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)

Question

From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable $ X $ denote the number of defective items in the sample. If the variance of $ X $ is $ \sigma^2 $, then $ 96\sigma^2 $ is equal to _________.

Answer 1

Think of picking an index $i$ uniformly at random from $1$ to $n$ and independently picking another index $j$ the same way, then looking at $x_i$ and $x_j$ as two independent draws $X$ and $Y$ from the same empirical distribution (the list of data values, each with probability $1/n$). Averaging $(x_i - x_j)^2$ over all $n^2$ ordered pairs $(i, j)$ is exactly the expected value $E[(X-Y)^2]$ for these two independent copies.

For any two independent random variables $X$ and $Y$ with the same distribution and mean $\mu$, a standard probability identity gives:

$$E[(X - Y)^2] = E[X^2] - 2E[X]E[Y] + E[Y^2] = 2E[X^2] - 2\mu^2 = 2\,\text{Var}(X)$$

where $\text{Var}(X) = E[X^2] - \mu^2$ is the population variance of the data, namely $\text{Var}(X) = \frac{1}{n}\sum_i (x_i - \bar{x})^2$.

So we have the relation:

$$\frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}(x_i - x_j)^2 = 2\,\text{Var}(X) = \frac{2}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2$$

Now plug in the given information. We are told $\frac{1}{2000}\sum_i\sum_j (x_i-x_j)^2 = 99$, so $\sum_i\sum_j(x_i-x_j)^2 = 198000$. With $n = 100$, $n^2 = 10000$, so:

$$\frac{198000}{10000} = 19.8 = \frac{2}{100}\sum_{i=1}^{n}(x_i-\bar{x})^2$$

Solve for the sum of squared deviations:

$$\sum_{i=1}^{n}(x_i-\bar{x})^2 = 19.8 \times \frac{100}{2} = 19.8 \times 50 = 990$$

Finally, divide by 99 as the question asks:

$$\frac{1}{99}\sum_{i=1}^{n}(x_i-\bar{x})^2 = \frac{990}{99} = 10$$ $$\boxed{10}$$

Answer 2