Median and Modal Class for Grouped Data

Modal Class for Grouped Data

When data are grouped into intervals, the table no longer contains each original observation. We can still identify the modal class, but the table may not identify one exact mode.

The modal class is the class interval with the highest frequency, so it contains more observations than any other interval.

Worked example:

Store A's shoe size data:

The highest frequency is $16$, so the modal class is $43\text{-}45$.

This tells us that the most frequently sold shoe sizes lie in that interval. Because the data are grouped, the table does not identify which individual size occurs most often.

Median for Grouped Data

The median splits ordered data into two equal halves. A grouped table lists intervals together with their frequencies. We therefore locate the median class and estimate the median's position within that class.

How to Find the Median for Grouped Data

1. Calculate the median position:

   Locate the middle of the data with $n/2$, where $n$ is the total frequency.

2. Create the cumulative frequency:

   Add a cumulative-frequency column ($F_k$). Each entry is the sum of all frequencies from the first class through the current class.

3. Determine the median class:

   The median class is the first class whose cumulative frequency is equal to or greater than $n/2$.

4. Use the median formula:

   Once the median class is known, estimate the median with:

   $$\operatorname{Me} = L + \left( \frac{\frac{n}{2} - F}{f_{\mathrm{Me}}} \right) \times c$$

   where:

   • $\operatorname{Me}$ = estimated median of the grouped data

   • $L$ = lower boundary of the median class. For adjacent integer-valued classes, calculate it with:

     $L$ is the lower limit of the median class minus $0.5$.

   • $n$ = total frequency

   • $F$ = cumulative frequency before the median class

   • $f_{\mathrm{Me}}$ = frequency of the median class

   • $c$ = class width, calculated with:

     $c$ is the upper class boundary minus the lower class boundary.

     Here, the $0.5$ correction applies because the shoe sizes are recorded as consecutive integers.

   The interpolation also assumes that observations are distributed uniformly within the median class. The result is therefore an estimate from the grouped intervals. The grouped table does not contain the original individual values.

Worked example:

Use Store A's shoe size data from the table below.

1. Median position: $n/2 = 30/2 = 15$. We need the observation at position $15$.

2. Cumulative frequency: The table already includes it.

3. Median class: The cumulative frequency reaches $13$ after the second class and $29$ after the third. Position $15$ therefore lies in $43\text{-}45$.

4. Formula components:

   • Median class: $43\text{-}45$

   • Lower boundary: $L = 43 - 0.5 = 42.5$

   • $n/2 = 30/2 = 15$

   • Cumulative frequency before the median class: $F = 13$

   • Median-class frequency: $f_{\mathrm{Me}} = 16$

   • Class width: $c = \text{Upper Boundary} - \text{Lower Boundary}$

     $$\begin{aligned} c &= (45 + 0.5) - (43 - 0.5) \\ &= 45.5 - 42.5 \\ &= 3 \end{aligned}$$

5. Substitute into the median formula:

   $$\operatorname{Me} = 42.5 + \left( \frac{15 - 13}{16} \right) \times 3$$

   $$\operatorname{Me} = 42.5 + \left( \frac{2}{16} \right) \times 3$$

   $$\operatorname{Me} = 42.5 + \left( \frac{1}{8} \right) \times 3$$

   $$\operatorname{Me} = 42.5 + \frac{3}{8}$$

   $$\operatorname{Me} = 42.5 + 0.375$$

   $$\operatorname{Me} = 42.875$$

The estimated median shoe size is $42.875$. It falls inside the median class $43\text{-}45$ once the continuous class boundaries are used.

This result is an estimate. The exact values inside each interval are unavailable, so the original middle observation cannot be recovered. The mean for grouped data has the same limitation.