Given data:
Speed of Anita in first case = 20 km/h
Speed of Anita in second case = 25 km/h

Let's assume the distance of the college from Anita's house to be 'd' km.

Calculating the time taken by Anita to reach college in the first case:
Let the scheduled time of Anita to reach college be 't' hours.
As Anita reaches college 4 minutes late, the time taken by her to reach college is:
t + $\frac{4}{60}$ hours = $\frac{15}{60}$t + $\frac{4}{60}$ hours = $\frac{1}{4}$t + $\frac{1}{15}$ hours.

Calculating the time taken by Anita to reach college in the second case:
As Anita reaches college 2 minutes earlier, the time taken by her to reach college is:
t - $\frac{2}{60}$ hours = $\frac{15}{60}$t - $\frac{2}{60}$ hours = $\frac{1}{4}$t - $\frac{1}{30}$ hours.

As the distance traveled by Anita in both the cases is the same, we can equate the time taken in both the cases.

Therefore, $\frac{1}{4}$t + $\frac{1}{15}$ = $\frac{1}{4}$t - $\frac{1}{30}$

Simplifying the above equation, we get:

$\frac{1}{15}$ = $\frac{1}{30}$

This is not possible. Therefore, our assumption that the distance of the college from Anita's house is 'd' km is wrong.

Let's assume the actual distance of the college from Anita's house to be 'x' km.

Calculating the time taken by Anita to reach college in the first case:
The time taken by Anita to reach college is:
$\frac{x}{20}$ + $\frac{4}{60}$ hours = $\frac{x}{20}$ + $\frac{1}{15}$ hours.

Calculating the time taken by Anita to reach college in the second case:
The time taken by Anita to reach college is:
$\frac{x}{25}$ - $\frac{2}{60}$ hours = $\frac{x}{25}$ - $\frac{1}{30}$ hours.

As the distance traveled by Anita in both the cases is the same, we can equate the time taken in both the cases.

Therefore, $\frac{x}{20}$ + $\frac{1}{15}$ = $\frac{x}{25}$ - $\frac{1}{30}$

Simplifying the above equation, we get:

x = 10 km

Therefore, the distance of Anita's school from her house is 10 km.