Quartiles for Grouped Data

How to Find Quartiles in Grouped Data

For single data, we just sort it and find the middle position. Now, if the data is grouped in a frequency table (like test scores grouped as 70-79, 80-89, etc.), the method is slightly different. We don't know the exact value of each data point, only how many data points are in each group (class interval).

Similar to the median for grouped data, to find quartiles ($Q_1$, $Q_2$, $Q_3$), we also use interpolation. Essentially, we "estimate" the quartile's position within the class interval where it falls.

We determine the position of the quartile using this formula:

• Position of $Q_1$ = the $\frac{1}{4}n$-th data point
• Position of $Q_2$ = the $\frac{2}{4}n$-th data point (or $\frac{1}{2}n$-th)
• Position of $Q_3$ = 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

Let's assume we have shoe sales data from Store A in a grouped frequency table format.

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:

Lower boundary = lower limit - 0.5

Upper boundary = upper limit + 0.5

Class width = Upper boundary - Lower boundary

Determine the Quartile Class Position

First, let's find the position of the data point for the quartile.

Total data ($n$) = 30.

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

  Look at the $F_k$ column. Which class contains the 7.5th data point? The first class has $F_k = 2$ (not enough). The second class has $F_k = 13$ (data points 3 through 13 are here). So, the 7.5th data point is in the 40-42 class.

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

  Look at $F_k$. The 15th data point is in the 43-45 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 22.5th data point is also in the 43-45 class (because the previous $F_k$ was 13, and this class's $F_k$ is 23).

Calculate the Quartile Value using the Interpolation Formula

Once we know the class, we use this formula to find the exact value:

Where:

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

Finding Q1 for Shoe Sales

Let's calculate $Q_1$ from the table above.

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

2. Class of $Q_1$: 40-42.

3. Let's gather the ingredients:

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

4. Plug into the formula:

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

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

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

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

   $$Q_1 = 39.5 + 1.5$$

   $$Q_1 = 41$$

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

Exercise

Try calculating $Q_3$ from the shoe sales data in the table above.

After getting the result, compare it with the method for finding quartiles for single data learned earlier. What's the difference, and why might the results be similar or different?

Answer Key

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

2. Class of $Q_3$: 43-45.

3. Gather the ingredients:

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

4. Plug into the formula:

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

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

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

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

   $$Q_3 = 42.5 + 2.85$$

   $$Q_3 = 45.35$$

So, the value of $Q_3$ is 45.35. This means about 75% of the shoes sold are size 45.35 or smaller (or 25% are sold in sizes larger than 45.35).

Comparison with Single Data:

Finding quartiles for grouped data uses interpolation because we don't know the exact value of each data point, only its range. The result is an estimated quartile value.

For single data, we can directly point to which data point is the quartile (or the average of two data points), so the result is more precise (if the data is indeed single). Quartiles for grouped data provide a good overview for large datasets that have already been grouped.