Quartiles for Grouped Data

Estimating Quartile Positions within Class Intervals

For ungrouped data, we can sort every observation and identify its position directly. A grouped frequency table provides only the intervals, such as $70\text{-}79$ and $80\text{-}89$, and the number of observations in each interval. It does not contain the individual values.

As with the median for grouped data, we therefore estimate $Q_1$, $Q_2$, and $Q_3$ by interpolation. The calculation places each quartile within the class interval that contains its target position.

We determine the position of the quartile using this formula:

• The position of $Q_1$ is the $\frac{1}{4}n$-th data point
• The position of $Q_2$ is the $\frac{2}{4}n$-th data point (or $\frac{1}{2}n$-th)
• The position of $Q_3$ is the $\frac{3}{4}n$-th data point

Where $n$ is the total number of data points.

Steps to Find the Value of Quartiles for Grouped Data

Consider the grouped shoe-sales data from Store A. The frequency table below gives each class interval and how many pairs fall in it, and the quartiles are estimated by interpolating inside the class that contains the position you want. Work through the steps with the table open, because every value you substitute comes from it.

Create a Cumulative Frequency Table

First, we need a frequency table with a cumulative frequency column ($F_k$). Cumulative frequency is the sum of frequencies from the first class up to that class. This is important to know which class the quartile falls into.

For example, here is the shoe sales table:

The lower boundary is the lower class limit minus $0.5$.

The upper boundary is the upper class limit plus $0.5$.

The class width is the upper boundary minus the lower boundary.

Determine the Quartile Class Position

First, locate each quartile position.

Total data ($n$) is $30$.

• Position of $Q_1$: the $\frac{1}{4} \times 30 = 7.5$-th data point.

  In the $F_k$ column, the first class ends at position $2$ and the second ends at position $13$. Positions $3$ through $13$ belong to the second class, so position $7.5$ lies in $40\text{-}42$.

• Position of $Q_2$ (Median): the $\frac{1}{2} \times 30 = 15$-th data point.

  Look at $F_k$. The $15$th data point is in the $43\text{-}45$ class (because the previous $F_k$ was $13$, and this class's $F_k$ is $23$).

• Position of $Q_3$: the $\frac{3}{4} \times 30 = 22.5$-th data point.

  Look at $F_k$. The $22.5$th data point is also in the $43\text{-}45$ class (because the previous $F_k$ was $13$, and this class's $F_k$ is $23$).

Estimate the Quartile Value by Interpolation

Once the quartile class is known, estimate the value within that interval with:

The quantities in this formula are:

• $Q_i$ = Value of the i-th Quartile (what we're looking for)
• $T_b$ = Lower boundary of the i-th quartile class
• $n$ = Total frequency
• $F_{\text{cum}}$ = cumulative frequency before the $i$-th quartile class
• $f_i$ = Frequency of the i-th quartile class
• $p$ = Class width

This interpolation assumes that observations are distributed uniformly within the selected class. It is a model-based estimate and cannot recover the original values that were omitted when the data were grouped.

Finding the Shoe Sales Quartile

Calculate $Q_1$ from the table above.

1. Position of $Q_1$: $7.5$th data point.

2. Class of $Q_1$: $40\text{-}42$.

3. Record the formula values:

   • Lower boundary of $Q_1$ class ($T_b$) is $39.5$
   • Total data ($n$) is $30$
   • Cumulative frequency before $Q_1$ class ($F_{\text{cum}}$) is $2$ (see $F_k$ for class $37\text{-}39$)
   • Frequency of $Q_1$ class ($f_1$) is $11$
   • Class width ($p$) is $3$

4. Plug into the formula:

   $$Q_1 = T_b + \left( \frac{\frac{1}{4}n - F_{\text{cum}}}{f_1} \right) p$$

   $$Q_1 = 39.5 + \left( \frac{7.5 - 2}{11} \right) 3$$

   $$Q_1 = 39.5 + \left( \frac{5.5}{11} \right) 3$$

   $$Q_1 = 39.5 + (0.5) \times 3$$

   $$Q_1 = 39.5 + 1.5$$

   $$Q_1 = 41$$

So, the value of $Q_1$ is $41$. This means about $25\%$ of the shoes sold are size $41$ or smaller.

Exercise

Calculate $Q_3$ from the shoe sales data in the table above.

Then compare grouped-data interpolation with the calculation of quartiles from ungrouped data. Why does the grouped table provide only an estimate?

Answer Key

1. Position of $Q_3$: $22.5$th data point.

2. Class of $Q_3$: $43\text{-}45$.

3. Gather the ingredients:

   • $T_b = 42.5$ (lower boundary of $Q_3$ class)
   • $n = 30$ (total data)
   • $F_{\text{cum}} = 13$ (see $F_k$ for class $40\text{-}42$)
   • $f_3 = 10$ (frequency of $Q_3$ class)
   • $p = 3$ (class width)

4. Plug into the formula:

   $$Q_3 = T_b + \left( \frac{\frac{3}{4}n - F_{\text{cum}}}{f_3} \right) p$$

   $$Q_3 = 42.5 + \left( \frac{22.5 - 13}{10} \right) 3$$

   $$Q_3 = 42.5 + \left( \frac{9.5}{10} \right) 3$$

   $$Q_3 = 42.5 + (0.95) \times 3$$

   $$Q_3 = 42.5 + 2.85$$

   $$Q_3 = 45.35$$

The estimated value of $Q_3$ is $45.35$. In this grouped model, about $75\%$ of the shoes sold have size $45.35$ or below. The remaining $25\%$ have a shoe size above $45.35$.

Comparison with Single Data:

Finding quartiles for grouped data uses interpolation because the table provides intervals and frequencies without the individual observations. The result is an estimated quartile value.

For ungrouped data, the ordered observations are known, so the declared quartile convention can select one value or interpolate directly between two known values. Grouped interpolation is less precise because the observations inside each class are no longer available.