Calculate the Mean (x).

To calculate the mean using the step-deviation method, we first find the class mark ($x_i$) for each interval, choose an assumed mean ($a$), calculate $u_i = \frac{x_i - a}{h}$, and then $f_i u_i$. The class size ($h$) is $20-15=5$. Let's choose the assumed mean $a = 37.5$ (the class mark of the 5th class).

Class Interval	$f_i$	$x_i$	$d_i = x_i - 37.5$	$u_i = \frac{d_i}{5}$	$f_i u_i$
15-20	2	17.5	-20	-4	-8
20-25	4	22.5	-15	-3	-12
25-30	18	27.5	-10	-2	-36
30-35	21	32.5	-5	-1	-21
35-40	33	37.5	0	0	0
40-45	11	42.5	5	1	11
45-50	3	47.5	10	2	6
50-55	6	52.5	15	3	18
55-60	2	57.5	20	4	8
Total	$\sum f_i = 100$				$\sum f_i u_i = -34$

Using the formula $\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h$:

$$\bar{x} = 37.5 + \frac{-34}{100} \times 5$$ $$\bar{x} = 37.5 - \frac{170}{100}$$ $$\bar{x} = 37.5 - 1.7$$ $$\bar{x} = 35.8$$

The mean is $\boxed{\text{35.8}}$.

First, we calculate the cumulative frequency (C.F.) and find the median class. The total frequency $n = \sum f_i = 100$, so $\frac{n}{2} = \frac{100}{2} = 50$.

Class Interval	$f_i$	C.F.
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

The cumulative frequency just greater than 50 is 78, which corresponds to the class interval 35-40. This is our median class.
From the median class:

• Lower limit ($l$) = 35
• Frequency ($f$) = 33
• Cumulative frequency of the class preceding the median class (C.F.) = 45
• Class size ($h$) = 5

Using the formula $M = l + \left[\frac{\frac{n}{2} - C.F.}{f}\right] \times h$:

$$M = 35 + \left[\frac{50 - 45}{33}\right] \times 5$$ $$M = 35 + \left[\frac{5}{33}\right] \times 5$$ $$M = 35 + \frac{25}{33}$$ $$M = 35 + 0.757575...$$ $$M \approx 35.76$$

The median is $\boxed{\text{35.76}}$.