Solution:
| Class interval | Mid value (x) | Frequency (f) | fx | Cumulative frequency |
| 100 – 120 | 110 | 12 | 1320 | 12 |
| 120 – 140 | 130 | 14 | 1820 | 26 |
| 140 – 160 | 150 | 8 | 1200 | 34 |
| 160 – 180 | 170 | 6 | 1000 | 40 |
| 180 – 200 | 190 | 10 | 1900 | 50 |
| N = 50 | Σfx = 7260 |
We know that,
Mean $= Σfx / N$
$= 7260/ 50$
$= 145.2$
Then,
We have, $N = 50$
⇒ N/2 $= 50/2 $= 25$
So, the cumulative frequency just greater than N/2 is $26$, then the median class is $120 – 140$
Such that $l = 120, h = 140 – 120 = 20, f = 14, F = 12$
$= 120 + 260/14$
$= 120 + 18.57$
$= 138.57$
From the data, its observed that maximum frequency is $14$, so the corresponding class $120 – 140$ is the modal class
And,
$l = 120, h = 140 – 120 = 20, f = 14, f1 = 12, f2 = 8$
$= 120 + 5$
$= 125$
Therefore, mean $= 145.2$, median $= 138.57$ and mode $= 125$