2019 AMC 10A Problem 12

Work out the three numbers one at a time, then put them in order.

Step 1: the modes and $d$.

1. A mode is a value that appears most often. The dates $1$ to $28$ each appear $12$ times (once per month). The dates $29$, $30$, $31$ appear fewer times. So the modes are $1, 2, \ldots, 28$.
2. $d$ is the median of these $28$ numbers: the average of the two middle ones, $14$ and $15$. So $d = 14.5$.

Step 2: the median $M$.

1. There are $365$ values. Lined up in order, the middle one is number $183$ (there are $182$ below it and $182$ above it).
2. The values $1$ through $15$ fill the first $15 \cdot 12 = 180$ places. The next $12$ places, numbers $181$ to $192$, are all $16$.
3. Place $183$ is among them, so $M = 16$.

Step 3: the mean $\mu$.

1. Imagine every date from $1$ to $31$ appeared $12$ times. The dates would be balanced around the middle date $16$, so the mean would be exactly $16$.
2. The real data is missing some values from that picture: one $29$, one $30$ and five $31$s. All of them are bigger than $16$. Removing big values pulls the mean down, so $\mu < 16$.
3. Is $\mu$ still bigger than $14.5$? Add everything up: $12 \cdot (1 + 2 + \cdots + 28) = 12 \cdot 406 = 4872$, then $11 \cdot 29 = 319$, $11 \cdot 30 = 330$ and $7 \cdot 31 = 217$. The total is $5738$, and $5738 \div 365$ is about $15.7$.

Step 4: order them.

$$d = 14.5 \;<\; \mu \approx 15.7 \;<\; M = 16.$$

The answer is $\boxed{\textbf{(E)}\ d < \mu < M}$.

Mean, median and mode each need a different kind of thinking. The median only needs counting to the middle place. The mean is easiest to judge by comparing with a perfectly balanced version of the data and asking which values are missing. Since the question only asks for the order, a rough value of the mean is enough.

An easier way to get the total in step 8. The balanced data (every date $12$ times) has $12 \cdot 31 = 372$ values with mean $16$, so its total is $16 \cdot 372 = 5952$. The real data is missing $29 + 30 + 5 \cdot 31 = 214$ from that, so its total is $5952 - 214 = 5738$. To compare with $14.5$, note that $14.5 \cdot 365 = 5292.5$, which is far below $5738$. So $\mu > 14.5$ without doing the division.

Modes. The values $1$ through $28$ each occur $12$ times, more than any other value, so they are all modes. Their median is the average of $14$ and $15$: $d = 14.5$.

Median. With $365$ values the median is the $183$rd in sorted order. The values $1$ through $15$ account for $15 \cdot 12 = 180$ entries, so entries $181$ through $192$ are all $16$. Thus $M = 16$.

Mean. If every date $1$ through $31$ occurred $12$ times, the mean would be exactly $16$. The actual data is missing one $29$, one $30$ and five $31$s from that ideal, and all the missing values exceed $16$, so $\mu < 16$. To see $\mu > 14.5$, compute: the total is $12(1 + \dots + 28) + 11 \cdot 29 + 11 \cdot 30 + 7 \cdot 31 = 4872 + 319 + 330 + 217 = 5738$, and $5738/365 \approx 15.7$.

So $d = 14.5 < \mu \approx 15.7 < M = 16$.

The answer is $\boxed{\textbf{(E)}\ d < \mu < M}$.