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

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==Solution==
 
==Solution==
We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. We take all the <math>3</math> letter names away from the list to see that the <math>3^{\mbox{rd}}</math> name in the new table is the desired length. Since there are <math>3</math> names that are <math>4</math> letters long, the median name length is <math>\text{(B)}</math> <math>4</math>.  
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We first notice that the median name will be the <math>10^{\mbox{th}}</math> name. The <math>10^{\mbox{th}}</math> name is <math>\boxed{\textbf{(B)}\ 4}</math>.  
  
 
{{AMC8 box|year=2016|num-b=5|num-a=7}}
 
{{AMC8 box|year=2016|num-b=5|num-a=7}}

Revision as of 23:20, 27 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)); label("name length", (4.5, -1)); [/asy]

Solution

We first notice that the median name will be the $10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{\textbf{(B)}\ 4}$.

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