[THOUGHT] We need to understand the experiment.

We have:

• Die: fair 6-sided die, outcomes 1 to 6.
• Parameter a: natural number, a ≤ 6.
• Experiment: roll die once.
  • If outcome ≤ a: toss coin 5 times, record number of heads.
  • If outcome > a: toss coin 3 times, record number of heads.

Repeat experiment 19200 times. Let X = number of times recorded value is 3 (i.e., exactly 3 heads). Given E(X) = 4800.

We need to compute P(X ≤ 4800 + 30a) = k, then compute 1000 × k.

We are given a standard normal table with z and P(0 ≤ Z ≤ z). We need to approximate using normal approximation (CLT) since n is large (19200). X is a sum of Bernoulli random variables (but not identical because probability of 3 heads differs depending on die outcome). However, X is the total number of successes (where success = recorded value = 3). We can compute p = probability that in a single experiment, the recorded value is 3.

Then X ~ Binomial(n=19200, p) approximately normal with mean μ = np = 4800, variance σ² = np(1-p). But wait: E(X) = 4800 given. So np = 4800. So p = 4800/19200 = 0.25 exactly. So p = 0.25.

But we must check if p = 0.25 is consistent with a? Because p depends on a. Let's compute p in terms of a.

Let die outcome D uniform on {1,...,6}.

Case 1: D ≤ a: then toss 5 coins, record number of heads H5 ~ Binomial(5, 0.5). Probability that H5 = 3 is C(5,3) * (0.5)^5 = 10/32 = 5/16 = 0.3125.

Case 2: D > a: then toss 3 coins, record H3 ~ Binomial(3, 0.5). Probability that H3 = 3 is (0.5)^3 = 1/8 = 0.125.

So overall p = P(D ≤ a) * (5/16) + P(D > a) * (1/8).

Since die uniform, P(D ≤ a) = a/6, P(D > a) = (6-a)/6.

Thus p = (a/6)(5/16) + ((6-a)/6)(1/8) = (a/6)(5/16) + ((6-a)/6)(2/16) [since 1/8 = 2/16].

Simplify: p = [a*5 + (6-a)2] / (616) = [5a + 12 - 2a] / 96 = (3a + 12) / 96 = (a + 4) / 32.

Given E(X) = np = 19200 * p = 4800, so p = 4800/19200 = 0.25 = 1/4.

Thus (a+4)/32 = 1/4 => a+4 = 32*(1/4) = 8 => a = 4.

So a = 4.

Now we need to compute P(X ≤ 4800 + 30a) = P(X ≤ 4800 + 120) = P(X ≤ 4920).

We have μ = 4800, σ² = np(1-p) = 19200 * 0.25 * 0.75 = 19200 * 0.1875 = 3600. So σ = sqrt(3600) = 60.

We use normal approximation: X ~ N(4800, 60²). Then standardize: Z = (X - μ)/σ.

We want P(X ≤ 4920) = P(Z ≤ (4920 - 4800)/60) = P(Z ≤ 120/60) = P(Z ≤ 2.0).

From the table, P(0 ≤ Z ≤ 2.0) = 0.477. Since normal distribution symmetric, P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977.

Thus k = 0.977.

Then 1000 × k = 977.

But we must check: the table gives P(0 ≤ Z ≤ z). For z=2.0, it's 0.477. So P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977 indeed.

Thus answer = 977.

We should verify with Python to ensure no rounding issues.

We'll compute p, a, then compute probability using normal approximation.

We'll also check if any other interpretation? The problem says "Using the standard normal table on the right, compute P(X ≤ 4800 + 30a)". That's exactly what we did.

Thus final answer: 977.

We'll write Python code to compute a, p, μ, σ, z, and probability.

But we must use the given table values. We'll just compute using those values.

We'll output 1000*k.

Let's do it. </think>

[THOUGHT] We need to compute a from given expected value, then compute the probability using normal approximation with the provided table.

Steps:

1. Compute p from E(X) = np = 4800, n=19200 → p = 4800/19200 = 0.25.
2. Express p in terms of a: p = (a+4)/32. Solve (a+4)/32 = 0.25 → a = 4.
3. Compute μ = 4800, σ = sqrt(np(1-p)) = sqrt(192000.250.75) = sqrt(3600) = 60.
4. Compute threshold: 4800 + 30a = 4800 + 120 = 4920.
5. Compute z = (4920 - μ)/σ = 120/60 = 2.0.
6. From table, P(0 ≤ Z ≤ 2.0) = 0.477.
7. So P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977.
8. Then 1000 × k = 977.

We'll verify with Python.

[PYTHON] import math

n = 19200 E_X = 4800 p = E_X / n # 0.25

Solve for a: p = (a+4)/32

a = p * 32 - 4 print("a =", a)

mu = E_X sigma = math.sqrt(n * p * (1 - p)) print("sigma =", sigma)

threshold = mu + 30 * a print("threshold =", threshold)

z = (threshold - mu) / sigma print("z =", z)

Table values: for z=2.0, P(0<=Z<=z) = 0.477

We'll store table as dict

table = { 0.5: 0.191, 1.0: 0.341, 1.5: 0.433, 2.0: 0.477, 2.5: 0.494, 3.0: 0.499 }

Find closest z in table (z is exactly 2.0)

if z in table: p_0_to_z = table[z] else: # Should not happen p_0_to_z = None print("P(0 <= Z <= z) from table =", p_0_to_z)

k = 0.5 + p_0_to_z print("k =", k)

result = 1000 * k print("1000 * k =", result) [/PYTHON]

[THOUGHT] The Python output confirms the calculation: a = 4, z = 2.0, k = 0.977, and 1000 × k = 977. So the final answer is 977.

\boxed{977}