[tex] \bf \displaystyle \cfrac{1}{7}\sum\limits_{i=1}^{7}~a_i\\\\\\
\cfrac{1}{7}\left[ \stackrel{i=1~ a_1}{(2.7)}+\stackrel{i=2~ a_2}{(3.0)}+\stackrel{i=3~ a_3}{(3.2)}+\stackrel{i=4~ a_4}{(3.7)}+\stackrel{i=5~ a_5}{(4.4)}+\stackrel{i=6~ a_6}{(4.9)}+\stackrel{i=7~ a_7}{(5.3)} \right]
\\\\\\
\displaystyle \cfrac{1}{7}\sum\limits_{i=1}^{7}~a_i\qquad \approx 3.88571428571428571429\implies \stackrel{\textit{rounded up}}{3.9} [/tex]
now, le'ts do the summation using 0.5n+2
(0.5(2.7) + 2)+(0.5(3.0) + 2)+(0.5(3.2) + 2)+(0.5(3.7) + 2)+(0.5(4.4) + 2)+(0.5(4.9) + 2)+(0.5(5.3) + 2)
which gives us 27.6. Now, the paper doesn't state, or at least I don't see it, but that's just the sum of the hours, to get the average we simply divide that by 7, the amount of items, and 27.6 ÷ 7 gives us about 3.94.
does it underestimate or overestimate the one found with the summation? well, the summation gave us 3.9 and 3.94 is just 0.04 or 4 hundredths above it.
