Solution: As the polynomial $q(x)$ is of degree $4$, we write \[ q(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4, \quad a_i \in \mathbb{R} . \] Now form the given data, we also have \begin{align*} & q(-2) = p(-2) \quad q(-1) = p(-1) \\ & q(1) = p(1) \quad q(3) = p(3). \end{align*} So, we can determine the polynomial $q$ completely. Take \[ q(x) = p(x) + k (x-1)(x+1)(x+2)(x-3). \] Then $q$ satisfies the given data at $x = -2,-1,1$ and $3$. We can determine $k$ by considering $q(0) = 2.5$. We get \begin{align*} q(0) = 2.5 & \implies p(0) + k (-1)(1)(2)(-3) = 2.5 \\ & \implies 2 + 6k = 2.5 \\ & \implies k = \frac{.5}{6} \\ & \implies k = \frac{1}{12}. \end{align*} Therefore, \[ q(x) = p(x) + \frac{1}{12}(x^2-1)(x+2)(x-3). \] Thus, \begin{align*} \frac{\mathrm{d}^4q}{\mathrm{d}x^4} & = \frac{\mathrm{d^4}}{\mathrm{d}x^4} \left( \frac{1}{12}(x^2-1)(x+2)(x-3) \right) \\ & = \frac{1}{12}\frac{\mathrm{d^4}}{\mathrm{d}x^4} \left( x^4 + r(x) \right), \end{align*} where $r(x)$ is a cubic polynomial. Hence, \[ \frac{\mathrm{d}^4q}{\mathrm{d}x^4} = 4!\times \frac{1}{12} = 2. \]