原本 ai ← i = 1 to n
平均 M = (1/n)Σai
調整 bi = 10√ai
平均 (1/n)Σbi = 65
標差 σ = √[(1/n)Σ(bi-65)²] = 15
先看 σ
√[(1/n)Σ(bi-65)²] = √[(1/n)Σ(bi²-2*65*bi+65²)]
= √{(1/n) [Σ(bi²) - 2*65*Σ(bi) + 65²n]}
= √{[(1/n)Σ(bi²)] - 2*65*[(1/n)Σ(bi)] + 65²}
= √{[(1/n)Σ(bi²)] - 2*65*65 + 65²}
= √{[(1/n)Σ(bi²)] - 65²} 題目很佛心，把推導結果給你了
= √{[(1/n)Σ(100ai)] - 65²} ← bi = 10√ai
= 15
(1/n)Σ(100ai) = 15² + 65²
100[(1/n)Σai] = 15² + 65²
100M = 15² + 65²
M = 5²(9 + 169) / 100
= 178/4 = 44.5 |
