1.1 Measures of central tendency mean, mode and median

A measure of central tendency is a single value that represents the centre or typical value of a data set. Instead of reading hundreds of figures, a manager can use one number to describe the whole set. For example, a county hospital can summarise a month of daily outpatient numbers with one typical daily figure. The three main measures are the mean (the arithmetic average), the median (the middle value when data are ordered) and the mode (the most frequent value). Each measure is calculated differently, answers a slightly different question and suits different kinds of data.

The arithmetic mean of ungrouped data

The arithmetic mean is the sum of all the values divided by the number of values. It uses every observation, so it is the most widely used measure for numerical data. A dairy SACCO, for instance, can use it to find the average daily milk delivery from its members.

Formula: mean = Σx ÷ n

where Σx is the sum of the values and n is the number of values.

Example: A dairy SACCO recorded milk deliveries (in litres) over seven days: 120, 135, 128, 142, 130, 125 and 140.

Σx = 120 + 135 + 128 + 142 + 130 + 125 + 140

Σx = 920

mean = 920 ÷ 7

mean = 131.43 litres

Properties of the mean:

• Uses all values: every observation contributes to the result, so a change in any one value changes the mean.
• Sum of deviations is zero: the deviations of all values from the mean add up to zero. In the dairy example, the values above 131.43 litres balance exactly those below it.
• Sensitive to extreme values: one very large or very small value can pull the mean away from where most of the data lie.
• Effect of adding a constant: if the same amount is added to every value, the mean increases by that amount.
• Effect of multiplying by a constant: if every value is multiplied by the same number, the mean is multiplied by that number. This is useful when converting units, for example from litres to millilitres.

Example: Five staff at a small hotel earn Ksh 25,000, 28,000, 30,000, 32,000 and 35,000 per month.

mean = (25,000 + 28,000 + 30,000 + 32,000 + 35,000) ÷ 5

mean = 150,000 ÷ 5

mean = Ksh 30,000

If the general manager, who earns Ksh 150,000, is added to the list:

mean = (150,000 + 150,000) ÷ 6

mean = 300,000 ÷ 6

mean = Ksh 50,000

The mean rises sharply, yet five of the six employees earn well below it. This shows how an extreme value can distort the mean.

The mean from a frequency table

When the same value occurs many times, the data are arranged in a frequency table. Each value x is multiplied by its frequency f, the products are added, and the total is divided by the total frequency. This saves time and avoids listing every observation.

Formula: mean = Σfx ÷ Σf

Example: A county office surveyed 50 households and recorded the number of children per household.

mean = Σfx ÷ Σf

mean = 98 ÷ 50

mean = 1.96 children per household

The answer need not be a whole number even though the data are counts. It describes the average, not an actual household.

The weighted mean

A weighted mean is used when some values matter more than others. Each value x is multiplied by a weight w that shows its importance, and the total is divided by the sum of the weights. It is used for average prices, course marks and index numbers.

Formula: weighted mean = Σwx ÷ Σw

Example: A building contractor bought 1,000 bags of cement in three lots: 200 bags at Ksh 700, 300 bags at Ksh 720 and 500 bags at Ksh 740.

Σwx = (200 × 700) + (300 × 720) + (500 × 740)

Σwx = 140,000 + 216,000 + 370,000

Σwx = 726,000

Σw = 200 + 300 + 500 = 1,000

weighted mean = 726,000 ÷ 1,000

weighted mean = Ksh 726 per bag

A simple average of the three prices would give Ksh 720, which wrongly treats a lot of 200 bags as equal to a lot of 500 bags.

The median

The median is the middle value of a data set arranged in order of size. Half the observations lie below it and half above it. Because it depends on position and not on the size of extreme values, it is a reliable measure when the data contain unusually large or small figures.

Median of ungrouped data:

1. Arrange the values in ascending order, so that positions can be counted from the smallest value.
2. Count the number of values, n.
3. Decide whether n is odd or even, because this determines how the middle is found.
4. If n is odd, the median is the value in position (n + 1) ÷ 2.
5. If n is even, the median is the mean of the values in positions n ÷ 2 and (n ÷ 2) + 1.
6. State the median with its units.

Example (odd n): The seven milk deliveries in order are 120, 125, 128, 130, 135, 140, 142.

position = (7 + 1) ÷ 2 = 4

median = 130 litres

Example (even n): The six hotel salaries in order are 25,000, 28,000, 30,000, 32,000, 35,000 and 150,000.

positions = 6 ÷ 2 = 3 and 4

median = (30,000 + 32,000) ÷ 2

median = Ksh 31,000

The median of Ksh 31,000 describes the typical employee far better than the mean of Ksh 50,000 found earlier.

Median from a frequency table:

Add the frequencies to form a cumulative frequency column, then find the position of the middle value.

Example: In the county office survey of 50 households, the cumulative frequencies are 5, 17, 35, 45 and 50 for 0, 1, 2, 3 and 4 children.

The middle lies between positions 25 and 26.

Both positions fall in the group with cumulative frequency 35, which is the group for 2 children.

median = 2 children

The mode

The mode is the value that occurs most often. It is the only measure of central tendency that can be used for non-numerical (categorical) data, for example the most common type of loan product at a bank. A data set may have one mode (unimodal), two modes (bimodal), more than two modes (multimodal) or no mode when every value occurs equally often.

Example (one mode): A shoe shop sold pairs in these sizes: 6, 7, 7, 8, 8, 8, 9, 9, 10. Size 8 occurs three times, more than any other, so the mode is size 8. This tells the owner which size to stock most.

Example (two modes): The values 2, 3, 3, 5, 7, 7, 9 have modes 3 and 7, so the data are bimodal.

Example (frequency table): In the county office survey, the highest frequency is 18, which belongs to 2 children, so the mode is 2 children.

Points to remember about the mode:

• Not affected by extremes: changing the largest or smallest value does not change the mode.
• May not be unique: a data set can have several modes, which limits its usefulness as a single summary.
• Works for categories: it can describe colours, brands, crop types or product types where no average exists.
• May not exist: if no value is repeated, the data have no mode.

Mean, median and mode of grouped data

When data are grouped into classes, the individual values are lost, so the measures are estimated. The class midpoint stands in for every value in the class.

Example data: A bank recorded the waiting time (in minutes) of 60 customers.

Mean of grouped data:

1. Find the midpoint of each class by adding the lower and upper limits and dividing by 2.
2. Multiply each midpoint by its class frequency to get fx.
3. Add the fx column to get Σfx.
4. Add the frequencies to get Σf.
5. Divide Σfx by Σf.
6. State the answer with its units, remembering that it is an estimate.

mean = Σfx ÷ Σf

mean = 760 ÷ 60

mean = 12.67 minutes

Median of grouped data:

1. Build the cumulative frequency column.
2. Calculate n ÷ 2 to locate the middle observation.
3. Find the first class whose cumulative frequency is at least n ÷ 2. This is the median class.
4. Note the lower boundary L of the median class, the cumulative frequency F before it, the class frequency f and the class width h.
5. Substitute into the formula: median = L + ((n ÷ 2 − F) ÷ f) × h.
6. Calculate and state the answer with its units.

n ÷ 2 = 60 ÷ 2 = 30

The median class is 10 to under 15, so L = 10, F = 20, f = 20 and h = 5.

median = 10 + ((30 − 20) ÷ 20) × 5

median = 10 + 2.5

median = 12.5 minutes

Mode of grouped data:

1. Identify the modal class, which is the class with the highest frequency.
2. Note its lower boundary L and class width h.
3. Find f1, the frequency of the modal class.
4. Find f0, the frequency of the class before it, and f2, the frequency of the class after it.
5. Substitute into the formula: mode = L + ((f1 − f0) ÷ ((f1 − f0) + (f1 − f2))) × h.
6. Calculate and state the answer with its units.

The modal class is 10 to under 15, so L = 10, h = 5, f1 = 20, f0 = 14 and f2 = 12.

mode = 10 + ((20 − 14) ÷ ((20 − 14) + (20 − 12))) × 5

mode = 10 + (6 ÷ 14) × 5

mode = 10 + 2.14

mode = 12.14 minutes

The assumed mean (coding) method

When midpoints or frequencies are large, calculation is easier if deviations from an assumed mean A are used. Choose A as a convenient midpoint near the centre of the data and let d = x − A for each class.

Formula: mean = A + (Σfd ÷ Σf)

Example: Using the bank waiting-time data with A = 12.5:

mean = 12.5 + (10 ÷ 60)

mean = 12.5 + 0.17

mean = 12.67 minutes

This agrees with the direct method, because the choice of A does not change the result.

Choosing the appropriate measure

The best measure depends on the type of data, the shape of the distribution and the purpose of the analysis.

The relationship between the three measures also shows the shape of the distribution:

• Symmetrical distribution: the mean, median and mode are equal or very close.
• Positively skewed distribution: a long tail of high values pulls the mean above the median, and the median above the mode. Household incomes in a county often behave this way.
• Negatively skewed distribution: a long tail of low values pulls the mean below the median, and the median below the mode. Marks in an easy examination may behave this way.

Example: A farm records the weight of tomato crates at harvest. If a few crates are much heavier than the rest, the mean weight will be above the median, and the median is the better description of a typical crate.