AMC10 2019 A

AMC10 2019 A · Q12

AMC10 2019 A · Q12. It mainly tests Averages (mean), Arithmetic misc.

Melanie computes the mean $\mu$, the median $M$, and modes of the 365 values that are the dates in the months of 2019. Thus her data consist of 12 1s, 12 2s, ..., 12 28s, 11 29s, 11 30s, and 7 31s. Let $d$ be the median of the modes. Which of the following statements is true?

Melanie 计算了 2019 年月份日期的 365 个值的均值 $\mu$、中位数 $M$ 和众数。因此她的数据包括 12 个 1、12 个 2、...、12 个 28、11 个 29、11 个 30 和 7 个 31。让 $d$ 为众数的中位数。以下哪个陈述是正确的？

(A) $\mu < d < M$ $\mu < d < M$

(B) $M < d < \mu$ $M < d < \mu$

(D) $d < M < \mu$ $d < M < \mu$

(E) $d < \mu < M$ $d < \mu < M$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Each of the values 1 through 28 is a mode, so $d=\frac{14+15}{2}=14.5$. There are $15\cdot12=180<\frac{365}{2}$ data values less than or equal to 15, and there are $16\cdot12=192>\frac{365}{2}$ values less than or equal to 16. Therefore more than half of the values are greater than or equal to 16 and more than half of the values are less than or equal to 16, so $M=16$. To see the relationship between $\mu$ and 16, note that if every month had 31 days, then there would be 12 of each value from 1 to 31, and the mean would be 16; because the actual data are missing some of the larger values, $\mu<16$. To see the relationship between $\mu$ and 14.5, note that if every month had 28 days, then there would be 12 of each value from 1 to 28, and the mean would be 14.5; because the actual data consist of all of these values together with some larger values, $\mu>14.5$. Therefore $d=14.5<\mu<16=M$.

答案（E）：从 1 到 28 的每个数都是众数，所以 $d=\frac{14+15}{2}=14.5$。小于或等于 15 的数据值有 $15\cdot12=180<\frac{365}{2}$ 个，而小于或等于 16 的数据值有 $16\cdot12=192>\frac{365}{2}$ 个。因此，超过一半的数值大于或等于 16，同时也有超过一半的数值小于或等于 16，所以 $M=16$。为了看出 $\mu$ 与 16 的关系，注意如果每个月都有 31 天，那么从 1 到 31 的每个数都会出现 12 次，均值将为 16；由于实际数据缺少一些较大的数值，所以 $\mu<16$。为了看出 $\mu$ 与 14.5 的关系，注意如果每个月都有 28 天，那么从 1 到 28 的每个数都会出现 12 次，均值将为 14.5；由于实际数据包含了所有这些数值并且还有一些更大的数值，所以 $\mu>14.5$。因此 $d=14.5<\mu<16=M$。

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