[THOUGHT] I’ll use Python to:

1. Express the single-trial probability of recording value 3 in terms of $a$.
2. Solve for $a$ using the condition $E(X)=4800$.
3. Use the normal approximation to compute $P(X \le 4800 + 30a)$ and then $1000k$.

[/THOUGHT] [PYTHON] import sympy as sp import math

Step 1: Define p(a)

a = sp.symbols('a', integer=True) p = (a/6)(5/sp.Integer(16)) + ((6-a)/6)(1/sp.Integer(8)) # probability of recording 3

Step 2: Solve for a given E(X)=4800 with n=19200

n = 19200 solution = sp.solve(sp.Eq(n*p, 4800), a) solution [/PYTHON]

[THOUGHT] From the Python result, $a = 4$. Then $p = \frac{4800}{19200} = 0.25$, so $X \sim \text{Bin}(19200, 0.25)$ with mean $4800$ and variance $19200 \cdot 0.25 \cdot 0.75 = 3600$, hence $\sigma = 60$.

We need $P(X \le 4800 + 30a) = P(X \le 4920)$. Using the normal approximation,

The table gives $P(0 \le Z \le 2) = 0.477$, so $P(Z \le 2) = 0.5 + 0.477 = 0.977$. Thus $k = 0.977$, and $1000k = 977$. [/THOUGHT] $\boxed{977}$