For the following frequency distribution, compute the percentile rank corresponding to the score of 66.

| Class Interval | 93-97 | 88-92 | 83-87 | 78-82 | 73-77 | 68-72 | 63-67 | 58-62 | 53-57 | 48-52 |
|---|---|---|---|---|---|---|---|---|---|---|
| Frequency | 4 | 7 | 5 | 8 |

Question

For the following frequency distribution, compute the percentile rank corresponding to the score of 66.

| Class Interval | 93-97 | 88-92 | 83-87 | 78-82 | 73-77 | 68-72 | 63-67 | 58-62 | 53-57 | 48-52 |
|---|---|---|---|---|---|---|---|---|---|---|
| Frequency | 4 | 7 | 5 | 8 | 3 | 6 | 7 | 10 | 5 | 4 |

Accepted Answer

> [!TLDR] For this grouped frequency distribution ($N=59$), the score $66$ falls in the class $63$-$67$ and its percentile rank works out to $\approx40.51$.

> [!FORMULA]
> $PR=\left[\dfrac{cf_L+\dfrac{X-L}{h}f}{N}ight]	imes100$
> where $L$ = lower boundary of the class containing score $X$, $cf_L$ = cumulative frequency of all classes below $L$, $f$ = frequency of the class containing $X$, $h$ = class width, $N$ = total frequency.

1. Reorder the classes ascending and build the cumulative-frequency table:

| CI | 48-52 | 53-57 | 58-62 | 63-67 | 68-72 | 73-77 | 78-82 | 83-87 | 88-92 | 93-97 |
|---|---|---|---|---|---|---|---|---|---|---|
| $f_i$ | 4 | 5 | 10 | 7 | 6 | 3 | 8 | 5 | 7 | 4 |
| CF | 4 | 9 | 19 | 26 | 32 | 35 | 43 | 48 | 55 | 59 |

2. $N=59$.
3. The score $X=66$ lies in the class $63$-$67$. Using continuity correction, its true class boundaries are $62.5$ to $67.5$, so $L=62.5$, $h=5$, and its frequency $f=7$.
4. $cf_L$ = cumulative frequency of all classes below $62.5$ = CF of $58$-$62$ = $19$. …

Score	Number of students
92	1
85	2
78	5
65	8
50	4

Concept understanding — Percentile Rank

What Does "Percentile Rank" Actually Mean?

The Intuition in One Sentence

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