AMC 8 · 2011 · #4

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

mean-median-mode-rangefraction-arithmetic identify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

median < mean < mode

(B)

mean < mode < median

(C)

mean < median < mode

(D)

median < mode < mean

(E)

mode < median < mean

AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Sort the nine catches from smallest to largest so the median and mode read straight off the list.

{2, 0, 1, 3, 0, 3, 3, 1, 2} ⟶ {0, 0, 1, 1, 2, 2, 3, 3, 3}

2STEP 2

Add the nine values to get 15 and divide by 9, so the mean is $\frac{5}{3}$ ≈ 1.67.

mean =

\frac{(0+0+1+1+2+2+3+3+3)}{9}

\frac{15}{9}

\frac{5}{3}

≈ 1.67

3STEP 3

With nine sorted values the median is the middle, 5th value: median = 2.

Sorted: 0, 0, 1, 1, 2, 2, 3, 3, 3 → median = 2

4STEP 4

Count how often each value appears; 3 shows up three times, more than any other, so mode = 3.

Counts: 0{:} 2, 1{:} 2, 2{:} 2, 3{:} 3 → mode = 3

5STEP 5

Line the three up: $\frac{5}{3}$ ≈ 1.67, then 2, then 3, which matches choice (C).

\frac{15}{9}

＜ 2 ＜ 3 ⟺ mean ＜ median ＜ mode → (C)

Answer

mean ＜ median ＜ mode

Quick sanity check: the data set has a few 3s clustered at the top, which pulls the mode up to 3, while the small values (0s and 1s) pull the mean down. So we expect mode ＞ median ＞ mean, exactly the chain in (C). Also, the mean

\frac{15}{9}

≈ 1.67 sits between the smallest value 0 and the largest value 3, as any mean must.

💡Key takeaway

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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