\frac{536}{25}}$