---
title: "Estimating missing precipitation at station B"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:missing-rainfall-four-stations"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/missing-rainfall-four-stations
site: BCT Engineering
---

# Estimating missing precipitation at station B

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2076 Chaitra, 6 marks
- Video: [Estimation of missing rainfall data | Engineering hydrology | Normal ratio method | civil engg |](https://www.youtube.com/watch?v=f0315_JmPuo) — CIVIL HUT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/missing-rainfall-four-stations
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

The normal annual rainfall of four stations A, B, C and D in a catchment area is 1150, 940, 1080 and 980 respectively. In a particular year the station B did not work and the rainfall amount of A, C and D were 1000, 970 and 890 respectively. Compute the missing precipitation at point B for that year.

## Short answer

Station normals differ >10%, so normal ratio method is used. Missing rainfall at B = (1/3)[(940/1150)×1000 + (940/1080)×970 + (940/980)×890] = 838.4 mm.

## Answer

The normal annual rainfall of the base stations A and C differs from that of B by more than 10% (A: 22%; C: 15%; D: 4%), so the **normal ratio method** is used:

$$
P_B = \frac{1}{n}\left[\frac{N_B}{N_A}P_A + \frac{N_B}{N_C}P_C + \frac{N_B}{N_D}P_D\right]
$$

with $n = 3$ stations, $N_A = 1150$, $N_B = 940$, $N_C = 1080$, $N_D = 980$ and $P_A = 1000$, $P_C = 970$, $P_D = 890$ mm.

$$
\begin{aligned}
\frac{N_B}{N_A}P_A &= \frac{940}{1150}\times1000 = 817.39\\
\frac{N_B}{N_C}P_C &= \frac{940}{1080}\times970 = 844.26\\
\frac{N_B}{N_D}P_D &= \frac{940}{980}\times890 = 853.67
\end{aligned}
$$

$$
P_B = \frac{817.39 + 844.26 + 853.67}{3} = 838.4\ \text{mm}
$$

**Answer: Missing precipitation at station B = 838.4 mm (about 838 mm).**

## Check yourself

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

1. Why is the normal ratio method used?
   A. Station B has no normal rainfall
   B. A base station's normal annual rainfall differs from B's by more than 10%
   C. There are only two neighbouring stations
   D. Rainfall data are daily
   Answer: B. A and C differ from B by 22% and 15%, over 10%, so simple averaging is not used. (Key point: Choice of method)

2. The normal ratio estimate for B is:
   A. Sum of PA, PC, PD divided by 3
   B. Average of (NB/Ni) x Pi over the base stations
   C. NB x sum of Pi
   D. Average of (Ni/NB) x Pi
   Answer: B. P_B = (1/n) sum of (N_B/N_i) P_i. (Key point: Formula)

3. What is the estimated rainfall at B?
   A. 838.4 mm
   B. 953.3 mm
   C. 1000 mm
   D. 817.4 mm
   Answer: A. (817.39+844.26+853.67)/3 = 838.4 mm. (Key point: Result)

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