Find the mean,median and mode of the following frequency distribution:

Class	$0-30$	$30-60$	$60-90$	$90-120$	$120-150$	$150-180$
Frequency	$8$	$15$	$16$	$20$	$12$	$9$

(N/A) $1$. Mean: The class marks $(x_i)$ are $15, 45, 75, 105, 135, 165$. The sum of frequencies $(\sum f_i)$ is $80$. The sum of products $(\sum f_i x_i)$ is $(8 \times 15) + (15 \times 45) + (16 \times 75) + (20 \times 105) + (12 \times 135) + (9 \times 165) = 120 + 675 + 1200 + 2100 + 1620 + 1485 = 7200$. Mean $= \frac{\sum f_i x_i}{\sum f_i} = \frac{7200}{80} = 90$.
$2$. Median: $N/2 = 40$. The cumulative frequencies are $8, 23, 39, 59, 71, 80$. The median class is $90-120$. Median $= l + \left( \frac{N/2 - cf}{f} \right) \times h = 90 + \left( \frac{40 - 39}{20} \right) \times 30 = 90 + (1/20) \times 30 = 90 + 1.5 = 91.5$.
$3$. Mode: The modal class is $90-120$ (highest frequency $20$). Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 90 + \left( \frac{20 - 16}{2(20) - 16 - 12} \right) \times 30 = 90 + \left( \frac{4}{40 - 28} \right) \times 30 = 90 + \left( \frac{4}{12} \right) \times 30 = 90 + 10 = 100$.