Nancy has classes three days a week. She attends classes three days a week 80% of the time, two days 15% of the time, one day 4% of the time, and no days 1% of the time. Suppose one week is randomly selected. Let X be the number of days Nancy attends class per week.
We begin by finding the probability that Nancy attends class $3$ days in a week. The probability that she attends class $3$ days per week is $80\%$, or $0.80$. For the days that she does not attend class at all, the probability is $1\%$, or $0.01$. Therefore, the probability that she attends class exactly 3 days a week is $0.80-0.01=0.79$.
Next, we find the probability that Nancy attends class $2$ days in a week. The probability that she attends class $2$ days per week is $15\%$, or $0.15$. The probability that she does not attend class at all is still $0.01$, as given in the problem. Therefore, the probability that she attends class exactly 2 days a week is $0.15-0.01=0.14$.
Finally, we find the probability that Nancy attends class $1$ day in a week. The probability that she attends class $1$ day per week is $4\%$, or $0.04$. As before, the probability that she does not attend class at all is $0.01$. Therefore, the probability that she attends class exactly 1 day a week is $0.04-0.01=0.03$.
The probability that Nancy attends class $0$ days in a week is $0.01$, as given in the problem.
Because these probabilities are mutually exclusive and cover all possible outcomes, they must sum to $1$. Therefore, \begin{align*}
0.79+0.14+0.03+0.01&=1\\
0.97+0.04&=1\\
1&=1,
\end{align*}as expected.
To calculate $\mathbb{E}(X)$, we multiply each outcome by its probability: \begin{align*}
0\cdot 0.01 +1\cdot 0.03+2\cdot 0.14+3\cdot 0.79&=3\cdot 0.79 + 2\cdot 0.14+0.03\\
&=2.37 +0.28 + 0.03\\
&=\boxed{2.68}.
\end{align*}