2019 AMC 10A Problem 12

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

Each statistic is computed separately from the frequency table. The median depends only on the count of entries, so a quick cumulative count locates it; the mean is best handled by comparing with a symmetric distribution (all values $12$ times, mean $16$) and noting which side the deviations fall on. Only the ordering of the three numbers matters, so an estimate of $\mu$ is enough once its bounds are clear.

• Taking the median of the data as the middle of 1 to 31 (16) without checking, or comparing the mean to 16 by feel instead of estimating it.
• Thinking the mode is a single value (there are $28$ of them), or taking $d$ as the mode of the whole data set rather than the median of the list of modes.