Question

下列 $4040$ 个数的中位数是多少？

$1, 2, 3, \ldots, 2020$ ， $1^2, 2^2, 3^2, \ldots, 2020^2$ 。

What is the median of the following list of $4040$ numbers?

$1, 2, 3, \ldots, 2020,$ $1^2, 2^2, 3^2, \ldots, 2020^2$

$1974.5$

$1975.5$

$1976.5$

$1977.5$

$1978.5$

中位数是第 $2020$ 小和第 $2021$ 小的值的平均。

不超过 $2020$ 的完全平方数为 $1^2, \ldots, 44^2$ （因为 $44^2 = 1936$ 且 $45^2 = 2025$ ），所以共有 $44$ 个。

在这个列表中，数值 $\le 1976$ 的项包括 $1976$ 个整数 $1, \ldots, 1976$ 以及这 $44$ 个平方数，总数为 $1976 + 44 = 2020$ 。

因此第 $2020$ 个值是 $1976$ ，第 $2021$ 个值是 $1977$ ，中位数为 $\dfrac{1976 + 1977}{2} = 1976.5$ 。

因此，正确答案是 C。

The median is the average of the $2020$ th and $2021$ st smallest values.

The perfect squares that are at most $2020$ are $1^2, \ldots, 44^2$ (since $44^2 = 1936$ and $45^2 = 2025$ ), so there are $44$ of them.

Among the list, the numbers $\le 1976$ are the $1976$ integers $1, \ldots, 1976$ together with those $44$ squares, totaling $1976 + 44 = 2020.$

Thus the $2020$ th value is $1976$ and the $2021$ st value is $1977,$ making the median $\dfrac{1976 + 1977}{2} = 1976.5.$

Thus, C is the correct answer.