Worksheet Grade 11 Math
Morwenna Creek at Westhorpe: Recurrence Intervals and Flood Odds
A grade 11 worksheet that ranks ten annual peak flows, computes recurrence intervals and exceedance probabilities, and builds an exponential ten-year risk function.
Objectives
- Students will rank annual peak flows and compute recurrence intervals from a ranking formula.
- Students will convert a recurrence interval into an annual exceedance probability.
- Students will build and evaluate an exponential function for risk across ten years.
Vocabulary
Key words: annual peak flow, recurrence interval, exceedance probability, stream gage, design flow
Teacher overview
The phrase "hundred-year flood" names an annual chance that a flow is reached or exceeded, although it is often read instead as a statement about how far apart such floods fall. This worksheet builds the idea from the ground up, using ten years of fictional annual peak flows from a gage on the fictional Morwenna Creek at Westhorpe. Students rank the peaks, apply the standard ranking formula for a recurrence interval, convert each interval into an annual chance, and then compound that chance across ten years. The misconception addressed here is that a flood with a small annual chance is also unlikely over a decade, which the final items disprove with a single exponential calculation.
Student Worksheet
A stream gage on the fictional Morwenna Creek at Westhorpe recorded the highest flow of each year for ten years. The flows are listed below in cubic feet per second.
| Year | Annual peak flow (cfs) |
|---|---|
| 1 | 1,150 |
| 2 | 2,400 |
| 3 | 890 |
| 4 | 4,200 |
| 5 | 1,760 |
| 6 | 640 |
| 7 | 3,050 |
| 8 | 1,420 |
| 9 | 2,050 |
| 10 | 980 |
Rank the flows from largest to smallest. The largest gets rank m = 1. Use the ranking formula T = (n + 1) divided by m, where n = 10 is the number of years and T is the recurrence interval in years.
- Give the rank m of the 2,050 cfs peak.
- Find the recurrence interval of the largest peak, to one decimal place.
- Find the recurrence interval of the 2,400 cfs peak, to one decimal place.
- The annual exceedance probability is p = 1 divided by T. Find p for the largest peak, to three decimal places.
- Find p for the 1,760 cfs peak, to three decimal places.
- In one sentence, state what a hundred-year flood means.
- Use your value of p from item 4. Find the chance that the largest peak is not exceeded in a single year, to three decimal places.
- Write a function P(t) for the chance that this peak is not exceeded in t years in a row.
- Evaluate P(10), to three decimal places.
- Find the chance that this peak is exceeded at least once in ten years, to three decimal places.
- Your answer to item 10 is much larger than the answer to item 4. Explain in one or two sentences why that happens.
- A culvert at Westhorpe is built for 3,000 cfs. Find how many of the ten recorded years had a peak above that flow.
- In one or two sentences, explain why a ten-year record is a weak basis for estimating a hundred-year flood.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 4.
Sorted from largest to smallest the peaks are 4,200, 3,050, 2,400, 2,050, 1,760, 1,420, 1,150, 980, 890, and 640 cubic feet per second, so 2,050 sits fourth.Answer: 11.0 years.
The largest peak has m = 1, and T = (10 + 1) divided by 1 = 11.0 years.Answer: 3.7 years.
The 2,400 peak has m = 3, and T = 11 divided by 3 = 3.667, which rounds to 3.7 years.Answer: 0.091.
Substituting gives p = 1 divided by 11.0 = 0.0909, which rounds to 0.091.Answer: 0.455.
The 1,760 peak has m = 5, so T = 11 divided by 5 = 2.2 years and p = 1 divided by 2.2 = 0.4545, which rounds to 0.455.Answer: A flow with a one percent chance of being reached or passed in any single year.
The label describes an annual chance, not a spacing in time, so two such floods can fall in consecutive years without contradicting the label.Answer: 0.909.
Subtracting gives 1 - 0.091 = 0.909.Answer: P(t) = 0.909 raised to the power t.
Each year carries the same independent chance of 0.909 that the peak is not exceeded, so t years in a row multiply that factor t times.Answer: 0.385.
Raising 0.909 to the tenth power gives 0.3852, which rounds to 0.385.Answer: 0.615.
Subtracting gives 1 - 0.385 = 0.615.Answer: Ten separate chances accumulate.
The 0.091 figure covers one year only, while ten years give the flood ten separate opportunities, and the chance that none of them produces an exceedance shrinks with every year added.Answer: 2 years.
Only year 4 at 4,200 cfs and year 7 at 3,050 cfs went above 3,000 cubic feet per second.Answer: The record is far shorter than the interval being estimated.
Ten years of data can only place a rank of 1 at an interval of 11 years, so any estimate of a hundred-year flow requires extending the pattern far past every flow ever measured at the gage.
Differentiation
Support
Give these students the ten fictional Morwenna Creek peaks already sorted with the rank column filled in, so their work begins at the formula. Provide items 7 through 10 as a guided sequence on one page, with the subtraction in item 7 and the exponent in item 9 shown as worked models using a different probability.
Extension
Ask these students to repeat items 4 and 9 for the fictional 3,050 cfs peak, then to find the number of years over which that peak's chance of at least one exceedance first passes one half. Have them explain why the answer is close to, but not equal to, the recurrence interval itself.
ELL supports
Pre-teach "peak flow," "rank," and "exceed" with a simple bar chart of the ten fictional years, tallest bar labeled first. Provide sentence frames for items 6, 11, and 13, such as "A hundred-year flood means the chance in one year is ___," and "Ten years give the flood ___ chances instead of one."
Extensions
Students can find their own county's flood maps and describe, in words alone, which zones use the one percent annual chance and how a homeowner would read that label. A second task is to explain why hydrologists prefer gages with fifty years or more of record, using the fictional Morwenna results to make the point concrete.