A distribution has the equal-width classes 500–520, 520–540, 540–560, 560–580, 580–600 with frequencies 14, 26, 30, 20, 10. Which method of finding the mean needs the least arithmetic here, and why?

**Assumed mean**: pick a value 'a' close to the centre. Compute deviations dᵢ = $x$ᵢ − a.

**Formula**: **$x$̄ = $a$ + Σ(fᵢdᵢ) / Σfᵢ**

This simplifies arithmetic when class marks are large.

**Step-deviation method**: when all class widths equal $h$, define uᵢ = $(x$ᵢ − a)/$h$.

**Formula**: **$x$̄ = $a+h$ × [Σ(fᵢuᵢ) / Σfᵢ]**

This turns class marks into small integers (e.g., −2, −1, 0, 1, 2) — much easier arithmetic.

**All three methods give the same mean** — choose by convenience.