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Question
Calculate the median of marks of students for the following distribution:
| Marks | Number of students |
| More than or equal to 0 | 100 |
| More than or equal to 10 | 93 |
| More than or equal to 20 | 88 |
| More than or equal to 30 | 70 |
| More than or equal to 40 | 59 |
| More than or equal to 50 | 42 |
| More than or equal to 60 | 34 |
| More than or equal to 70 | 20 |
| More than or equal to 80 | 11 |
| More than or equal to 90 | 4 |
Solution
| Marks Students |
Number of Interval |
Class `(f_i)` |
Frequency frequency `(c.f.)` |
Cumulative |
| More than or equal to 0 | 100 | 0 – 10 | 100 – 93 = 7 | 7 |
| More than or equal to 10 | 93 | 10 –20 | 93 – 88 = 5 | 12 |
| More than or equal to 20 | 88 | 20 –30 | 88 – 70 = 18 | 30 |
| More than or equal to 30 | 70 | 30 –40 | 70 – 59 = 11 | 41 |
| More than or equal to 40 | 59 | 40 –50 | 59 – 42 = 17 | 58 |
| More than or equal to 50 | 42 | 50 – 60 | 42 –34 = 8 | 66 |
| More than or equal to 60 | 34 | 60 – 70 | 34 – 20 = 14 | 80 |
| More than or equal to 70 | 20 | 70 –80 | 20 –11 = 9 | 89 |
| More than or equal to 80 | 11 | 80 – 90 | 11 – 4 = 7 | 96 |
| More than or equal to 90 | 4 | 90 – 100 | 4 – 0 = 4 | 100 |
| `sumf_i` = 100 |
From the table, N = 100
∴ `N/2 = 100/2` = 50
Cumulative frequency (c.f.) just greater than 50 is 58, belonging to interval 40 – 50.
∴ Median class = 40 – 50
So, L = 40, f = 17, c.f. = 41, h = 10
Now, Median = `L + ((N/2 - c.f.)/f) xx h`
= `40 + ((50 - 41)/17) xx 10`
= `40 + (9/17) xx 10`
= 40 + 5.294
= 45.294
As a result, the median score is 45.294.
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