Let $$x_{1},x_{2},...x_{10}$$ be ten observations such that $$\sum_{i=1}^{10}(x_{i}-2)=30,\sum_{i=1}^{10}(x_{i}-\beta)^{2}=98,\beta > 2$$, and their variance is $$\frac{4}{5}$$. If $$\mu$$ and $$\sigma^{2}$$ are respectively the mean and the variance of $$2(x_{1}-1)+4\beta, 2(x_{2}-1)+4\beta,....,2(x_{10}-1)+4\beta$$, then $$\frac{\beta \mu}{\sigma^{2}}$$ is equal to :

Given ten observations $$x_1, x_2, \ldots, x_{10}$$ with:

$$\sum_{i=1}^{10} (x_i - 2) = 30, \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 98, \quad \beta > 2,$$

and variance $$\frac{4}{5}$$. We need to find $$\frac{\beta \mu}{\sigma^2}$$, where $$\mu$$ and $$\sigma^2$$ are the mean and variance of the new data set $$y_i = 2(x_i - 1) + 4\beta$$.

First, recall that variance is given by $$\frac{1}{n} \sum (x_i - \bar{x})^2$$. With $$n = 10$$ and variance $$\frac{4}{5}$$:

$$\frac{1}{10} \sum_{i=1}^{10} (x_i - \bar{x})^2 = \frac{4}{5}$$

Multiplying both sides by 10:

$$\sum_{i=1}^{10} (x_i - \bar{x})^2 = 8 \quad \text{(Equation 1)}$$

From $$\sum_{i=1}^{10} (x_i - 2) = 30$$:

$$\sum_{i=1}^{10} (x_i - 2) = \sum x_i - 20 = 30$$

So:

$$\sum x_i = 50 \quad \text{(Equation 2)}$$

The mean $$\bar{x}$$ is:

$$\bar{x} = \frac{50}{10} = 5 \quad \text{(Equation 3)}$$

Using Equation 1 and $$\bar{x} = 5$$:

$$\sum_{i=1}^{10} (x_i - 5)^2 = \sum (x_i^2 - 10x_i + 25) = \sum x_i^2 - 10 \sum x_i + 250 = 8$$

Substituting $$\sum x_i = 50$$:

$$\sum x_i^2 - 10 \times 50 + 250 = 8 \implies \sum x_i^2 - 500 + 250 = 8 \implies \sum x_i^2 = 258 \quad \text{(Equation 4)}$$

Given $$\sum_{i=1}^{10} (x_i - \beta)^2 = 98$$:

$$\sum (x_i^2 - 2\beta x_i + \beta^2) = \sum x_i^2 - 2\beta \sum x_i + 10\beta^2 = 98$$

Substituting Equations 2 and 4:

$$258 - 2\beta \times 50 + 10\beta^2 = 98 \implies 258 - 100\beta + 10\beta^2 = 98$$

Rearranging:

$$10\beta^2 - 100\beta + 160 = 0 \implies \beta^2 - 10\beta + 16 = 0$$

Solving the quadratic equation:

Discriminant $$D = (-10)^2 - 4 \times 1 \times 16 = 100 - 64 = 36$$

$$\beta = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2}$$

So $$\beta = 8$$ or $$\beta = 2$$. Given $$\beta > 2$$, we have $$\beta = 8$$.

Now, define the new data set:

$$y_i = 2(x_i - 1) + 4\beta = 2x_i - 2 + 4\beta$$

Substituting $$\beta = 8$$:

$$y_i = 2x_i - 2 + 4 \times 8 = 2x_i + 30$$

So $$y_i = 2x_i + 30$$.

For a linear transformation $$y_i = a x_i + b$$, the new mean $$\mu = a \bar{x} + b$$ and new variance $$\sigma^2 = a^2 \times (\text{original variance})$$.

Here, $$a = 2$$, $$b = 30$$, $$\bar{x} = 5$$, original variance $$= \frac{4}{5}$$.

Thus:

$$\mu = 2 \times 5 + 30 = 10 + 30 = 40$$

$$\sigma^2 = (2)^2 \times \frac{4}{5} = 4 \times \frac{4}{5} = \frac{16}{5}$$

Now compute:

$$\frac{\beta \mu}{\sigma^2} = \frac{8 \times 40}{\frac{16}{5}} = \frac{320}{\frac{16}{5}} = 320 \times \frac{5}{16} = \frac{1600}{16} = 100$$

Therefore, $$\frac{\beta \mu}{\sigma^2} = 100$$.