---
title: "Arithmetic mean and Thiessen average precipitation"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:arithmetic-thiessen-numerical"
source: https://www.bctengineer.com/q/semester_6/hydro_power/arithmetic-thiessen-numerical
site: BCT Engineering
---

# Arithmetic mean and Thiessen average precipitation

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2082 Kartik, 1+3 marks
- Video: [Finding Mean Rainfall || Arithmetic Mean Method and Thiessen Method](https://www.youtube.com/watch?v=AQg0q9znOw4) — EduCare Foundation
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/arithmetic-thiessen-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

Calculate the average precipitation using the arithmetic mean method and the Thiessen polygon method based on the following data.

| Station No. | Precipitation (mm) | Area (km²) |
|---|---|---|
| 1 | 30.8 | 45 |
| 2 | 34.6 | 40 |
| 3 | 32.0 | 30 |
| 4 | 24.6 | 38 |

## Short answer

Arithmetic mean = (30.8+34.6+32.0+24.6)/4 = 30.50 mm. Thiessen polygon mean = (Σ P_i A_i)/Σ A_i = 4664.8/153 = 30.49 mm.

## Answer

#### Arithmetic mean method
$$
\bar{P} = \frac{30.8 + 34.6 + 32.0 + 24.6}{4} = \frac{122.0}{4} = 30.50\ \text{mm}
$$

#### Thiessen polygon method
Each station's rainfall is weighted by its polygon area.

| Station | $P_i$ (mm) | $A_i$ (km²) | $P_iA_i$ (mm·km²) |
|---|---|---|---|
| 1 | 30.8 | 45 | 1386.0 |
| 2 | 34.6 | 40 | 1384.0 |
| 3 | 32.0 | 30 | 960.0 |
| 4 | 24.6 | 38 | 934.8 |
| **Total** | | **153** | **4664.8** |

$$
\bar{P} = \frac{\sum P_iA_i}{\sum A_i} = \frac{4664.8}{153} = 30.49\ \text{mm}
$$

**Answer:** Arithmetic mean = 30.50 mm; Thiessen mean = 30.49 mm. The two are nearly equal here because the station areas are fairly similar; the Thiessen value is slightly lower mainly because station 3 (32.0 mm) has the smallest area and so gets less weight.

## Check yourself

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

1. What is the arithmetic mean precipitation for 30.8, 34.6, 32.0 and 24.6 mm?
   A. 30.49 mm
   B. 31.0 mm
   C. 30.50 mm
   D. 29.6 mm
   Answer: C. The sum is 122.0 mm; dividing by 4 gives 30.50 mm. (Key point: Arithmetic mean)

2. How is the Thiessen mean computed from the table?
   A. Sum of the rainfalls divided by the number of stations
   B. Sum of Pi*Ai divided by the sum of Ai
   C. Sum of Ai divided by the sum of Pi
   D. Largest Pi times the total area
   Answer: B. P = sum(PiAi)/sum(Ai) = 4664.8/153 = 30.49 mm. (Key point: Thiessen method)

3. Why are the two means (30.50 and 30.49 mm) nearly equal here?
   A. Thiessen always equals the arithmetic mean
   B. Only one station was used
   C. The station areas are fairly similar, so weights are near equal
   D. The rainfall is the same at all stations
   Answer: C. When polygon areas are fairly similar, area weighting changes little from a simple average. (Key point: Why results are close)

## Related in Hydro Power

- [Thiessen polygon method with rainfall data](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Athiessen-polygon-explain-numerical) — Hydro Power, chapter 2

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