find the sum of the average monthly rainfalls...

# Explanation: ## Step1: Count the number of 'x's for each value For $\frac{1}{8}$, there are 3 'x's; for $\frac{1}{4}$, there are 2 'x's; for $\frac{3}{8}$, there is 1 'x'; for $\frac{1}{2}$, there is 1 'x'; for $\frac{5}{8}$, there is 1 'x'; for $\frac{3}{4}$, there is 1 'x'; for $\frac{7}{8}$, there is 1 'x'; for 1, there is 1 'x'; for $1\frac{1}{8}$, there is 1 'x'; for $1\frac{1}{4}$, there is 1 'x'; for $1\frac{3}{8}$, there is 1 'x'; for $1\frac{1}{2}$, there is 1 'x'; for $1\frac{5}{8}$, there is 1 'x'; for $1\frac{3}{4}$, there is 1 'x'; for $1\frac{7}{8}$, there is 1 'x'; for 2, there is 1 'x'. ## Step2: Calculate the sum of the products of each value and its frequency \[ \begin{align*} &3\times\frac{1}{8}+ 2\times\frac{1}{4}+1\times\frac{3}{8}+1\times\frac{1}{2}+1\times\frac{5}{8}+1\times\frac{3}{4}+1\times\frac{7}{8}+1\times1+1\times1\frac{1}{8}+1\times1\frac{1}{4}+1\times1\frac{3}{8}+1\times1\frac{1}{2}+1\times1\frac{5}{8}+1\times1\frac{3}{4}+1\times1\frac{7}{8}+1\times2\\ =&\frac{3}{8}+\frac{2}{4}+\frac{3}{8}+\frac{1}{2}+\frac{5}{8}+\frac{3}{4}+\frac{7}{8}+1+\frac{9}{8}+\frac{5}{4}+\frac{11}{8}+\frac{3}{2}+\frac{13}{8}+\frac{7}{4}+\frac{15}{8}+2\\ =&\frac{3 + 4+3 + 4+5+6+7+8+9+10+11+12+13+14+15+16}{8}\\ =&\frac{128}{8}\\ =&16 \end{align*} \] # Answer: 16