Grouped Data - Percentile worksheet - Free Printable

Let’s solve this step by step to find the 85th percentile (P85) of the heights of 40 junior high school students.

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Step 1: Complete the table with Lower Boundaries (LB) and Cumulative Frequency (cf)

We are given:

| Height in cm | Frequency (f) |
|--------------|---------------|
| 166 – 170 | 3 |
| 161 – 165 | 8 |
| 156 – 160 | 9 |
| 151 – 155 | 11 |
| 146 – 150 | 3 |
| 141 – 145 | 6 |

Total N = 3 + 8 + 9 + 11 + 3 + 6 = 40

Now, let’s compute:

#### Lower Boundaries (LB):
To get lower boundary, subtract 0.5 from the lower limit of each class.

- 166–170 → LB = 165.5
- 161–165 → LB = 160.5
- 156–160 → LB = 155.5
- 151–155 → LB = 150.5
- 146–150 → LB = 145.5
- 141–145 → LB = 140.5

#### Cumulative Frequency (cf):
Add frequencies from bottom to top (since we usually list classes from lowest to highest, but here they’re listed from highest to lowest — so we’ll reverse order for cumulative frequency).

Actually, looking at the table, it's listed from highest height to lowest. But for percentile calculation, we need cumulative frequency from the *lowest* class upward. So let’s reorder the table from smallest to largest height first to avoid confusion.

Reordered table (from lowest to highest):

| Height in cm | f | LB | cf |
|--------------|----|--------|------|
| 141 – 145 | 6 | 140.5 | 6 |
| 146 – 150 | 3 | 145.5 | 9 |
| 151 – 155 | 11 | 150.5 | 20 |
| 156 – 160 | 9 | 155.5 | 29 |
| 161 – 165 | 8 | 160.5 | 37 |
| 166 – 170 | 3 | 165.5 | 40 |

✔ Now cumulative frequency is correct: starts at 6, then 6+3=9, 9+11=20, etc., up to 40.

Also, class width i = upper limit - lower limit + 1? Wait — actually, for continuous data, class width is difference between lower boundaries.

Check: 145.5 to 140.5 = 5 → so i = 5.

Yes, all classes have width 5.

So:
i = 5

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Step 2: Compute the Pₖ class (for k = 85)

Formula:
Position = (k × N) / 100 = (85 × 40) / 100 = 3400 / 100 = 34

So we need the class where the 34th value falls.

Look at cumulative frequency:

- Up to 156–160: cf = 29 → not enough
- Next class: 161–165 → cf = 37 → includes positions 30 to 37 → so 34 is in this class.

✔ So P85 class is 161 – 165

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Step 3: Find values needed for formula

From the P85 class (161–165):

- Lower Boundary (LB) = 160.5
- Frequency of Pₖ class (f_Pk) = 8
- Cumulative frequency before Pₖ class (cf_b) = cumulative freq of previous class = 29 (from 156–160)
- i = 5
- kN/100 = 34 (already calculated)

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Step 4: Plug into formula

Formula:

Pₖ = LB + [ (kN/100 - cf_b) / f_Pk ] × i

So:

P85 = 160.5 + [ (34 - 29) / 8 ] × 5
= 160.5 + (5 / 8) × 5
= 160.5 + (25 / 8)
= 160.5 + 3.125
= 163.625

Round off to two decimal places → 163.63

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Step 5: Interpretation

Therefore, 85% of the students has a height of less than or equal to 163.63 cm.

(Note: Percentile means “value below which X% of observations fall”. So P85 = 163.63 means 85% of students are ≤ 163.63 cm tall.)

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Final Answer:
163.63