The annual rainfall (in mm) was recorded for Cherrapunji, Meghalaya: | Year | Rainfall (in cm) | | --- | --- | | 2001 | 1.2 | | 2002 | 1.9 | | 2003 | 2 | | 2004 | 1.4 | | 2005 | 2.1 | | 2006 | 1.3 | | 2007 | 1.8 | | 2008 | 1.1 | | 2009 | 1.3 | Determine the trend of rainfall by 3-year moving averages

> [!TLDR] > Averaging each year with its immediate neighbour on either side (3-year moving average) smooths the rainfall series into the trend values 1.70 → 1.40 cm for 2002–2008. > [!FORMULA] > **3-year moving average** placed against the middle year: > $$\text{MA}_t=\frac{y_{t-1}+y_t+y_{t+1}}{3}$$ > where $y_t$ is the rainfall in year $t$. (The prompt's "mm" is a typo — the table header reads cm, which I use.) **Steps** 1. Form the 3-year moving totals (each year plus its two neighbours), then divide by 3. | Year | Rainfall $y$ (cm) | 3-yr moving total | 3-yr moving average (trend) | | --- | --- | --- | --- | | 2001 | 1.2 | — | — | | 2002 | 1.9 | $1.2+1.9+2.0=5.1$ | $5.1/3=1.70$ | | 2003 | 2.0 | $1.9+2.0+1.4=5.3$ | $5.3/3=1.77$ | | 2004 | 1.4 | $2.0+1.4+2.1=5.5$ | $5.5/3=1.83$ | | 2005 | 2.1 | $1.4+2.1+1.3=4.8$ | $4.8/3=1.60$ | | 2006 | 1.3 | $2.1+1.3+1.8=5.2$ | $5.2/3=1.73$ | | 2007 | 1.8 | $1.3+1.8+1.1=4.2$ | $4.2/3=1.40$ | | 2008 | 1.1 | $1.8+1.1+1.3=4.2$ | $4.2/3=1.40$ | | 2009 | 1.3 | — | — | 2. The first (2001) and last (2009) years have no complete 3-year window, so they carry no moving average. 3. The smoothed column is the trend: it removes year-to-year jumps, showing rainfall hovering near 1.6–1.8 cm and settling to 1.40 by 2007–2008. …

The annual rainfall (in mm) was recorded for Cherr – Class 12 Applied Mathematics

Practice Exercise · Q10

The annual rainfall (in mm) was recorded for Cherrapunji, Meghalaya:

Year	Rainfall (in cm)
2001	1.2
2002	1.9
2003	2
2004	1.4
2005	2.1
2006	1.3
2007	1.8
2008	1.1
2009	1.3

Determine the trend of rainfall by 3-year moving averages

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Concept understanding — Moving Average

Moving Average: Seeing the Signal Through the Noise

Imagine you're tracking the daily temperature in your city. One day it's 35°C, the next it's 28°C, then 32°C, then a sudden 40°C. The numbers jump around wildly. Is the weather getting hotter? Hard to tell from individual readings — each day's number is just a snapshot, full of random ups and downs.

What you really want is the trend: is the general temperature rising or falling? That's exactly what a moving average does. It smooths out the daily jitter so you can see the underlying pattern.

The Core Intuition

A moving average replaces each data point with the average of itself and its neighbours. Instead of asking "what was the value on Tuesday?", you ask "what was the typical value around Tuesday?" By averaging a small window of consecutive points, you cancel out the random noise and reveal the smoother, more meaningful shape of the data.

Note

The "moving" part means the window slides forward one step at a time. You compute an average for day 1–3, then day 2–4, then day 3–5, and so on — the window "moves" across the data.

The Precise Definition

Given a sequence of numbers $x_1, x_2, x_3, \dots, x_n$ , a simple moving average of window size $k$ (where $k$ is a positive integer) produces a new sequence $y_t$ defined as:

y_t = \frac{x_t + x_{t-1} + x_{t-2} + \dots + x_{t-k+1}}{k}

for $t = k, k+1, \dots, n$ .

Each $y_t$ is the arithmetic mean of the $k$ most recent observations ending at time $t$ . The first $k-1$ terms of the original sequence don't have enough predecessors to form a full window, so they are usually omitted from the smoothed output.

\text{SMA}_t = \frac{1}{k} \sum_{i=0}^{k-1} x_{t-i}

A Concrete Example

Take the temperature data: 35, 28, 32, 40, 36, 31 (in °C). Let's use a window of size $k=3$ .

For $t=3$ : $y_3 = \frac{35 + 28 + 32}{3} = \frac{95}{3} \approx 31.7$
For $t=4$ : $y_4 = \frac{28 + 32 + 40}{3} = \frac{100}{3} \approx 33.3$
For $t=5$ : $y_5 = \frac{32 + 40 + 36}{3} = \frac{108}{3} = 36.0$
For $t=6$ : $y_6 = \frac{40 + 36 + 31}{3} = \frac{107}{3} \approx 35.7$

The smoothed sequence is 31.7, 33.3, 36.0, 35.7. Notice how the wild 40°C spike is now part of a gentler rise and fall — the moving average has done its job.

Why It Matters

Moving averages are everywhere in exams and real life: …

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Averaging each year with its immediate neighbour on either side (3-year moving average) smooths the rainfall series into the trend values 1.70 → 1.40 cm for 2002–2008.

\text{MA}_t=\frac{y_{t-1}+y_t+y_{t+1}}{3}

Steps

Form the 3-year moving totals (each year plus its two neighbours), then divide by 3.

Year	Rainfall $y$ (cm)	3-yr moving total	3-yr moving average (trend)
2001	1.2	—	—
2002	1.9	$1.2+1.9+2.0=5.1$	$5.1/3=1.70$
2003	2.0	$1.9+2.0+1.4=5.3$	$5.3/3=1.77$
2004	1.4	$2.0+1.4+2.1=5.5$	$5.5/3=1.83$
2005	2.1	$1.4+2.1+1.3=4.8$	$4.8/3=1.60$
2006	1.3	$2.1+1.3+1.8=5.2$	$5.2/3=1.73$
2007	1.8	$1.3+1.8+1.1=4.2$	$4.2/3=1.40$
2008	1.1	$1.8+1.1+1.3=4.2$	$4.2/3=1.40$
2009	1.3	—	—

The first (2001) and last (2009) years have no complete 3-year window, so they carry no moving average.
The smoothed column is the trend: it removes year-to-year jumps, showing rainfall hovering near 1.6–1.8 cm and settling to 1.40 by 2007–2008. …

Start your 14-day free trial to unlock the full solution →

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