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]](http://latex.artofproblemsolving.com/1/3/b/13b970ebbe901f54d387fa83765bf69ef8046d0e.png)
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 
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
