2018 AMC 8 Problems/Problem 8

Problem 8

Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.

[asy] size(8cm); void drawbar(real x, real h) {   fill((x-0.15,0.5)--(x+0.15,0.5)--(x+0.15,h)--(x-0.15,h)--cycle,gray); } draw((0.5,0.5)--(7.5,0.5)--(7.5,5)--(0.5,5)--cycle); for (real i=1; i<5; i=i+0.5) {   draw((0.5,i)--(7.5,i),gray); } drawbar(1.0,1.0); drawbar(2.0,2.0); drawbar(3.0,1.5); drawbar(4.0,3.5); drawbar(5.0,4.5); drawbar(6.0,2.0); drawbar(7.0,1.5); for (int i=1; i<8; ++i) {   label("$"+string(i)+"$",(i,0.25)); } for (int i=1; i<9; ++i) {   label("$"+string(i)+"$",(0.5,0.5*(i+1)),W); } label("Number of Days of Exercise",(4,-0.1)); label(rotate(90)*"Number of Students",(-0.1,2.75)); [/asy] What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?

$\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00$

Solution

The mean, or average number of days is the total number of days divided by the number of students. The total number of days is $1\cdot 1+2\cdot 3+3\cdot 2+4\cdot 6+5\cdot 8+6\cdot 3+7\cdot 2=109$. The total number of students is $1+3+2+6+8+3+2=25$. Hence, $\frac{109}{25}=\boxed{\textbf{(C) } 4.36}$

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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

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