Error propagation and effect of averaging

(a) Error propagation

For $V = \dfrac{\pi d^2 h}{4}$ the relative error depends on each variable as

$$\frac{\delta V}{V} = 2\frac{\delta d}{d} + \frac{\delta h}{h} \quad (\text{maximum}), \qquad \frac{\delta V}{V} = \sqrt{\left(2\frac{\delta d}{d}\right)^2 + \left(\frac{\delta h}{h}\right)^2} \quad (\text{probable})$$ $$V = \frac{\pi (25)^2 (60)}{4} = 29\,452\ \text{mm}^3$$

• $2\,\delta d/d = 2(0.02/25) = 0.0016$
• $\delta h/h = 0.05/60 = 0.000833$

Maximum error: $\delta V = 29452 \times (0.0016 + 0.000833) = 71.7\ \text{mm}^3$ (0.243 %).

Probable error: $\delta V = 29452\sqrt{0.0016^2 + 0.000833^2} = 29452 \times 0.001804 = 53.1\ \text{mm}^3$ (0.180 %).

The diameter contributes most of the error because it is squared.

Answer: $V = 29\,452 \pm 72$ mm³ (maximum), $\pm 53$ mm³ (probable).

(b) Averaging

Number of readings $n = 8$.

$$\bar d = \frac{25.02+24.98+25.01+25.03+24.99+25.00+25.02+25.01}{8} = \frac{200.06}{8} = 25.0075\ \text{mm}$$

Sum of squares $= 1950\times10^{-6}$ mm².

$$s = \sqrt{\frac{1950\times10^{-6}}{n-1}} = \sqrt{278.6\times10^{-6}} = 0.0167\ \text{mm}$$ $$\sigma_{\bar d} = \frac{s}{\sqrt{n}} = \frac{0.0167}{\sqrt 8} = 0.0059\ \text{mm}$$

Since the standard error varies as $1/\sqrt n$, halving it needs $n$ four times larger: $4 \times 8 = 32$ readings. Averaging reduces random error but not systematic error.

Answer: mean = 25.0075 mm, $s$ = 0.0167 mm, standard error = 0.0059 mm; 32 readings for half the standard error.