2019 AMC 10A Problem 12

P11-15~4 minPrint

Statistics & Data Bound the quantity above/below or estimate to pin it down Set up the equation/formula and compute; no special trick needed

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

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Key insight

The modes are 1 to 28, so d = 14.5. The 183rd value is 16, so M = 16. Missing large dates pull the mean just below 16.

Solution

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

Step 1: the modes and $d$ .

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 $1, 2, \ldots, 28$ .
$d$ 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 $M$ .

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

Step 3: the mean $\mu$ .

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 $16$ .
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 $\mu < 16$ .
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 $\boxed{\textbf{(E)}\ d < \mu < M}$ .

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Prealgebra, Chapter 13: Data and Statistics