Question

For a given data set \(\{x_1, x_2, \ldots, x_n\}\), where \(n = 100\), it is known that
\[ \frac{1}{2000}\sum_{i=1}^{n}\sum_{j=1}^{n}(x_i - x_j)^2 = 99 \]
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 _______. (Answer in integer)

Hint:

Use the identity \(\sum_i\sum_j (x_i-x_j)^2 = 2n\sum_i (x_i-\bar{x})^2\) to convert the given double sum into the sum of squared deviations from the mean.

Answer 1

Correct Answer: 10

Step 1: Recall the identity linking the pairwise sum to the variance-type sum.
There is a standard identity connecting the double sum of squared pairwise differences to the sum of squared deviations from the mean: \[ \sum_{i=1}^{n}\sum_{j=1}^{n}(x_i - x_j)^2 = 2n\sum_{i=1}^{n}(x_i - \bar{x})^2 \]
We can see why this holds by expanding the left side. Write \(S_2 = \sum x_i^2\) and \(S_1 = \sum x_i\). \[ \sum_i\sum_j (x_i - x_j)^2 = \sum_i\sum_j (x_i^2 - 2x_ix_j + x_j^2) = n S_2 - 2S_1^2 + n S_2 = 2nS_2 - 2S_1^2 \]
On the other hand, the sum of squared deviations from the mean expands as \(\sum_i (x_i - \bar{x})^2 = S_2 - \frac{S_1^2}{n}\), so \(2n\sum_i(x_i-\bar{x})^2 = 2nS_2 - 2S_1^2\), which matches the left side exactly. This confirms the identity.

Step 2: Find the value of the double sum from the given information.
We are given \[ \frac{1}{2000}\sum_{i}\sum_{j}(x_i - x_j)^2 = 99 \] Multiply both sides by 2000: \[ \sum_{i}\sum_{j}(x_i - x_j)^2 = 99 \times 2000 = 198000 \]

Step 3: Use the identity to find the sum of squared deviations from the mean.
From Step 1, with \(n = 100\): \[ 198000 = 2n\sum_{i}(x_i-\bar{x})^2 = 2(100)\sum_i(x_i-\bar{x})^2 = 200\sum_i(x_i-\bar{x})^2 \] Divide both sides by 200: \[ \sum_{i=1}^{n}(x_i - \bar{x})^2 = \frac{198000}{200} = 990 \]

Step 4: Compute the requested quantity. \[ \frac{1}{99}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \frac{990}{99} = 10 \]

Final Answer:
The value asked for is 10. \[ \boxed{10} \]