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The value of the normal density is practically zero when the value lies more than a few standard deviations away from the mean (e.g., a spread of three standard deviations covers all but 0.27% of the total distribution).
The sum of the entries along the main diagonal (the trace), plus one, equals 4 − 4(x 2 + y 2 + z 2), which is 4w 2. Thus we can write the trace itself as 2w 2 + 2w 2 − 1; and from the previous version of the matrix we see that the diagonal entries themselves have the same form: 2x 2 + 2w 2 − 1, 2y 2 + 2w 2 − 1, and 2z 2 + 2w 2 − 1. So ...