Translation preserves mean absolute deviation

Proved

WorkbookSource.problem_13814

by wamlart · Sep 12, 2026 · Mathlib 0df444a (Lean v4.33.1)

lean-workbooksource-checked

Let's say the set is $S_{1}=[a,b,c,d]$ let $m=\frac{a+b+c+d}{4}$ . From the information given, the standard deviation is the average of the differences between the elements of the set and $m$ . Something like $D_{1}=\frac{1}{4}(|a-m|+|b-m|+|c-m|+|d-m|)$ and that equals 6. Now, the new set should be $S_{2}=[a+3,b+3,c+3,d+3]$ let $n=\frac{a+3+b+3+c+3+d+3}{4}=m+3$ . $D_{2}=\frac{1}{4}(|a+3-(m+3)|+|b+3-(m-3)|+|c+3-(m+3)|+|d+3-(m+3)|)$ . This should hold no matter how many elements the sets contain. We can see that the $3s$ cancel out so that $D_{1}=D_{2}=6$ .

Source: InternLM Lean-Workbook, record lean_workbook_13814 (Apache-2.0). The complete source proposition and explicit variable declarations are preserved. The displayed absolute-deviation formula measures mean absolute deviation, although the source calls it standard deviation.

Preamble

import Mathlib
open Real Nat

Formal statement

theorem WorkbookSource.problem_13814  (a b c d m n : ℝ)
  (h₀ : m = (a + b + c + d) / 4)
  (h₁ : n = (a + 3 + b + 3 + c + 3 + d + 3) / 4)
  (h₂ : abs (a - m) + abs (b - m) + abs (c - m) + abs (d - m) = 24) :
  abs (a + 3 - n) + abs (b + 3 - n) + abs (c + 3 - n) + abs (d + 3 - n) = 24  :=  by sorry

Source

https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_13814; Apache-2.0

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