What happens with my variance if I multiply observations by 5? Suppose a dataset with observation values:
2
3
4
4
2

Variance of this observation, assuming this is the entire population, would be:

Where the average is: 3.  So the variance equals: 0.8.
Now, i read around that if I multiply the observation values by 5, the variance should increase by 25.  But that doesn't seem to be the case, am I interpreting something erroneously?
10
15
20
20
10

The average is: 15, so the variance equals: 20.
 A: The variance increases by a factor of 25 (multiplication), it does not increase by 25 (addition). All of your calculations are correct. In sample 1, variance is 0.8 and in sample 2 variance is 20, which is 25 times larger than 0.8, i.e. 20=25*0.8.
In general, multiplying all observations of a random variable $X$ by a constant $c$ scales the variance up by $c^2$. Let $V(X)$ denote the variance operator.
$$V(cX)=c^2V(X).$$
To see this more easily, note that the mean of your new sample is $c\mu$. Replace $X_i$ in your formula with $cX_i$ and $\mu$ with $c\mu$:
$$\sum_{i=1}^N \frac{(cX_i-c\mu)^2}{N}=\sum_{i=1}^N \frac{c^2(X_i-\mu)^2}{N} =c^2 \sum_{i=1}^N \frac{(X_i-\mu)^2}{N} = c^2V(X).$$
In your example, $c=5$.
A: Thinking about units may help in making it clear what to expect.
Suppose you have a set of measurements in kilograms:
2
3
4
4
2

with mean 3 kg, and standard deviation (computed from the variance) of 0.8944 kg
If you send these objects to a friend in the United States and she measures the weights, they will be in pounds. The numbers will all be multiplied by 2.2 lb/kg.
4.4
6.6
8.8
8.8
4.4

What's the average of these? Well, it had better be the number you get from converting the known average, 3 kg, to pounds, by multiplying by 2.2 lb/kg: 6.6 lb. And it is.
What's the standard deviation? Well, it had better be the number you get from converting the known standard deviation, 0.8944 kg, to pounds, by multiplying by 2.2 lb/kg: 1.968 lb. And it is.
The variance is harder to think about because it's squared, but it should be the number you get by squaring the standard deviation and so it should be the number you get by multiplying 0.8 by $(2.2\ \textrm{lb}/\textrm{kg})^2$. And so it is: $3.872\ \textrm{lb}^2$
(As an additional irrelevant note: if this is a teaching example then I think you were expected to use the $N-1$ version of the variance, because that way the variance is 1, a nice round number.)
