1. Find the mean of the original observations

Given $\sum_{i=1}^{10}(x_i-2)=30$ So, $\sum_{i=1}^{10}x_i - 20 = 30$ $\sum_{i=1}^{10}x_i = 50$ Hence the mean is $\bar x = \mu_x = \frac{50}{10}=5$

2. Use the given variance of the original observations

The variance of $x_1,x_2,\dots,x_{10}$ is $\frac{1}{10}\sum_{i=1}^{10}(x_i-\mu_x)^2 = \frac{4}{5}$ Therefore, $\sum_{i=1}^{10}(x_i-5)^2 = 10\cdot \frac45 = 8$

3. Use the identity for sum of squares about another point

We are given $\sum_{i=1}^{10}(x_i-\beta)^2 = 98$ Using $\sum_{i=1}^{n}(x_i-a)^2 = \sum_{i=1}^{n}(x_i-\bar x)^2 + n(\bar x-a)^2$ we get $98 = 8 + 10(5-\beta)^2$ So, $10(5-\beta)^2 = 90$ $(5-\beta)^2 = 9$ Thus, $\beta = 2 \text{ or } 8$ But given $\beta>2$, hence $\beta=8$

4. Form the new observations and find their mean

The new observations are $y_i = 2(x_i-1)+4\beta = 2x_i-2+4\beta$ Since $\beta=8$, $y_i = 2x_i-2+32 = 2x_i+30$ For a linear transformation $y=ax+b$, the mean becomes $\mu = a\mu_x+b$ Hence, $\mu = 2\cdot 5 + 30 = 40$

5. Find the variance of the new observations

For $y=ax+b$, variance becomes $\sigma^2 = a^2\cdot \text{Var}(x)$ So, $\sigma^2 = 2^2\cdot \frac45 = 4\cdot \frac45 = \frac{16}{5}$

6. Compute the required value

We need $\frac{\beta\mu}{\sigma^2}$ Substitute $\beta=8$, $\mu=40$, $\sigma^2=\frac{16}{5}$: $\frac{\beta\mu}{\sigma^2} = \frac{8\cdot 40}{16/5} = 320\cdot \frac{5}{16} = 20\cdot 5 = 100$

So the correct option is $\boxed{100}$ which is Option A.