AMC 8 · 2016 · #6
Grade 6 arithmeticProblem
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
Pick an answer.
AMC 8 2016 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: A bar graph shows how many of $19$ people have a name of each length from $3$ to $7$ letters. The bars give the counts $7, 3, 1, 4, 4$ for lengths $3, 4, 5, 6, 7$. We need the median name length — the middle value once all $19$ name lengths are lined up in order.
Givens: $19$ people in total; From the bar graph: length $3 \to 7$ people, length $4 \to 3$, length $5 \to 1$, length $6 \to 4$, length $7 \to 4$; Total checks out: $7 + 3 + 1 + 4 + 4 = 19$; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The median of the $19$ name lengths
Understand
Restated: A bar graph shows how many of $19$ people have a name of each length from $3$ to $7$ letters. The bars give the counts $7, 3, 1, 4, 4$ for lengths $3, 4, 5, 6, 7$. We need the median name length — the middle value once all $19$ name lengths are lined up in order.
Givens: $19$ people in total; From the bar graph: length $3 \to 7$ people, length $4 \to 3$, length $5 \to 1$, length $6 \to 4$, length $7 \to 4$; Total checks out: $7 + 3 + 1 + 4 + 4 = 19$; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The data come from a picture, so Tool #1 (Draw a Diagram) first — read each bar and write down the count per name length. Then Tool #7 (Identify Subproblems) splits the task into three small pieces: (a) confirm the total is $19$, (b) find which position is the median, (c) walk the counts until we reach that position. Tool #2 (Make a Systematic List) is the workhorse: build a cumulative tally — "through length $3$ we have $7$, through length $4$ we have $10$, ..." — and the median position drops out by inspection. No algebra needed.
Execute — Answer: B
3.MD.B.3 Step 1 - Read the bar graph.
- The horizontal axis labels name lengths $3, 4, 5, 6, 7$ and each bar's height is the number of people with that length.
- The bar heights are $7, 3, 1, 4, 4$.
💡 Reading a scaled bar graph and pulling out the count for each category is a Grade 3 data-display skill.
6.SP.B.5 Step 2 - Confirm the total.
- Add the five bar heights to make sure they sum to the stated $19$ people.
💡 Reporting the number of observations is the first habit of summarizing a data set.
6.SP.A.3 Step 3 - Locate the median position.
- With $19$ values (odd), the median is the single middle one — the $\tfrac{19+1}{2} = 10$th value when the lengths are written in order from smallest to largest.
💡 The median is a measure of center that names a position in the ordered list, not a computed average.
6.SP.B.5 Step 4 - Build a cumulative tally in order of length.
- After the length-$3$ group we have used positions $1$ through $7$; the length-$4$ group then fills positions $8, 9, 10$.
- So the $10$th value sits inside the length-$4$ group.
💡 A cumulative-count list lets you locate any ranked position without writing all $19$ names.
6.SP.A.3 Step 5 - Read off the answer.
- The $10$th value is in the length-$4$ block, so the median name length is $4$.
💡 The position you found lands inside one category, and that category's value is the median.
3.MD.B.3 Read the bar graph. The horizontal axis labels name lengths $3, 4, 5, 6, 7$ and 6.SP.B.5 Confirm the total. Add the five bar heights to make sure they sum to the stated 6.SP.A.3 Locate the median position. With $19$ values (odd), the median is the single mid 6.SP.B.5 Build a cumulative tally in order of length. After the length-$3$ group we have 6.SP.A.3 Read off the answer. The $10$th value is in the length-$4$ block, so the median Review
Reasonableness: The smallest length ($3$) already accounts for $7$ of $19$ people — more than a third — so the median should sit near the low end. Length $4$ is the very next category, which pushes the cumulative count from $7$ to $10$. Since $10$ is exactly the median position for $n = 19$, the median has to be $4$. A median of $5$ or $6$ would require fewer short names at the front, which the graph contradicts.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: the $7$ length-$3$ names take positions $1$-$7$, so positions $8, 9, 10$ are inside the length-$4$ block. That immediately kills (A) $3$ (too early), and (C), (D), (E) (too late, since the cumulative count only reaches $11$ at length $5$). Only (B) $4$ contains the $10$th position.
CCSS standards used (min grade 6)
3.MD.B.3Draw and interpret scaled picture graphs and bar graphs (Reading the heights of the five bars to extract the counts $7, 3, 1, 4, 4$ for name lengths $3, 4, 5, 6, 7$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Using the definition of the median as the middle value of an ordered data set, and identifying it as the $10$th value when $n = 19$.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Reporting the total count ($n = 19$) and building a cumulative tally to locate the median position inside the grouped data.)
⭐ Even with $19$ names, you never have to write them all out — a short cumulative-count list shows the $10$th name lands in the length-$4$ group.
⭐ Even with $19$ names, you never have to write them all out — a short cumulative-count list shows the $10$th name lands in the length-$4$ group.
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