原平均数与实际平均数的差是哪个减哪个
2个回答
设各标志值为 X1,X2,...,XN.
则原平均数为
U = [X1 + X2 + ...+ XN]/N.
1609 = [(X1-2)^2 + (X2-2)^2 + ...+ (XN-2)^2]/N
9 = [(X1-U)^2 + (X2-U)^2 + ...+ (XN-U)^2]/N
= [(X1-2+2-U)^ + (X2-2+2-U)^ + ...+ (XN-2+2-U)^2]/N
= {[(X1-2)^2 + 2(X1-2)(2-U) + (2-U)^2] + [(X2-2)^2 + 2(X2-2)(2-U) + (2-U)^2] + ...+ [(XN-2)^2 + 2(XN-2)(2-U) + (2-U)^2]}/N
= [(X1-2)^2 + (X2-2)^2 + ...+ (XN-2)^2]/N + 2(2-U)[(X1-2) + (X2-2) + ...+ (XN-2)]/N + (2-U)^2
= 1609 + 2(2-U)[X1 + X2 + ...+ XN]/N + 2(2-U)[-2] + (2-U)^2
= 1609 + 2(2-U)U -4(2-U) + (2-U)^2
= 1609 + 4U - 2U^2 - 8 + 4U + 4 - 4U + U^2
= 1609 - U^2 + 4U - 4
= 1609 - (U-2)^2,
(U-2)^2 = 1600
U-2 = 400,或者,U-2 = -400.
U = 402,或者,U = -398.
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