---
title: "Flood statistics and exceedance probability, 2000-2009"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:flood-stats-2000-2009"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-stats-2000-2009
site: BCT Engineering
---

# Flood statistics and exceedance probability, 2000-2009

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2078 Bhadra, 6+4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-stats-2000-2009
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Using given data for annual discharge in a section of a river from year 2000 to 2009.

(i) Calculate sample mean, sample standard deviation and sample coefficient of skewness.

(ii) What would be the probability of the flood of magnitude 250 m$^3$/s occurring in the next year?

| Year | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 |
|---|---|---|---|---|---|---|---|---|---|---|
| Discharge (m$^3$/s) | 33.9 | 31.7 | 31.5 | 59.6 | 50.5 | 38.6 | 43.4 | 28.7 | 32.0 | 51.8 |

Reduced mean $\bar{y}_n = 0.4952$, Reduced Standard Deviation $(\sigma_n) = 0.9497$.

## Short answer

Sample mean = 40.17 m³/s, standard deviation = 10.63 m³/s, coefficient of skewness = 0.75. Probability of a 250 m³/s flood next year is ≈4.4×10⁻⁹ (practically zero).

## Answer

#### (i) Statistics (n = 10)
| Year | x (m³/s) | x − mean | (x − mean)² | (x − mean)³ |
|---|---|---|---|---|
| 2000 | 33.9 | -6.27 | 39.31 | -246.5 |
| 2001 | 31.7 | -8.47 | 71.74 | -607.6 |
| 2002 | 31.5 | -8.67 | 75.17 | -651.7 |
| 2003 | 59.6 | 19.43 | 377.52 | 7335.3 |
| 2004 | 50.5 | 10.33 | 106.71 | 1102.3 |
| 2005 | 38.6 | -1.57 | 2.46 | -3.9 |
| 2006 | 43.4 | 3.23 | 10.43 | 33.7 |
| 2007 | 28.7 | -11.47 | 131.56 | -1509.0 |
| 2008 | 32.0 | -8.17 | 66.75 | -545.3 |
| 2009 | 51.8 | 11.63 | 135.26 | 1573.0 |
| Sum | 401.7 | 0.00 | 1016.92 | 6480.3 |

$$
\bar x=\frac{\sum x}{n}=\frac{401.7}{10}=40.17 \text{ m}^3/\text{s}
$$

$$
s=\sqrt{\frac{\sum (x-\bar x)^2}{n-1}}=\sqrt{\frac{1016.92}{9}}=10.63 \text{ m}^3/\text{s}
$$

$$
C_s=\frac{n\sum (x-\bar x)^3}{(n-1)(n-2)\,s^3}=\frac{10\times6480.3}{9\times 8\times1201.1}=0.75
$$

**Answer: mean = 40.17 m³/s, standard deviation = 10.63 m³/s, skewness coefficient = 0.75** (positively skewed).

#### (ii) Probability of a 250 m³/s flood next year (Gumbel)
$$
K=\frac{250-40.17}{10.63}=19.74
$$

$$
y_T=\bar y_n+K\sigma_n = 0.4952 + 19.74\times0.9497=19.24
$$

$$
P(X\ge 250)=1-e^{-e^{-y_T}}=1-e^{-e^{-19.24}} \approx 4.4\times10^{-9}
$$

The return period is $T=1/P \approx 2.3\times10^{8}$ years.

**Answer: probability ≈ $4.4\times10^{-9}$ (about $4.4\times10^{-7}$ %), i.e. practically zero.** The value is far outside the range of the 10-year record (maximum 59.6 m³/s), so it is a very large extrapolation.

## Check yourself

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

1. The sample standard deviation uses a divisor of:
   A. n
   B. n + 1
   C. n^2
   D. n - 1
   Answer: D. s = sqrt(sum(x-mean)^2/(n-1)) = 10.63. (Key point: Standard deviation)

2. The sample skewness coefficient is:
   A. Cs = n sum(x-mean)^3 / ((n-1)(n-2) s^3)
   B. sum(x-mean)^3 / n
   C. sum(x-mean)^2 / s
   D. n sum(x-mean)^2/ s^3
   Answer: A. This gives about 0.75, positively skewed. (Key point: Skewness)

3. The probability of a 250 m3/s flood next year is:
   A. About 0.25
   B. Practically zero (about 4.4e-09)
   C. About 0.01
   D. About 0.5
   Answer: B. K = 19.7 gives y_T = 19.24 so P is about 4.4e-9; it is far outside the data. (Key point: Probability result)

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