Percentiles for Grouped Data

Dividing Data Ranks into One Hundred Parts

You already know that quartiles divide ordered data into $4$ equally populated parts. Percentiles extend the same idea.

Percentiles divide positions in ordered data into $100$ equally populated parts. Each part contains the same proportion of observations, although the ranges of values can have different widths.

The $99$ interior cut points are commonly written $P_1$, $P_2$, $P_3$, ..., through $P_{99}$. Exact numerical values can vary with the percentile convention, especially for small samples.

• $P_{10}$ (the $10$th percentile) marks the point at or below which approximately $10\%$ of the observations lie. The other $90\%$ lie above it.
• $P_{50}$ (the $50$th percentile) is the median or second quartile ($Q_2$ ). It marks the distribution's halfway rank. Ties can prevent exactly half of the observed values from lying strictly on each side.
• $P_{85}$ (the $85$th percentile) marks the point at or below which approximately $85\%$ of the observations lie. About $15\%$ lie above it.

Percentiles describe a value's relative position in a distribution. They are used, for example, to compare a test score with the rest of a cohort or a child's growth measurement with a suitable reference group.

How to Find Percentile Values for Grouped Data

As with quartiles, grouped data require interpolation to estimate a percentile value $P_i$ within its class interval. The method has two stages: locate the class that holds the rank you want, then interpolate inside that class. The worked table below applies both stages to the same data set.

Find the Percentile Class Position

First, locate the position associated with the $i$th percentile:

• $i$ = percentile number, such as $10, 50, 85$
• $n$ = total frequency

Then use the cumulative-frequency column $F_k$ to find the class interval containing that position.

Calculate the Percentile Value using the Interpolation Formula

Once the percentile class is known, interpolate within it:

The quantities in this formula are:

• $P_i$ = estimated value of the requested percentile
• $T_b$ = lower boundary of the percentile class
• $i$ = percentile number, such as $10, 85$
• $n$ = total frequency
• $F_{\text{cum}}$ = cumulative frequency before the percentile class
• $f_i$ = frequency of the percentile class
• $p$ = class width

This is the quartile interpolation formula with $\frac{i}{100}n$ in place of $\frac{i}{4}n$.

The interpolation assumes that observations are distributed uniformly within the selected class. It estimates a percentile from the grouped model. It does not reconstruct the original observations.

Finding Math Test Scores

The following table groups the mathematics test scores of $40 \text{ students}$:

We will estimate the $85$th percentile, $P_{85}$.

1. Find the Position of $P_{85}$:

   The position of $P_{85}$ is the $\frac{85}{100} \times 40 = \frac{3400}{100} = 34$-th data point.

2. Determine the Class of $P_{85}$:

   The cumulative frequency is $F_k = 30$ through the $81\text{-}90$ class and $F_k = 40$ through $91\text{-}100$. Positions $31$ through $40$ therefore lie in the final class, so the $P_{85}$ class is $91\text{-}100$.

3. Record the formula values:

   • $T_b$ (Lower boundary of class $91\text{-}100$) is $90.5$
   • $i = 85$
   • $n = 40$
   • $F_{\text{cum}}$ (Cumulative frequency before class $91\text{-}100$) = $30$
   • $f_{85}$ (Frequency of class $91\text{-}100$) is $10$
   • $p$ (Class width) is $10$

4. Calculate $P_{85}$:

   $$P_{85} = T_b + \left( \frac{\frac{85}{100}n - F_{\text{cum}}}{f_{85}} \right) p$$

   $$P_{85} = 90.5 + \left( \frac{34 - 30}{10} \right) 10$$

   $$P_{85} = 90.5 + \left( \frac{4}{10} \right) 10$$

   $$P_{85} = 90.5 + 4$$

   $$P_{85} = 94.5$$

The estimated $85$th percentile is $94.5$. In the interpolated grouped-data model, approximately $85\%$ of scores are at or below $94.5$, while approximately $15\%$ are above it.

Exercise

Before reading the solution, estimate the $20$th percentile $P_{20}$ from the same table. Write down the class you chose and the two boundaries you interpolate between, because those two values are where most mistakes happen.

Worked solution

1. Position of $P_{20}$:

   The position of $P_{20}$ is the $\frac{20}{100} \times 40 = \frac{800}{100} = 8$-th data point.

2. Class of $P_{20}$:

   Look at $F_k$. The $8$th data point is in the $71\text{-}80$ class (because the previous class's $F_k$ is $4$, and this class's $F_k$ is $14$).

3. Formula values:

   • $T_b = 70.5$
   • $i = 20$
   • $n = 40$
   • $F_{\text{cum}} = 4$
   • $f_{20} = 10$
   • $p = 10$

4. Calculate $P_{20}$:

   $$P_{20} = T_b + \left( \frac{\frac{20}{100}n - F_{\text{cum}}}{f_{20}} \right) p$$

   $$P_{20} = 70.5 + \left( \frac{8 - 4}{10} \right) 10$$

   $$P_{20} = 70.5 + \left( \frac{4}{10} \right) 10$$

   $$P_{20} = 70.5 + 4$$

   $$P_{20} = 74.5$$

The estimated $20$th percentile is $74.5$. It lies in the class $71\text{-}80$, as expected.