Can an expected value (mean) be higher than the values used to create it?

The expected value cannot be larger than any of the possible values, but it can certainly be larger than any of the probabilities. If you roll a fair die, each of the possible values ($1,2,3,4,5$, and $6$) occurs with probability $\frac16$, so the expected value is $$(1)\frac16+(2)\frac16+(3)\frac16+(4)\frac16+(5)\frac16+(6)\frac16=3.5\;,$$ far bigger than $\frac16$.

Now imagine that the die has a $6$ on every face. The probability that it comes up $6$ is $1$, and the expected value is clearly $6$: you can’t get anything else!

The expected value can be thought of as the long run average value of the possible values, each of them weighted by its probability of occurrence; in principle it can be anywhere between the smallest possible value and the largest possible value. It is affected by the probabilities only to the extent that they affect the weighting of the possible values.

The expectation of your distribution is exactly $$ 0.2 x_0 + 0.25 x_1 + 0.3 x_2 + 0.15 x_3 + 0.1 x_4. $$ According to the frequentist interpretation, the expected value is the value that you get when you perform a large number of experiments and compute the average. So if for example your random variable is constant, say always equal to $10$, then the expectation is $10$. There's no problem with the expectation being bigger than $1$. However, since the expectation is a weighted average of the values of the random variable, it always lies between the minimal value and the maximal value. So if your random variable always gets a value betewen $3$ and $7$, then the expectation is also going to be between $3$ and $7$.