Difference between revisions of "2016 AMC 8 Problems/Problem 6"

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The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
 
The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?
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<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math>
 
<math>\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7</math>
 
<asy>
 
<asy>

Revision as of 16:17, 23 November 2016

The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7$ [asy] unitsize(0.9cm); draw((-0.5,0)--(10,0), linewidth(1.5)); draw((-0.5,1)--(10,1)); draw((-0.5,2)--(10,2)); draw((-0.5,3)--(10,3)); draw((-0.5,4)--(10,4)); draw((-0.5,5)--(10,5)); draw((-0.5,6)--(10,6)); draw((-0.5,7)--(10,7)); label("frequency",(-0.5,8)); label("0", (-1, 0)); label("1", (-1, 1)); label("2", (-1, 2)); label("3", (-1, 3)); label("4", (-1, 4)); label("5", (-1, 5)); label("6", (-1, 6)); label("7", (-1, 7)); filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black); filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black); filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black); filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black); label("3", (0.5, -0.5)); label("4", (2.5, -0.5)); label("5", (4.5, -0.5)); label("6", (6.5, -0.5)); label("7", (8.5, -0.5)); [/asy]

Solution

We first notice that the median name will be the $10^{\mbox{th}}$ name. We take all the $3$ letter names away from the list to see that the $3^{\mbox{rd}}$ name in the new table is the desired length. Since there are $3$ names that are $4$ letters long, the median name length is $\text{(B)}$ $4$.

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions