The median is the middle value of a set of data when the data are arranged in ascending or descending order. It divides the dataset into two equal halves.

a) Median for ungrouped data

Case 1: Odd number of values

If the number of data values, $N$, is odd, the median is the value at the position:

$\text{Median} = \left( \frac{N + 1}{2} \right)^\text{th} \text{ value}$

Case 2: Even number of values

If $N$ is even, the median is the average of the two middle values at positions:

$\text{Median} = \frac{\left(\frac{N}{2}\right)^\text{th} \text{ value} + \left(\frac{N}{2} + 1\right)^\text{th} \text{ value}}{2}$

Example

Find the median of the values: 27, 25, 19, 23, 31, 42, 51, 17, 22

Solution:

Total number of values: $N = 9$ (odd)

Arrange the data in ascending order: 17, 19, 22, 23, 25, 27, 31, 42, 51

Median = 5th value = 25

Example

Find the median of the values: 2, 5, 5, 7, 12, 13, 15, 15, 16, 19

Solution:

$N = 10$ (even)

5th value = 12, 6th value = 13

$\text{Median} = \frac{12 + 13}{2} = 12.5$

Example

Given the following frequency table, find the median:

Solution:

• Total frequency, $N = 1 + 3 + 2 + 5 + 8 = 19$
• Median position: $\frac{19 + 1}{2} = 10^\text{th}$ value
• Cumulative frequencies: 1, 4, 6, 11, 19

10th value lies in the 4th score (which is 4), next is 11th value (score = 5)

$\text{Median} = \frac{4 + 5}{2} = 4.5$

b) Median for grouped data

The median of grouped data is found using the formula:

$\text{Median} = L + \left( \frac{\frac{N}{2} - f_b}{f_w} \right) \times c$

Where:

• $L$ = lower boundary of the median class
• $N$ = total frequency
• $f_b$ = cumulative frequency before the median class
• $f_w$ = frequency of the median class
• $c$ = class width

Example: Table of children's ages

Solution:

• Total frequency: $N = 140$
• Median position: $\frac{140}{2} = 70^\text{th}$ value
• Cumulative frequencies:
  • 10–15: 30
  • 15–20: 72
• Median class: 15–20
• $L = 14.5$, $f_b = 30$, $f_w = 42$, $c = 5$

$\text{Median} = 14.5 + \left( \frac{70 - 30}{42} \right) \times 5 = 14.5 + 4.76 = 19.26$

Therefore, the median age is approximately 19.26 months.

The mode is the value that occurs most frequently in a data set. A distribution may have:

1. One mode (unimodal)
2. Two modes (bimodal)
3. Three or more modes (multimodal)

The mode is particularly useful when the most typical or common value is needed. It is also applicable for categorical (nominal) data such as religious preference, gender, or political affiliation. In business, the mode can help identify the most preferred commodity by customers in a specific location.

a) Mode of ungrouped data

Example

a) Find the mode of the values:

9, 8, 10, 10, 10, 12, 6, 11, 12, 10, 8, 8

The value with the highest frequency is 10, occurring 4 times.

∴ Mode = 10

b) Find the mode of the values:

110, 120, 70, 90, 90, 100, 80, 130, 40

The value with the highest frequency is 90.

∴ Mode = 90

b) Mode of grouped data

To find the mode for grouped data, we use the following formula:

$\text{Mode} = L + \left( \frac{d_1}{d_1 + d_2} \right) \times c$

Where:

• $L$ = lower boundary of the modal class
• $d_1 = f_m - f_{1}$: frequency of modal class minus frequency of class before
• $d_2 = f_m - f_{2}$: frequency of modal class minus frequency of class after
• $c$ = class width

Example

Find the mode from the following frequency distribution:

Modal class = 20 – 24, with frequency = 11

$L = 19.5$, $f_1 = 6$, $f_2 = 9$, $c = 5$

$\text{Mode} = 19.5 + \left( \frac{11 - 6}{(11 - 6) + (11 - 9)} \right) \times 5 = 19.5 + \left( \frac{5}{7} \right) \times 5 = 19.5 + 3.57 = \boxed{23.07}$

Example

Find the mode from the following data:

Modal class = 341 – 360, with frequency = 42

$L = 340.5$, $f_1 = 30$, $f_2 = 22$, $c = 20$

$\text{Mode} = 340.5 + \left( \frac{42 - 30}{(42 - 30) + (42 - 22)} \right) \times 20 = 340.5 + \left( \frac{12}{32} \right) \times 20 = 340.5 + 7.5 = \boxed{348.0}$

Estimation of mode using histogram

The mode of data can be estimated from the drawn histogram as follows:

1. Draw a straight line from top left corner of the tallest rectangle to the top left corner of the rectangle in the right of it.
2. Draw a straight line from top right corner of the tallest rectangle to the top right corner of the rectangle which is in the left hand side.
3. Observe where these lines intersect.
4. Mode is the point of intersection which can be read on the horizontal axis.

a) The quartiles

Quartiles are statistical values that divide a given data set into four equal parts or quarters after the data has been arranged in ascending order. These divisions help us understand the distribution and spread of the data by identifying key positional values. The three main quartiles are:

• First quartile ($Q_1$): Also called the lower quartile, it marks the 25th percentile of the data.
• Second quartile ($Q_2$): The median of the data, corresponding to the 50th percentile.
• Third quartile ($Q_3$): Also called the upper quartile, it marks the 75th percentile.

Each quartile divides the data into segments such that:

• 25% of the data lies below $Q_1$, and 75% lies above it.
• 50% of the data lies below and above $Q_2$ (the median).
• 75% of the data lies below $Q_3$, and 25% lies above it.

Quartiles are not only measures of position but also useful indicators of central tendency and variability within the data set. They are foundational for understanding the spread and are often used to construct box plots.

i. Quartile calculation for ungrouped data

Suppose you have $N$ observations arranged in ascending order. To find the position of the quartiles (which may not always be integers), use the following formulas:

$Q_1 = \text{value at position } \frac{N + 1}{4}$

$Q_2 = \text{value at position } \frac{2(N + 1)}{4} = \frac{N + 1}{2}$

$Q_3 = \text{value at position } \frac{3(N + 1)}{4}$

If the calculated position is not an integer, the quartile value is obtained by interpolation: that is, by taking the weighted average between the values at the closest ranks. For example, if $Q_1$ lies between the 4th and 5th values, then

$Q_1 = x_4 + (p)(x_5 - x_4)$

where $p$ is the fractional part of the position.

Detailed interpretation of quartiles

1. First quartile $Q_1$: The median of the lower half of the data (excluding the overall median if the number of observations is odd). It separates the lowest 25% of the data from the rest.
2. Second quartile $Q_2$: This is the overall median. For an odd number of data points, it is the middle value. For even number of points, it is the mean of the two central values.
3. Third quartile $Q_3$: The median of the upper half of the data (excluding the overall median if odd). It separates the lowest 75% from the highest 25%.

In practice, the quartiles can also be viewed as quantiles dividing the distribution into four parts, each containing approximately an equal number of data points.

Example. Given the data set:

3, 7, 8, 5, 12, 14, 21, 13, 18

Arrange the data in ascending order:

3, 5, 7, 8, 12, 13, 14, 18, 21

Find the median $Q_2$: Since $N = 9$ (odd), the median is the middle value (position $\frac{9+1}{2} = 5^{th}$ value):

$Q_2 = 12$

Identify the lower half and upper half for quartiles calculation:

• Lower half (values before median): 3, 5, 7, 8
• Upper half (values after median): 13, 14, 18, 21

Find the first quartile $Q_1$ (median of lower half): Since the lower half has 4 values (even),

$Q_1 = \frac{5 + 7}{2} = 6$

Find the third quartile $Q_3$ (median of upper half): Similarly,

$Q_3 = \frac{14 + 18}{2} = 16$

Result:

$Q_1 = 6, \quad Q_2 = 12, \quad Q_3 = 16$

ii. Interquartile range (I.Q.R)

The interquartile range measures the spread or variability of the middle 50% of the data. It is the difference between the third and first quartiles:

$\text{I.Q.R} = Q_3 - Q_1$

In the previous example,

$\text{I.Q.R} = 16 - 6 = 10$

The I.Q.R is a robust measure of dispersion, less sensitive to outliers than the full range.

iii. Semi-interquartile range (S.I.Q.R)

This is half of the interquartile range and provides an average deviation of the middle 50% from the median:

$\text{S.I.Q.R} = \frac{1}{2} \times (Q_3 - Q_1) = \frac{\text{I.Q.R}}{2}$

Using the example data:

$\text{S.I.Q.R} = \frac{10}{2} = 5$

Example: Quartiles and I.Q.R for another dataset

Data: 14, 23, 39, 4, 11, 19, 45, 31, 7, 30, 42, 17

Arrange data in ascending order:

4, 7, 11, 14, 17, 19, 23, 30, 31, 39, 42, 45

$N = 12$

• Median position: $\frac{12}{2} = 6^\text{th}$ value = 19 (or average of 6th and 7th: $\frac{19 + 23}{2} = 21$)
• Lower half: 4, 7, 11, 14, 17, 19
• Upper half: 23, 30, 31, 39, 42, 45
• $Q_1$ = median of lower half = $\frac{11 + 14}{2} = 12.5$
• $Q_3$ = median of upper half = $\frac{31 + 39}{2} = 35$

$\text{I.Q.R} = Q_3 - Q_1 = 35 - 12.5 = 22.5$

$\text{S.I.Q.R} = \frac{22.5}{2} = 11.25$

iv. Quartiles for grouped data

Quartiles for grouped data are determined using cumulative frequencies and the following formula:

$Q_i = L_i + \left( \frac{\frac{iN}{4} - f_b}{f_w} \right) \times c, \quad \text{for } i = 1, 2, 3$

Where:

• $Q_i$: The $i$th quartile ($Q_1$, $Q_2$, $Q_3$)
• $L_i$: Lower class boundary of the quartile class
• $N$: Total number of frequencies (observations)
• $f_b$: Cumulative frequency before the quartile class
• $f_w$: Frequency of the quartile class
• $c$: Width of the class interval

Example

Given the following frequency distribution:

Total frequency $N = 100$.

a. Calculate the lower quartile ($Q_1$)

First, compute the cumulative frequencies:

To find $Q_1$, calculate the position:

$\frac{1}{4} \times 100 = 25\text{th position}$

This lies in the class 11–16. So,

• $L_1 = 10.5$
• $f_b = 18$
• $f_w = 26$
• $c = 6$

$Q_1 = 10.5 + \left( \frac{25 - 18}{26} \right) \times 6 = 10.5 + \left( \frac{7}{26} \right) \times 6$

$Q_1 = 10.5 + 1.615 \approx 12.12$

b. Calculate the upper quartile ($Q_3$)

$\frac{3}{4} \times 100 = 75\text{th position}$

This lies in the class 23–28. So,

• $L_3 = 22.5$
• $f_b = 64$
• $f_w = 12$
• $c = 6$

$Q_3 = 22.5 + \left( \frac{75 - 64}{12} \right) \times 6 = 22.5 + \left( \frac{11}{12} \right) \times 6$

$Q_3 = 22.5 + 5.5 = 28$

v) Estimating quartiles using cumulative frequency curve (ogive)

Quartiles can also be estimated graphically from an ogive (cumulative frequency curve):

1. Plot cumulative frequency on the vertical axis and upper class boundaries on the horizontal axis.
2. Locate:
   • $\frac{1}{4}N$ for $Q_1$
   • $\frac{1}{2}N$ for $Q_2$ (Median)
   • $\frac{3}{4}N$ for $Q_3$
3. Draw horizontal lines from these cumulative frequency values to the ogive, then drop perpendiculars to the x-axis to estimate the quartiles.

Note: This method is approximate and useful for visual estimation.

b. Percentiles

Percentiles divide a set of data into 100 equal parts. While quartiles split data into four equal parts, percentiles provide a more detailed positional measure, particularly useful in education and health fields. They help identify the position of a value in a data set and give insights into how values are spread across the entire range.

i. Percentiles for ungrouped data

To compute the $k^{th}$ percentile $P_k$, the rank (or position) is determined using the formula:

$R_k = \frac{k}{100} \times N$

Where:

• $k$ is the percentile (1 ≤ k ≤ 99)
• $N$ is the total number of data values
• $R_k$ is the rank or position of the $k^{th}$ percentile

Example: Given $N = 120$, find the rank of the 55$^{th}$ percentile.

Using the formula:

$R_{55} = \frac{55}{100} \times 120 = 66$

Therefore, the 55$^{th}$ percentile is the 66$^{th}$ value in the ordered list.

ii. Percentiles for grouped data

To compute the $k^{th}$ percentile $P_k$ in grouped data, the formula used is:

$P_k = L_k + \left( \frac{\frac{k}{100}N - f_b}{f_w} \right) \times c$

Where:

• $L_k$: Lower class boundary of the class containing the $k^{th}$ percentile
• $f_b$: Cumulative frequency before the $k^{th}$ percentile class
• $f_w$: Frequency of the $k^{th}$ percentile class
• $c$: Class width
• $N$: Total frequency

Example: The frequency distribution of history scores is given below:

a. Find the 85$^{th}$ percentile

Total frequency $N = 80$.

Rank: $R_{85} = \frac{85}{100} \times 80 = 68$

The 85$^{th}$ percentile lies in the class 60 – 69 (cumulative frequency just reaches or exceeds 68).

Using:

• $L_k = 59.5$
• $f_b = 59$
• $f_w = 10$
• $c = 10$

$P_{85} = 59.5 + \left( \frac{68 - 59}{10} \right) \times 10 = 59.5 + 9 = 68.5$

∴ The 85$^{th}$ percentile is 68.5.

b. Find the 50$^{th}$ percentile (median)

Rank: $R_{50} = \frac{50}{100} \times 80 = 40$

The 50$^{th}$ percentile lies in the class 40 – 49.

Using:

• $L_k = 39.5$
• $f_b = 21$
• $f_w = 20$
• $c = 10$

$P_{50} = 39.5 + \left( \frac{40 - 21}{20} \right) \times 10 = 39.5 + 9.5 = 49$

∴ The 50$^{th}$ percentile (median) is 49.