---
title: "Additional rain gauges for 5% error, seven stations"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:optimum-raingauges-7stn-130"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/optimum-raingauges-7stn-130
site: BCT Engineering
---

# Additional rain gauges for 5% error, seven stations

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2070 Asar, 4 marks; 2068 Chaitra, 4 marks (asked 2 times)
- 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-7stn-130
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

A catchment has seven raingauge stations. In a year the annual rainfall in cm recorded by the gauges are as follows: 130, 142.1, 118.2, 108.5, 165.2, 102.1, 146.9. For a 5% error in the estimation of the mean rainfall, calculate the minimum number of additional stations required to be established in the catchment.

## Short answer

Mean is 130.43 cm, sample standard deviation 22.545 cm, Cv=17.29%. Optimum number N=(17.29/5)²≈12. With 7 existing stations, 5 additional are required.

## Answer

The optimum number of rain gauges for a permissible error $\varepsilon$ in the mean rainfall is

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

**Data (cm):** 130, 142.1, 118.2, 108.5, 165.2, 102.1, 146.9; existing stations $n = 7$.

- Mean: $\bar P = 913/7 = 130.43$ cm
- Standard deviation (sample): $\sigma_{n-1} = 22.545$ cm
- Coefficient of variation: $C_v = \dfrac{100 \times 22.545}{130.43} = 17.29\%$

For $\varepsilon = 5\%$:

$$
N = \left(\frac{17.29}{5}\right)^2 = 11.95 \approx 12
$$

Additional stations $= 12 - 7 = 5$.

**Answer: Total 12 stations are needed, so at least 5 additional rain gauges must be established.**

## Check yourself

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

1. Optimum number of gauges:
   A. N = (Cv/ε)^2
   B. N = Cv/ε
   C. N = Cv x ε
   D. N = (ε/Cv)^2
   Answer: A. Cv in percent. (Key point: Formula)

2. Cv is:
   A. σ / n
   B. 100 x σ(n-1) / mean
   C. mean / σ
   D. 100 x mean
   Answer: B. Cv = 17.29%. (Key point: Cv)

3. N is:
   A. 17.29/5 = 3.5
   B. 5 x 17.29
   C. (17.29/5)^2 = 11.95, about 12
   D. 7
   Answer: C. Rounded up. (Key point: Computation)

4. Additional stations:
   A. 12
   B. 7
   C. 19
   D. 12 - 7 = 5
   Answer: D. Total minus existing. (Key point: Additional)

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