Calculating averages for discrete and grouped data

The median is the middle value after the data has been put in order. With an odd number of observations, there is one middle value, at position $(n+1)/2$. With an even number, find the values at positions $n/2$ and $n/2+1$, then take their arithmetic mean.

The arithmetic mean is the total of the values divided by their number:

$$\bar{x}=\frac{\sum x}{n}$$

Here, $\bar{x}$ means the arithmetic mean, $n$ is the number of observations and $\sum x$ means ‘the sum of the values’. The mean need not be a value that actually occurs in the data.

For example, 15 students’ times for solving a problem, in seconds, are:

$12,10,15,14,17,11,12,13,9,21,14,20,19,16,23$.

Their total is $226$ seconds, so the mean is $226\div15=15.1$ seconds to three significant figures. In order, the times are:

$9,10,11,12,12,13,14,14,15,16,17,19,20,21,23$.

The middle is the eighth value, giving a median of $14$ seconds. Both $12$ and $14$ occur twice, more often than any other value, so these are the two modes. Each average answers a different question about the same observations.

Calculating from a discrete frequency table

Discrete data takes separate values, such as a number of pets or a shoe size. A frequency table records each value alongside the number of times it occurs. Unlike grouping values into intervals, this does not lose the individual values: a frequency of six for shoe size nine represents six separate observations of nine.

For the mean, multiply each value $x$ by its frequency $f$. This gives the contribution of that row to the total. Then add the products and divide by the total frequency:

$$\bar{x}=\frac{\sum fx}{\sum f}$$

Consider this shoe-size data:

The mean shoe size is $278.5\div31=8.98$ to three significant figures.

The modal shoe size is $10$, because it has the greatest frequency, $11$.

There are $31$ observations, so the median is the sixteenth value. Cumulative frequency is a running total: the table shows that the first $11$ observations have shoe sizes up to eight, and the first $17$ have sizes up to nine. The sixteenth observation therefore has shoe size nine, making the median $9$.

Weighting an average

A weighted mean allows some values to contribute more than others. Its calculation is:

$$\bar{x}_w=\frac{\sum wx}{\sum w}$$

Here, $w$ is the weight attached to value $x$. Multiply each value by its weight, add these products, then divide by the total weight. A frequency-table mean is a weighted mean in which the frequencies are the weights.

This also explains how to combine group means. A group’s total is its mean multiplied by its number of members. Add the group totals and divide by the combined number of members:

$$\text{combined mean}=\frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}$$

The group sizes $n_1$ and $n_2$ act as weights. Simply averaging two group means would give the groups equal influence, even when one group contains many more observations.

Geometric mean: averaging multiplicative change

A geometric mean combines positive values by multiplication rather than addition. Multiply all the values together, then take the root corresponding to how many values there are:

$$G=\sqrt[n]{x_1x_2\cdots x_n}=(x_1x_2\cdots x_n)^{1/n}$$

Here, $n$ is the number of values. For two values, take a square root; for three, take a cube root. For example, the geometric mean of $2$, $4$ and $8$ is:

$$G=\sqrt[3]{2\times4\times8}=\sqrt[3]{64}=4$$

The result is the single value which, multiplied by itself three times, gives the same product as the original three values.

This makes the geometric mean useful for growth multipliers, because successive changes combine by multiplication. If a quantity doubles in one year and halves in the next, its multipliers are $2$ and $0.5$. Their product is $1$, so the quantity returns to its starting value. The geometric mean multiplier is $\sqrt{2\times0.5}=1$, representing no average annual change. The arithmetic mean multiplier, , would incorrectly suggest sustained growth.

When changes are given as percentages, first convert them to multipliers. An increase of $12\%$ has multiplier $1.12$; a decrease of $5\%$ has multiplier $0.95$. Find the geometric mean of the multipliers. If the resulting mean multiplier is $G$, the equivalent average percentage change per period is $(G-1)\times100\%$. A negative result represents a decrease.

Why grouping changes the calculation

A grouped frequency table places observations into class intervals. This is useful for large data sets and for continuous measurements, such as times and masses, which can take any value within a range.

For example, $10\leq t<20$ includes times from $10$ minutes up to, but not including, $20$ minutes. A time of exactly $20$ minutes belongs in the next class. Carefully written intervals avoid gaps and overlaps.

Grouping loses information. Knowing how many times lie between $10$ and $20$ minutes does not reveal their exact values. Consequently, the grouped mean and median are usually estimates, not exact answers.

For a grouped table, report the modal class, the interval with the greatest frequency, rather than claiming to know one exact modal value.

Estimating the grouped arithmetic mean

Represent every observation in a class by its midpoint:

$$m=\frac{\text{lower endpoint}+\text{upper endpoint}}{2}$$

Multiply each midpoint by its class frequency, add the products and divide by the total frequency:

$$\text{estimated mean}=\frac{\sum fm}{\sum f}$$

The midpoint represents the centre of its interval. If observations are evenly spread within a class, values above and below the midpoint balance one another. If they cluster near one end instead, the midpoint is a less accurate representative.

The method works for equal or unequal class widths. Calculate each midpoint from that class’s own endpoints. For example, the midpoint of $10\leq t<20$ is $15$, whereas the midpoint of $20\leq t<40$ is $30$. The wider class does not change the rule: its frequency still tells you how many observations its midpoint represents.

Estimating the grouped median by linear interpolation

The grouped median is the point halfway through the distribution, at cumulative frequency $n/2$. First calculate cumulative frequencies and identify the class containing this halfway point.

You then need to estimate where the median lies inside that class. Linear interpolation assumes the observations are evenly spread across it.

Let:

• $L$ be the lower boundary of the median class;
• $c$ be the cumulative frequency before that class;
• $f$ be the frequency of that class;
• $h$ be its width;
• $n$ be the total frequency.

Then:

$$\text{estimated median}=L+\frac{n/2-c}{f}\times h$$

The expression $n/2-c$ counts how far into the median class you must go to reach halfway through all observations. Dividing by $f$ gives the fraction of that class’s observations you have passed. Under the even-spread assumption, you move the same fraction of the distance across its interval, then add this distance to $L$.

For instance, if the halfway point is one quarter of the way through the observations in the median class, estimate the median one quarter of the way along that interval. This is why the width belongs in the calculation: the same fraction corresponds to a larger distance in a wider class.

Linear interpolation is required here for both equal and unequal class widths. As with the grouped mean, the answer is an estimate because the individual values within the class are unknown.

Class boundaries for rounded measurements

Some intervals appear to have gaps because measurements have been rounded. For measurements recorded to the nearest kilogram, a stated class of $10$–$19$ kg represents true masses from $9.5$ kg up to, but not including, $19.5$ kg. These are its class boundaries, and its width is $10$ kg.

Use these boundaries when interpolating the median. They describe the complete interval occupied by the measurements, rather than just the smallest and largest recorded integers.

Extending both ends equally leaves the midpoint unchanged: the midpoint of this class is $14.5$ kg using either its stated endpoints or its true boundaries. For other recording precisions, extend each end by half the rounding unit.

Mean seasonal variation: averaging a repeating effect

A time series records a quantity over time, usually at equal intervals. Its trend describes the general direction of change, such as a gradual rise or fall. Seasonal variation is a pattern that repeats at corresponding points in a regular cycle. The cycle might be a year, with quarterly observations, or a week, with daily observations: ‘seasonal’ does not only mean spring, summer, autumn and winter.

To measure the seasonal variation at a particular point, compare the actual observation with the trend value at the same time:

$$\text{seasonal variation}=\text{actual value}-\text{trend value}$$

A positive variation means the actual value is above the trend; a negative variation means it is below. The variation has the same units as the original quantity.

The mean seasonal variation, also called estimated mean seasonal variation or average seasonal effect, is the arithmetic mean of the variations for the same season in different cycles:

$$\text{mean seasonal variation}=\frac{\text{sum of variations for that season}}{\text{number of those variations}}$$

For a constructed illustration, suppose a shop’s fourth-quarter sales are compared with their trend values over three years. All sales and trend values below are numbers of items.

The mean fourth-quarter seasonal variation is $(20+25+15)\div3=20$ items. This means fourth-quarter sales are, on average, $20$ items above the underlying trend. Sales have risen across the three years, but subtracting each year’s trend value separates that general rise from the repeating fourth-quarter effect.

Calculate a separate mean for each season. Averaging fourth-quarter variations together estimates the fourth-quarter effect; mixing them with other quarters would answer a different question.