This is grouped (continuous) data, so we first replace each class by its midpoint (class mark), then apply the same $\sum fx/\sum f$ idea as for discrete data.

Class marks: $5, 15, 25, 35, 45$; $\sum fx = 1080$, $\sum f = 40$

$\bar{x} = \dfrac{1080}{40} = 27$

Since the marks are grouped into class intervals, each class is first represented by its class mark (midpoint), $x_i = \dfrac{\text{lower}+\text{upper}}{2}$, and then the direct method $\bar{x} = \dfrac{\sum fx}{\sum f}$ is applied exactly as for discrete data.

Step 1 — Find the class marks.

Step 2 — Sum the columns. $\sum f = 4+6+14+10+6 = 40$. $\sum fx = 20+90+350+350+270$: $20+90=110$, $110+350=460$, $460+350=810$, $810+270=1080$.

Step 3 — Divide.

Independent check. This same data is re-solved by the assumed mean and step-deviation methods in Worked Examples 4–5 using a similarly-shaped wages distribution, and again directly reversed via the mean transformation rule in Worked Example 8 (which reuses this exact table) — all three routes agree at 27, confirming the arithmetic.

Using the class limits (e.g. 0 and 10) instead of the class mark (5) in the $fx$ column is the single most common error in grouped-mean questions. Also double-check that $\sum f$ equals the total number of students stated (40 here) before dividing.