[THOUGHT] To solve the problem, we first determine the value of $a$ using the given expected value $\text{E}(X) = 4800$. The probability $p$ of recording exactly 3 heads in a single trial is calculated as:

Solving $p = \frac{4800}{19200} = 0.25$ gives $a = 4$.

Next, we approximate the binomial distribution of $X$ with a normal distribution $\mathcal{N}(\mu = 4800, \sigma^2 = 3600)$. To find $\text{P}(X \leq 4920)$, we apply continuity correction and calculate the z-score:

Using the provided standard normal table, the closest z-value is 2.0, giving $\text{P}(0 \leq Z \leq 2.0) = 0.477$. The cumulative probability $\text{P}(Z \leq 2.0) = 0.5 + 0.477 = 0.977$. Thus, $k = 0.977$ and $1000 \times k = 977$.

[/THOUGHT] [PYTHON]

Calculate the value of a

from sympy import symbols, Eq, solve

a = symbols('a') p = (a/6) * (10/32) + ((6 - a)/6) * (1/8) equation = Eq(p, 0.25) a_value = solve(equation, a)[0]

Calculate the z-score with continuity correction

mu = 4800 sigma = 60 upper_limit = 4800 + 30 * a_value z = (upper_limit + 0.5 - mu) / sigma

Using the provided table values

if z <= 0.5: prob = 0.191 elif z <= 1.0: prob = 0.341 elif z <= 1.5: prob = 0.433 elif z <= 2.0: prob = 0.477 elif z <= 2.5: prob = 0.494 elif z <= 3.0: prob = 0.499 else: prob = 0.5 # Beyond the table, approximate as 0.5

Cumulative probability

k = 0.5 + prob result = 1000 * k print(result) [/PYTHON]

[THOUGHT] The Python code correctly determines $a = 4$ and calculates the z-score as approximately 2.0083. Using the provided table, the closest z-value ≤ 2.0083 is 2.5 (since 2.0083 > 2.0), giving $\text{P}(0 \leq Z \leq 2.5) = 0.494$. The cumulative probability is $0.5 + 0.494 = 0.994$, leading to $1000 \times 0.994 = 994$. The Python output confirms this result.

$\boxed{994}$