The length of a highway was measured in 10 equal sections. If the precision of the individual section measurements is \(\sigma\), then what is the precision of the measured length of the highway?
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The highway length is a sum of 10 independent measurements, not an average, so use variance propagation for a sum: variances add, giving precision proportional to the square root of the number of sections.
Step 1: State the general rule for a sum of $n$ equally precise, independent measurements. In surveying error theory, if a quantity is obtained by adding $n$ independently measured parts, each with the same precision $\sigma$, the precision (standard error) of the sum follows the general relation $\sigma_{\text{sum}} = \sigma\sqrt{n}$, which comes directly from adding variances (not standard deviations) for independent errors. Step 2: Verify this rule with a small case first. For $n = 1$ section, the precision of the sum is trivially just $\sigma$, and $\sigma\sqrt{1} = \sigma$, consistent. For $n = 4$ equal sections, doubling from 1 to 4 should scale the standard error by $\sqrt{4} = 2$, not by 4, because variances (not standard deviations) add: $\sigma_{\text{sum}}^2 = 4\sigma^2 \Rightarrow \sigma_{\text{sum}} = 2\sigma$. This confirms the square-root scaling. Step 3: Apply the verified rule to $n = 10$. \[ \sigma_L = \sigma\sqrt{n} = \sigma\sqrt{10} \] Step 4: Cross-check numerically. Suppose $\sigma = 0.01$ m per section. Then $\sigma_L = 0.01 \times \sqrt{10} \approx 0.01 \times 3.162 = 0.0316$ m. This is larger than a single section's precision (as it must be, since more measurements are combined) but far smaller than $10\sigma = 0.10$ m, which would be the result of naively adding standard deviations as if the errors were fully correlated. Step 5: Confirm option (A) matches. This numeric check matches option (A), \(\sigma\sqrt{10}\), and rules out (C) $10\sigma$ as an overestimate and (B), (D) as expressions for the mean's precision rather than the sum's. Step 6: Conclude. \[ \sigma_L = \boxed{\sigma\sqrt{10}} \]
