Determining net charge of amino acid at given pH
You are right in your reasoning: at any pH for any titratable group, there is a distribution between the protonated and deprotonated species. We can calculate this distribution and therefore the average charge of the species with the Henderson-Hasselbalch equation:

$$\mathrm{pH = pKa + log\left(\frac{[B]}{[A]}\right)}$$

Or, rearranged:

$$\mathrm{\frac{[B]}{[A]} = 10^{pH - pKa}}$$

Where A is the conjugate acid and B is the conjugate base. For carboxyls and primary amines, this can also be written as $\ce{\frac{[COO-]}{[COOH]}}$ and $\ce{\frac{[NH2]}{[NH3+]}}$, respectively. As a simple example, if the pH was 2.28, the ratio between the protonated and deprotonated form of the carboxyl group (pKa = 2.28) would be $\ce{\frac{COO-}{COOH} = 10^0 = 1}$. This means $\ce{[COO-] = [COOH]}$ and the average charge of the carboxyl is thus -0.5.

Now, for cases where $\mathrm{pH \neq pKa}$, we need to determine the mole fraction, $\chi$, for each charged species. Given that $\mathrm{\frac{[B]}{[A]} = \frac{\chi_B}{\chi_A}}$ and $\ce{\chi_A + \chi_B = 1}$, we can derive these equations:

$$\mathrm{\chi_{COO^-} = \frac{1}{1 + 10^{pKa - pH}}}$$
$$\mathrm{\chi_{NH_3^+} = \frac{1}{1 + 10^{pH - pKa}}}$$

Plugging in the numbers, we get:

$$\mathrm{\chi_{COO^-} \approx 0.999981}$$

$$\mathrm{\chi_{NH_3^+} \approx 0.993872}$$

To state it another way, ~99.9981% of the carboxyl groups are in the deprotonated (negative) state and ~99.3872% of the amino groups are in the protonated (positive) state. The average charge of methionine at pH 7 is simply the sum of these mole fractions, accounting for the charge of each species, and is approximately equal to -0.006109. I assume your textbook is rounding.



You can plot these functions to see how charge varies with pH: