AMC 8 · 2011 · #4
Easy mode Grade 6Problem
Last summer Tyler went fishing 9 times. Here is how many fish he caught each time:
2,0,1,3,0,3,3,1,2.
Look at the mean, the median, and the mode of this list.
Which order is correct?
Pick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Tyler's nine fishing outings yielded the catches $2, 0, 1, 3, 0, 3, 3, 1, 2$. Compute the mean, median, and mode of this data set, then pick the inequality that correctly orders all three.
Givens: Data set of $9$ values: $\{2, 0, 1, 3, 0, 3, 3, 1, 2\}$; Three summary statistics to compare: mean, median, mode; Answer choices are five different orderings of those three statistics
Unknowns: Which of the five orderings (A)-(E) is the true inequality chain
Understand
Restated: Tyler's nine fishing outings yielded the catches $2, 0, 1, 3, 0, 3, 3, 1, 2$. Compute the mean, median, and mode of this data set, then pick the inequality that correctly orders all three.
Givens: Data set of $9$ values: $\{2, 0, 1, 3, 0, 3, 3, 1, 2\}$; Three summary statistics to compare: mean, median, mode; Answer choices are five different orderings of those three statistics
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Re-arrange / Reorder
The question asks one thing — an ordering — but answering it cleanly needs three separate calculations. Tool #7 (Identify Subproblems) says: handle mean, median, and mode one at a time, then compare the three numbers at the end. Tool #2 (Re-arrange) does the upfront work that makes the median and mode obvious: sorting the nine values from smallest to largest lets us read the middle value off the list and spot the most frequent value at a glance.
Execute — Answer: C
6.SP.B.4 Step 1 - Sort the data from smallest to largest.
- This is the Tool #2 (Re-arrange) move — it does not change the data, but it makes the median and mode easy to read.
💡 Putting the values in order is the Grade 6 data-display habit that exposes the shape of the data.
6.SP.B.5 Step 2 - Find the mean.
- Add all nine values, then divide by $9$.
- Since the list is sorted, pair the values mentally if you like ($0{+}0$, $1{+}1$, $2{+}2$, $3{+}3{+}3$) to keep the addition clean.
💡 Computing the mean as "sum divided by count" is the Grade 6 definition of average.
6.SP.B.5 Step 3 - Find the median.
- With $9$ sorted values, the median is the $5$th value (four below it, four above it).
💡 The middle position of an odd-length sorted list is the median by definition.
6.SP.B.5 Step 4 - Find the mode.
- Scan the sorted list and pick the value that appears most often.
- $0$ appears twice, $1$ appears twice, $2$ appears twice, $3$ appears three times.
💡 The mode is just the tallest bar in a frequency count.
5.NF.B.3 Step 5 - Compare the three numbers.
- $\tfrac{15}{9} = \tfrac{5}{3} \approx 1.67$, which is less than $2$, which is less than $3$.
💡 Reading $\tfrac{15}{9}$ as $15 \div 9 = 1.67$ shows it is below $2$, so the chain is settled — a Grade 5 "fraction as division" check.
6.SP.B.4 Sort the data from smallest to largest. This is the Tool #2 (Re-arrange) move — 6.SP.B.5 Find the mean. Add all nine values, then divide by $9$. Since the list is sorted 6.SP.B.5 Find the median. With $9$ sorted values, the median is the $5$th value (four bel 6.SP.B.5 Find the mode. Scan the sorted list and pick the value that appears most often. 5.NF.B.3 Compare the three numbers. $\tfrac{15}{9} = \tfrac{5}{3} \approx 1.67$, which is Review
Reasonableness: Quick sanity check: the data set has a few $3$s clustered at the top, which pulls the mode up to $3$, while the small values ($0$s and $1$s) pull the mean down. So we expect mode $>$ median $>$ mean, exactly the chain in (C). Also, the mean $\tfrac{15}{9} \approx 1.67$ sits between the smallest value $0$ and the largest value $3$, as any mean must.
Alternative: Tool #4 (Eliminate Possibilities): once you know the mode is $3$ (the highest value in the data set), the mode must be the largest of the three statistics, since no statistic can exceed the maximum data value. That rules out (A), (B), (D), and (E), which all put a different statistic on top. Only (C) keeps the mode at the top, so (C) must be correct.
CCSS standards used (min grade 6)
6.SP.B.4Display numerical data in plots on a number line, including dot plots, histograms, and box plots (Sorting the nine catches into increasing order so the middle value and the most frequent value can be read off directly.)6.SP.B.5Summarize numerical data sets in relation to their context, including measures of center (mean, median, mode) (Computing the mean ($\tfrac{15}{9}$), median ($2$), and mode ($3$) of the data set as the three measures of center to be compared.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reading $\tfrac{15}{9}$ as $15 \div 9 \approx 1.67$ so it can be compared with the integers $2$ and $3$.)
⭐ This AMC 8 problem only needs the Grade 6 ideas of mean, median, and mode — compute each one, then line them up.
⭐ This AMC 8 problem only needs the Grade 6 ideas of mean, median, and mode — compute each one, then line them up.
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