, 2025

Reading from the graph, there is $1$ student who is $14$, $0$ students who are $15$, $3$ students who are $16$, and $2$ students who are $17$. The total number of students in the Art Club is $1+0+3+2=6$. Using the work from part (a), the mean (average) age of the students in the Art Club is $\dfrac{(1\times14)+(3\times16)+(2\times17)}{6}=\dfrac{96}{6}=16$. From part (b), the sum of the ages of the $6$ students currently in the Art Club is $96$. If $n$ $15$-year old students join the club, the number of students in the club will be $6+n$, and the sum of their ages will be $96+15n$. When $n$ $15$-year old students join the club, the mean age of all students in the club is $15.5$. Solving for $n$, we get $$\begin{align*} \dfrac{96+15n}{6+n}&=15.5\\ 96+15n&=15.5(6+n)\\ 96+15n&=93+15.5n\\ 3&=0.5n\end{align*}$$ and so $n=\dfrac{3}{0.5}=6$. Therefore, the number of $15$-year old students that must join the Art Club so that the mean age of the students in the club is $15.5$ is $6$.