For a given data set $\{X_1,X_2,\dots ,X_n\}$ where $n=100$

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Let us denote\overline{\mathrm{ \; x}}=\frac{1}{n}\sum _{i=1}^n{x}_i

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The value of\frac{\mathrm{ 1}}{\mathrm{ \; 99}}\sum _{i=1}^n(x_i-\overline{x}{)}^2  is ______.

The correct answer is 10.

Step-by-Step Derivation and Explanation:

We are given a data set $\{x_1,x_2,\dots ,x_n\}$ with $n=100$ observations.

The mean of the data set is defined as:

$$\overline{x}=\frac{1}{n}\sum _{i=1}^n{x}_i$$

First, let us find a relationship between the double summation term $\sum _{i=1}^n\sum _{j=1}^n{(x_i-x_j)}^2$ and the sum of squared deviations from the mean $\sum _{i=1}^n{(x_i-\overline{x})}^2$.

We can rewrite the term $(x_i-x_j)$ by adding and subtracting the mean $\overline{x}$:

$$x_i-x_j=(x_i-\overline{x})-(x_j-\overline{x})$$

Squaring both sides gives:

$${(x_i-x_j)}^2={(x_i-\overline{x})}^2+{(x_j-\overline{x})}^2-2(x_i-\overline{x})(x_j-\overline{x})$$

Now, we take the double summation over $i=1$ to $n$ and $j=1$ to $n$:

$$\sum _{i=1}^n\sum _{j=1}^n{(x_i-x_j)}^2=\sum _{i=1}^n\sum _{j=1}^n{(x_i-\overline{x})}^2+\sum _{i=1}^n\sum _{j=1}^n{(x_j-\overline{x})}^2-2\sum _{i=1}^n\underset{j=1}{\overset{n}{���}}(x_i-\overline{x})(x_j-\overline{x})$$

Let's simplify each of the three terms on the right-hand side:

1. First term:
Since ${(x_i-\overline{x})}^2$ does not depend on $j$, summing over $j$ gives $n$ times the expression:
$$\sum _{i=1}^n\sum _{j=1}^n{(x_i-\overline{x})}^2=n\sum _{i=1}^n{(x_i-\overline{x})}^2$$

2. Second term:
Similarly, since ${(x_j-\overline{x})}^2$ does not depend on $i$, summing over $i$ gives $n$ times the expression:
$$\sum _{i=1}^n\sum _{j=1}^n{(x_j-\overline{x})}^2=n\sum _{j=1}^n{(x_j-\overline{x})}^2$$

3. Third term:
We can factor the double summation as the product of two independent sums:
$$\sum _{i=1}^n\sum _{j=1}^n(x_i-\overline{x})(x_j-\overline{x})=\left(\sum _{i=1}^n(x_i-\overline{x})\right)\left(\sum _{j=1}^n(x_j-\overline{x})\right)$$

Because $\sum _{i=1}^n(x_i-\overline{x})=0$ (since the sum of deviations from the mean is always zero), this entire third term vanishes:
$$-2\left(0\right)\left(0\right)=0$$

Combining these three results, we get the classic algebraic identity:

$$\sum _{i=1}^n\sum _{j=1}^n{(x_i-x_j)}^2=2n\sum _{i=1}^n{(x_i-\overline{x})}^2$$

Given that $n=100$, we substitute $2n=2\left(100\right)=200$:

$$\sum _{i=1}^n\sum _{j=1}^n{(x_i-x_j)}^2=200\sum _{i=1}^n{(x_i-\overline{x})}^2$$

We are given the following equation in the question:

$$\frac{1}{2000}\sum _{i=1}^n\sum _{j=1}^n{(x_i-x_j)}^2=99$$

Substitute our identity into this given equation:

$$\frac{1}{2000}\left[200\sum _{i=1}^n{(x_i-\overline{x})}^2\right]=99$$

Simplify the fraction $\frac{200}{2000}=\frac{1}{10}$:

$$\frac{1}{10}\sum _{i=1}^n{(x_i-\overline{x})}^2=99$$

Multiplying both sides by 10, we find the sum of squared deviations:

$$\sum _{i=1}^n{(x_i-\overline{x})}^2=990$$

Finally, we need to calculate the value of:

$$\frac{1}{99}\sum _{i=1}^n{(x_i-\overline{x})}^2$$

Substitute $\sum _{i=1}^n{(x_i-\overline{x})}^2=990$ into this expression:

$$\frac{1}{99}\times 990=10$$

Thus, the value of the expression is indeed 10.