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For the distribution 0–10: 5, 10–30: 12, 30–35: 9, 35–60: 4 the class widths are 10, 20, 5 and 25. Which statement is correct?

Class 10StatisticsStatistics and Probability
  • AThe step-deviation method is still best, since h may be chosen freely
  • BThe mean cannot be found at all when the widths are unequal
  • CThe step-deviation method does not apply in the usual way, so use the direct or the assumed-mean method
  • DOnly the mode can be found from such a table

Step-by-Step Solution

needs one common width . With widths 10, 20, 5 and 25 there is no common , so the step-deviation shortcut breaks down. The direct method (Σfx/Σf) and the assumed-mean method still work, because they use the class marks themselves.
Answer: C) The step-deviation method does not apply in the usual way, so use the direct or the assumed-mean method

Key idea: Mean by assumed mean and step-deviation

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

**Formula**: **̄ = + Σ(fᵢdᵢ) / Σfᵢ**

This simplifies arithmetic when class marks are large.

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

**Formula**: **̄ = × [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.

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