---
title: "Optimum rain gauges for 10% error, six stations"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:optimum-raingauges-6stn-240"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/optimum-raingauges-6stn-240
site: BCT Engineering
---

# Optimum rain gauges for 10% error, six stations

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2079 Bhadra, 4 marks
- Video: [Optimum number of rain gauges || Hydrology #civilengineering #hydrology](https://www.youtube.com/watch?v=POAAn-vg7hs) — CIVIL HUT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/optimum-raingauges-6stn-240
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

A catchment has six rain gauge stations. In a year, the annual rainfall in cm recorded by the gauges are as follows: 240, 252.2, 228.3, 218.6, 275.3, 212.2. For a 10% error in the estimation of the mean rainfall, calculate the optimum number of stations in the catchment.

## Short answer

Mean rainfall = 237.77 cm, standard deviation = 23.38 cm, Cv = 9.83%. Optimum number N = (9.83/10)² = 0.97 ≈ 1; the existing 6 stations are adequate for 10% error.

## Answer

The optimum number of gauges for a permissible error $\varepsilon$ is

$$
N = \left(\frac{C_v}{\varepsilon}\right)^2, \qquad C_v = \frac{100\,\sigma_{n-1}}{\bar P}
$$

**Data (cm):** 240, 252.2, 228.3, 218.6, 275.3, 212.2; $n = 6$.

- Mean: $\bar P = 1426.6/6 = 237.77$ cm
- Standard deviation: $\sigma_{n-1} = 23.38$ cm
- $C_v = \dfrac{100\times23.38}{237.77} = 9.83\%$

For $\varepsilon = 10\%$:

$$
N = \left(\frac{9.83}{10}\right)^2 = 0.97 \approx 1
$$

**Answer: The optimum number is about 1 gauge. The existing 6 stations are more than enough for a 10% error, so no additional gauge is needed.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The optimum number of gauges for permissible error e is:
   A. N = Cv/e
   B. N = (Cv/e)^2
   C. N = e/Cv
   D. N = (e/Cv)^2
   Answer: B. N = (Cv/e)^2 with Cv the coefficient of variation in percent. (Key point: Formula)

2. Cv is found as:
   A. 100 x mean / sigma(n-1)
   B. sigma(n-1) / n
   C. 100 x sigma(n-1) / mean
   D. mean - sigma(n-1)
   Answer: C. Cv = 100 sigma/mean = 100 x 23.38/237.77 = 9.83%. (Key point: Coefficient of variation)

3. N comes out about 1 for 10% error. What does this mean?
   A. Six gauges are too few; add more
   B. The existing 6 gauges are more than enough; none needed
   C. Remove all but one gauge for being unsafe
   D. The error must be reduced to 1%
   Answer: B. Optimum N = 0.97 is less than the 6 existing stations, so no additional gauge is required. (Key point: Interpretation)

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