What Averaging Buys, and What It Cannot
What Averaging Buys, and What It Cannot
Ada re-derives this chapter’s own numbers step by step, at full precision
ADA · CALCULATION AUDIT
What Averaging Buys, and What It Cannot
Five raw temperature readings of 24.8°C, 25.2°C, 24.9°C, 25.1°C, and 25.0°C carry noise of 0.14°C, and the chapter’s 5-sample moving average brings that down to 0.063°C — roughly a 2.24x improvement. This audit asks the question that improvement invites: does averaging away that random noise also fix a systematic offset in the readings, or only make the wrong answer look steadier?
Companion to the chapter Sensor Data Processing — every number here comes from that chapter.
See the relationship before changing it
The figure reads from left to right. The blue card is averaged samples. The middle card applies this page's rule. The green card is random spread. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only averaged samples, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 5 samples.
- 2
Name the relationship. spread = 0.141421 deg C / sqrt(sample count)
- 3
Substitute with units. 0.141421 / sqrt(5) = 0.063246 deg C
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change averaged samples
Try Predict the direction of spread = 0.141421 deg C / sqrt(sample count). Test another averaged samples, then compare random spread.
Observe Averaging independent noise gives diminishing square-root improvement and leaves bias untouched. Reset averaged samples to 5 and compare random spread.
Explain Averaging independent noise gives diminishing square-root improvement and leaves bias untouched.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
Average the five readings 24.8, 25.2, 24.9, 25.1, and 25.0 C; use Check audit after finding sigma.
The mean holds at 25.00 C, sigma reads 0.141421 C, and averaging 5 independent samples reduces random spread to about 0.063246 C.
Averaging cancels zero-mean fluctuations by the square-root law but leaves a shared +2 C calibration bias unchanged.
Ada: The section above takes five readings and reports noise of 0.14 C falling to 0.063 C after averaging – a 2.24x gain. Let me rebuild that from the raw numbers, then name the one error averaging can never touch.
- Five readings 24.8, 25.2, 24.9, 25.1, 25.0 have mean
125.0 / 5 = 25.00 C. - Squared deviations:
0.04 + 0.04 + 0.01 + 0.01 + 0 = 0.10; variance0.10 / 5 = 0.02; sosigma = sqrt(0.02) = 0.141421 C(the chapter’s 0.14). - Averaging N independent samples divides that by
sqrt(N):0.141421 / sqrt(5) = 0.063246 C, an improvement ofsqrt(5) = 2.2361x. Both match.
The sqrt(N) law is a law of diminishing returns: halving the noise again needs N to quadruple, because sqrt(4N) = 2 sqrt(N). But the sharper point is what stays fixed – averaging attacks only the random spread. If all five readings shared a +2 C offset, the mean would still land 2 C high; sqrt(N) shrinks the jitter around the wrong answer without moving it. That is the dividing line the chapter draws next: filtering earns precision, and only calibration can earn accuracy.
Every number above is taken from the chapter’s own material and re-derived step by step.
Technical boundaries: The square-root reduction assumes independent zero-mean samples; it does not represent autocorrelation, outliers, drift, fixed calibration bias, changing process temperature, or quantisation floors.