Measures of Central Tendency

Throughout your journey in statistical analysis, you will encounter a variety of measures that are utilized to make inferences from a data set. The most simple of these are noted as the measures of central tendency, which inherently describe the center or most likely value of a data set.

These are:

1. Arithmetic Mean

2. Median

3. Mode

1. Arithmetic Mean

The arithmetic mean, or mean, is the average of the total sum of all the data values in a data set. It is important to note that this measure essentially incorporates every single value of a data set, excluding none. The sample arithmetic mean is denoted by the symbol $\bar{x}$, whereas the population arithmetic mean is denoted by the symbol $\bar{X}$.

The arithmetic mean can be calculated as such:

Ungrouped Data

$$\text{Mean, } \bar{x} = \frac{\sum x}{n}$$

In which,

$$\begin{align*} n &= \text{total number of data values} \\ x &= \text{a single data value} \\ \sum x &= \text{sum of all data values} \\ \bar{x} &= \text{mean of the data set} \\ \end{align*}$$

For example, given the data values: $3, 4, 10, 12, 15, 18$.

In this case, the total number of data values, $n = 6$. As there are six countable values amongst the raw data values.

As well as,

The total sum of all the data values, ${\sum x} = 62$.

Hence,

$$\begin{align*} \text{Mean, } \bar{x} &= \frac{62}{6} \\ \text{Mean, } \bar{x} &= 10.3 \text{ (to 3 s.f.)}\\ \end{align*}$$

Grouped Data

$$\text{Mean, } \bar{x} = \sum \frac{fx}{f}$$

In which,

$$\begin{align*} \sum{f} &= \text{the total frequency of the data values} \\ x &= \text{a single data value} \\ \sum fx &= \text{sum of all data values multiplied by frequency} \\ \bar{x} &= \text{mean of the data set} \\ \end{align*}$$

For example, given:

[ \begin{array}{|c|c|} \hline \text{Marks } & \text{Frequency, } f \\ \hline 1-10 & 3 \\ \hline 20-30 & 5 \\ \hline 30-40 & 2 \\ \hline 40-50 & 4 \\ \hline \end{array} ]

Whereby, we must then calculate the following: ${\sum fx}$ and ${\sum f}$.

This can be done by calculating the midpoint, $x$ which is found by:

$$\text{Midpoint, } x = UCB-LCB$$

In which, $$\begin{align*} \text{UCB} &= \text{the upper class boundary} \\ \text{LCB} &= \text{the lower class boundary}\\ \end{align*}$$

Furthermore, it is optimal for the resulting calculations to be added to the table:

[ \begin{array}{|c|c|c|c|} \hline \text{Marks} & \text{Frequency, }f & \text{Midpoint, }x & fx \\ \hline 1-10 & 3 & 5.5 & 16.5 \\ \hline 20-30 & 5 & 25 & 125 \\ \hline 30-40 & 2 & 35 & 70 \\ \hline 40-50 & 4 & 45 & 180 \\ \hline & \sum{f}=14 & & \sum{fx}=391.5 \\ \hline \end{array} ]

From the table, we can see that: ${\sum fx} = 391.5$ and ${\sum f} = 14$.

Hence,

$$\begin{align*} \text{Mean, } \bar{x} &= \frac{391.5}{14} \\ \text{Mean, } \bar{x} &= 27.96 \\ \text{Mean, } \bar{x} &= 28.0 \text{ (to 3 s.f)} \\ \end{align*}$$

2. Median

The median of a data set is the middle of center value. It is not $50%$ of the data set but rather it is the $50^{\text{th}}$ percentile of the data set. This means that the median is the value at which $50%$ of the data lies below it. In order to calculate the median of a data set, the data must be arrayed.

Ungrouped Data

$$\text{Median } (Q_{\text{2}}) = \frac{n+1}{2}$$

In which,

$$\begin{align*} n &= \text{total number of data values} \\ \end{align*}$$

For example, given the data set: $3, 7, 4, 10, 15, 12, 18$. We are to find the Median $(Q_{\text{2}})$.

Firstly, we must array the data set as such: $3, 4, 7, 10, 12, 15, 18.$

In this case, the total number of data values, $n = 6$. As there are six countable values amongst the raw data values.

Secondly, we calculate the position of the median:

$$\begin{align*} \text{Median, } (Q_{\text{2}}) &= \frac{n+1}{2} \\ \text{Median, } (Q_{\text{2}}) &= \frac{6+1}{2} \\ \text{Median, } (Q_{\text{2}}) &= \frac{7}{2} \\ \text{Median, } (Q_{\text{2}}) &= 3.5^{\text{th}} \text { position} \end{align*}$$

Since the $\text{Median, } (Q_{\text{2}}) \notin \mathbb{Z}^+$, we find the values at the $3^{\text{rd}}$ and $4^{\text{th}}$ positions and take the average of it. If the $\text{Median, } (Q_{\text{2}}) \in \mathbb{Z}^+$, the value at that position would be taken.

Finally, from our data set,

$$\begin{align*} \text{Median, } (Q_{\text{2}}) &= \frac{7 + 10}{2} \\ \text{Median, } (Q_{\text{2}}) &= \frac{17}{2} \\ \text{Median, } (Q_{\text{2}}) &= 8.5 \\ \end{align*}$$

Grouped Data

Utilizing Linear Interpolation,

$$\text{Median } (Q_{\text{2}}) = l + \left( \frac{\frac{n}{2} - F}{f} \right) \times h$$

In which,

$$\begin{align*} n &= \text{total number of data values} \\ l &= \text{lower limit of the median class} \\ F &= \text{cumulative frequency of classes preceding the median class} \\ f &= \text{frequency of the median class} \\ h &= \text{class width} \end{align*}$$

Utilizing graph estimation for continuous data,

$$\text{Median } (Q_{\text{2}}) = \frac{n}{2}$$

For example, given the table:

[ \begin{array}{|c|c|} \hline \text{Mass, g} & \text{Frequency, } f \\ \hline 0-10 & 2 \\ \hline 10-20 & 5 \\ \hline 20-30 & 8 \\ \hline 30-40 & 4 \\ \hline 40-50 & 1 \\ \hline \end{array} ] We must identify the Median class utilizing the previous method:

$$\begin{align*} \text{Median, } (Q_{\text{2}}) &= \frac{n+1}{2} \\ \text{Median, } (Q_{\text{2}}) &= \frac{20+1}{2} \\ \text{Median, } (Q_{\text{2}}) &= \frac{21}{2} \\ \text{Median, } (Q_{\text{2}}) &= 10.5^{\text{th}} \text { position} \end{align*}$$

Hence, the median class is: $20-30$. This was found by cumulatively adding up the frequencies until the $10.5^{\text{th}}$ location is determined to fall within a respective class.

Now, $$\begin{align*} n &= \text{20} \\ l &= \text{20} \\ F &= \text{7} \\ f &= \text{8} \\ h &= \text{10} \end{align*}$$

Finally, we can apply these values to the formula given above:

$$\begin{align*} \text{Median } (Q_{\text{2}}) &= 20 + \left( \frac{\frac{20}{2} - 7}{8} \right) \times 10 \\ \text{Median } (Q_{\text{2}}) &= 23.75 \\ \text{Median } (Q_{\text{2}}) &= 23.8 \text{ (to 3 s.f.)} \\ \end{align*}$$

3. Mode

The mode of a data set is the most frequent or repeated value. In any given data set, there can be multiple modes, no mode or a singular mode. Statistically, the mode simply gives us an idea of the data value that is most picked. For example, the most popular colour, car, brand, chips etc.

Unrouped Data

As aforementioned, the mode is simply the most frequent value.

Therefore, for example, for a given data set: $3, 3, 4, 3, 5, 6, 3, 4, 3$.

The mode can be identified to be $3$. This is because it was repeated $5$ times in comparison to the other values in the data set.

Grouped Data

$$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$

In which,

$$\begin{align*} f_1 &= \text{frequency of the modal class} \\ f_0 &= \text{frequency of the class preceding the modal class} \\ f_2 &= \text{frequency of the class after the modal class} \\ l &= \text{lower limit of the modal class} \\ h &= \text{class width} \end{align*}$$

For example, given the table:

[ \begin{array}{|c|c|} \hline \text{Age, years} & \text{Frequency, } f \\ \hline 0-10 & 4 \\ \hline 10-20 & 12 \\ \hline 20-30 & 3 \\ \hline 30-40 & 6 \\ \hline \end{array} ] Firstly, the modal class can be easily identified as it is the class with the highest frequeuncy. Therefore, the modal class can be denoted as: $10-20$.

Now, from the table:

$$\begin{align*} f_1 &= \text{12} \\ f_0 &= \text{4} \\ f_2 &= \text{3} \\ l &= \text{10} \\ h &= \text{10} \end{align*}$$

Therefore, applying the formula stated above:

$$\begin{align*} \text{Mode} &= 10 + \left( \frac{12 - 4}{2(12) - 4 - 3} \right) \times 10 \\ \text{Mode} &= 14.70588... \\ \text{Mode} &= 14.7 \text{ (to 3 s.f.)} \\ \end{align*}$$

Exercise

Try:

1. Calculate the mean and median of the data set: $3, 10, 22, 44, 31, 25, 10, 12, 13, 41, 24, 26, 32, 40, 150, 198, 210$.

2. The heights of students of Remuria Krai Secondary School were recorded in the table designated below:

[ \begin{array}{|c|c|} \hline \text{Height, cm} & \text{Frequency, } f \\ \hline 0-5 & 20 \\ \hline 5-10 & 15 \\ \hline 10-15 & 18 \\ \hline 15-20 & 32 \\ \hline 20-25 & 17 \\ \hline \end{array} ]

Calculate: (i) Mean, (ii) Median, (iii) Mode

3. The number of days that 100 employees were early for work in the last financial period is displayed below:

[ \begin{array}{|c|c|} \hline \text{Number of Days Early} & \text{Frequency, } f \\ \hline 0-15 & 8 \\ \hline 16-20 & 16 \\ \hline 21-30 & 24 \\ \hline 41-50 & 26 \\ \hline 61-70 & 26 \\ \hline \end{array} ] Calculate the (i) mean, (ii) median and (iii) modal number of days that the workers were early. (iv) Describe an advantage of the mean and median.