AMC 8 · 2011 · #4

Easy mode Grade 6

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Problem

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.

(A)

median < mean < mode

(B)

mean < mode < median

(C)

mean < median < mode

(D)

median < mode < mean

(E)

mode < median < mean

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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

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

#2 Re-arrange / Reorder 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.

$$\{2, 0, 1, 3, 0, 3, 3, 1, 2\} \;\longrightarrow\; \{0, 0, 1, 1, 2, 2, 3, 3, 3\}$$

💡 Putting the values in order is the Grade 6 data-display habit that exposes the shape of the data.

#7 Identify Subproblems 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.

$$\text{mean} = \dfrac{0+0+1+1+2+2+3+3+3}{9} = \dfrac{15}{9} = \dfrac{5}{3} \approx 1.67$$

💡 Computing the mean as "sum divided by count" is the Grade 6 definition of average.

#7 Identify Subproblems 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).

Sorted: $0, 0, 1, 1, \underline{2}, 2, 3, 3, 3 \;\Rightarrow\; \text{median} = 2$

💡 The middle position of an odd-length sorted list is the median by definition.

#7 Identify Subproblems 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.

Counts: $0{:}\,2,\; 1{:}\,2,\; 2{:}\,2,\; 3{:}\,3 \;\Rightarrow\; \text{mode} = 3$

💡 The mode is just the tallest bar in a frequency count.

#7 Identify Subproblems 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 $3$.

$$\dfrac{15}{9} < 2 < 3 \;\Longleftrightarrow\; \text{mean} < \text{median} < \text{mode} \;\Rightarrow\; \textbf{(C)}$$

💡 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.

[1] #2 6.SP.B.4 Sort the data from smallest to largest. This is the Tool #2 (Re-arrange) move —

[2] #7 6.SP.B.5 Find the mean. Add all nine values, then divide by $9$. Since the list is sorted

[3] #7 6.SP.B.5 Find the median. With $9$ sorted values, the median is the $5$th value (four bel

[4] #7 6.SP.B.5 Find the mode. Scan the sorted list and pick the value that appears most often.

[5] #7 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.4 Display 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.5 Summarize 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.3 Interpret 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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