- Ah = 8, and the class 142–150 has u = 0
- Bh = 4, and the class 134–142 has u = 0
- Ch = 8, and the class 134–142 has u = 0
- Dh = 8, and the class 126–134 has u = 0
Step-by-Step Solution
The common width is h=126−118=8 (not 4, which is only half the width). u=(x−a)/h is zero when the class mark equals a, and the class mark of 134–142 is (134+142)/2=138=a, so that class carries u=0.
Answer: C) h = 8, and the class 134–142 has u = 0
Key idea: Mean by assumed mean and step-deviation
**Assumed mean**: pick a value 'a' close to the centre. Compute deviations dᵢ = xᵢ − a.
**Formula**: **x̄ = a + Σ(fᵢdᵢ) / Σfᵢ**
This simplifies arithmetic when class marks are large.
**Step-deviation method**: when all class widths equal h, define uᵢ = (xᵢ − a)/h.
**Formula**: **x̄ = 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.