C.1 Handling data

Significant figures communicate measurement precision. A calculated result must not claim finer precision than the least accurate measurement used to obtain it.

A rectangle measured as $2.4\,\mathrm{cm}$ by $3.68\,\mathrm{cm}$ has calculator area $8.832\,\mathrm{cm^2}$. The $2.4\,\mathrm{cm}$ reading is quoted to 2 significant figures, so report $A=8.8\,\mathrm{cm^2}$, not $8.832\,\mathrm{cm^2}$.

Leading zeros are not significant, but zeros between non-zero digits and stated trailing zeros are: $0.00450$ has 3 significant figures. Scientific notation makes the intended precision explicit: $4.50\times10^{-3}$.

For $n$ repeated readings of the same quantity, the arithmetic mean is $\bar{x}=(x_1+x_2+\cdots+x_n)/n$. It reduces the influence of random variation but does not remove a systematic error.

For times $5.23$, $5.89$, $5.66$ and $5.01\,\mathrm{s}$, $\bar{t}=(5.23+5.89+5.66+5.01)/4=5.4475\,\mathrm{s}$, reported as $5.45\,\mathrm{s}$.

If each reading is for several cycles, first average the repeated total times and then divide by the number of cycles. For example, a mean time of $7.55\,\mathrm{s}$ for $5T$ gives $T=7.55/5=1.51\,\mathrm{s}$.

A probability lies from 0 to 1. In radioactive decay it describes the chance that a nucleus decays during a stated interval; identical nuclei have the same chance, but the time at which any one nucleus decays is unpredictable.

If each of $500$ nuclei has probability $0.020$ of decaying in a short interval, the expected number of decays is $Np=500\times0.020=10$. A particular observation need not contain exactly 10 decays.

A larger sample gives a more stable fraction of decays, while individual counts still fluctuate randomly. Probability predicts the behaviour of many repeated trials or a large population, not a deterministic result for one trial.

An order of magnitude is a power-of-ten scale. Write each variable as a coefficient times $10^n$, perform the exponent arithmetic separately, then normalise the result to identify its scale.

If $y=ab/c$, with $a\approx3\times10^6$, $b\approx2\times10^{-4}$ and $c\approx5\times10^2$, then $y\approx(6/5)\times10^{6-4-2}=1.2\times10^0$. Its order is $10^0$.

Round inputs enough to reveal the scale, but keep the coefficient while combining terms because it may shift the normalised power of ten. Compare the estimate with the calculator result to catch an incorrect exponent or prefix.

For the simple worst-case treatment used here, first express each measurement with an absolute or fractional uncertainty. The operation that combines the measured values determines how their uncertainties combine.

If $L_1=(12.4\pm0.1)\,\mathrm{cm}$ and $L_2=(8.2\pm0.1)\,\mathrm{cm}$, then $L_1+L_2=20.6\,\mathrm{cm}$ and $\Delta L=0.1+0.1=0.2\,\mathrm{cm}$. Report $(20.6\pm0.2)\,\mathrm{cm}$.

For $Q=A^2B/C$, $\Delta Q/Q=2(\Delta A/A)+(\Delta B/B)+(\Delta C/C)$. After adding the fractional terms, multiply by the calculated $Q$ to convert back to an absolute uncertainty when required.