Question

For a given data set \({X$_1$, X$_2$, ..., X$_n$}\) where n = 100 
$\frac{1}{2000} \sum_{i=1}^{n} \sum_{j=1}^{n} (x_i - x_j)^2 = 99$ 
Let us denote $\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$ 
The value of $\frac{1}{99} \sum_{i=1}^{n} (x_i - \bar{x})^2$ is __________.
 

Hint:

The identity $\sum_{i,j} (x_i - x_j)^2 = 2n \sum_{i} (x_i - \bar{x})^2$ is a very useful formula in statistics, connecting the sum of all pairwise squared distances to the variance of the data. Memorizing it can provide a quick solution to problems of this type.

Answer 1

Correct Answer: 10

Step 1: Understanding the Question:
We are given an equation involving a double summation of squared differences between all pairs of data points. We need to use this information to find the value of an expression that is proportional to the sample variance.
Step 2: Key Formula or Approach:
The key is to simplify the double summation term. There is a standard identity that relates the sum of squared pairwise differences to the sum of squared differences from the mean:
\[ \sum_{i=1}^{n} \sum_{j=1}^{n} (x_i - x_j)^2 = 2n \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
Let's quickly derive this identity:
$\sum_{i,j} (x_i - x_j)^2 = \sum_{i,j} (x_i^2 - 2x_i x_j + x_j^2)$
$= \sum_{i} \sum_{j} x_i^2 - 2 \sum_{i} \sum_{j} x_i x_j + \sum_{i} \sum_{j} x_j^2$
$= \sum_{i} n x_i^2 - 2 (\sum_{i} x_i)(\sum_{j} x_j) + \sum_{j} n x_j^2$
$= n \sum x_i^2 - 2 (n\bar{x})(n\bar{x}) + n \sum x_j^2$
$= 2n \sum x_i^2 - 2n^2 \bar{x}^2 = 2n(\sum x_i^2 - n\bar{x}^2)$
Also, $\sum (x_i - \bar{x})^2 = \sum (x_i^2 - 2x_i\bar{x} + \bar{x}^2) = \sum x_i^2 - 2n\bar{x}^2 + n\bar{x}^2 = \sum x_i^2 - n\bar{x}^2$.
Thus, the identity is proven.
Step 3: Detailed Explanation:
Now, we substitute this identity into the given equation. Given: \[ \frac{1}{2000} \sum_{i=1}^{n} \sum_{j=1}^{n} (x_i - x_j)^2 = 99 \] Substitute the identity: \[ \frac{1}{2000} \left( 2n \sum_{i=1}^{n} (x_i - \bar{x})^2 \right) = 99 \] We are given n = 100. \[ \frac{2 \times 100}{2000} \sum_{i=1}^{100} (x_i - \bar{x})^2 = 99 \] \[ \frac{200}{2000} \sum_{i=1}^{100} (x_i - \bar{x})^2 = 99 \] \[ \frac{1}{10} \sum_{i=1}^{100} (x_i - \bar{x})^2 = 99 \] Multiply both sides by 10 to isolate the sum: \[ \sum_{i=1}^{100} (x_i - \bar{x})^2 = 99 \times 10 = 990 \] Now, we can find the value of the expression asked for in the question: \[ \frac{1}{99} \sum_{i=1}^{100} (x_i - \bar{x})^2 \] Substitute the value of the sum we just found: \[ \frac{1}{99} \times 990 = 10 \] Step 4: Final Answer:
The value of the expression is 10.