1995 AJHSME Problems/Problem 19

Problem

The graph shows the distribution of the number of children in the families of the students in Ms. Jordan's English class. The median number of children in the family for this distribution is

[asy] unitsize(12); for(int i = 1; i <= 7; ++i) { draw((0,i)--(19,i),dotted); draw((-0.5,i)--(0.5,i)); } for(int i = 0; i <= 5; ++i) { draw((3*i+2,0)--(3*i+2,-0.5)); } fill((1,0)--(1,2)--(3,2)--(3,0)--cycle,white); fill((4,0)--(4,1)--(6,1)--(6,0)--cycle,white); fill((7,0)--(7,2)--(9,2)--(9,0)--cycle,white); fill((10,0)--(10,2)--(12,2)--(12,0)--cycle,white); fill((13,0)--(13,6)--(15,6)--(15,0)--cycle,white); draw((0,9)--(0,0)--(19,0)); draw((1,0)--(1,2)--(3,2)--(3,0)); draw((4,0)--(4,1)--(6,1)--(6,0)); draw((7,0)--(7,2)--(9,2)--(9,0)); draw((10,0)--(10,2)--(12,2)--(12,0)); draw((13,0)--(13,6)--(15,6)--(15,0)); label("$1$",(2,-0.5),S); label("$2$",(5,-0.5),S); label("$3$",(8,-0.5),S); label("$4$",(11,-0.5),S); label("$5$",(14,-0.5),S); label("$6$",(17,-0.5),S); label("$2$",(-0.5,2),W); label("$4$",(-0.5,4),W); label("$6$",(-0.5,6),W); label("$\textbf{Number of Children}$",(9,-1.5),S); label("$\textbf{in the Family}$",(9,-2.5),S); label("$\textbf{Number}$",(-1.5,6),W); label("$\textbf{of}$",(-3,5),W); label("$\textbf{Families}$",(-1.5,4),W);[/asy]

$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$

Solution

Counting, there are thirteen total families. The middle number is 7th in either direction, and it is easy to see from the right side that this number is 4 $\text{(D)}$

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions