GATE 2026 DA – Question 62
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*)
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Correct answer: 10
Explanation
Expanding the double sum, $\sum_i\sum_j (x_i - x_j)^2 = 2n\sum_i (x_i - \bar{x})^2$, because the cross terms cancel when we use $\sum_i (x_i - \bar{x}) = 0$. So with $n = 100$, $\frac{1}{2000} \times 200\sum_i (x_i - \bar{x})^2 = \frac{1}{10}\sum_i(x_i - \bar{x})^2 = 99$, which gives $\sum_i (x_i - \bar{x})^2 = 990$. Then $\frac{990}{99} = 10$.