To solve the problem, we begin by analyzing the structure of the experiment and how it relates to the expectation $\text{E}(X) = 4800$.

Step 1: Determine the value of $a$

We are told that:

• A die is rolled.
• If the outcome is less than or equal to $a$, we toss a coin 5 times and record the number of heads.
• If the outcome is greater than $a$, we toss a coin 3 times and record the number of heads.
• This entire process is repeated $n = 19200$ times.
• Let $X$ be the number of times the recorded value is exactly 3 (i.e., 3 heads in the coin tosses).

We're given that $\text{E}(X) = 4800$. Our goal is to compute P(X $$\leq$$ 4800 + 30a), which will be some value $k$, and then find $1000k$.

Step 2: Compute the probability $p$ of success in one trial

Let’s compute the probability $p$ that in a single trial, the number of heads is exactly 3.

• With probability $$\frac{a}{6}$$, the die roll is $$\leq$$ a, so we toss the coin 5 times.
  • The probability of getting exactly 3 heads in 5 tosses is:

• With probability $$\frac{6 - a}{6}$$, the die roll is $> a$, so we toss the coin 3 times.
  • The probability of getting exactly 3 heads in 3 tosses is:

Thus, the total probability $p$ of recording a 3 in one trial is:

Get a common denominator (96):

But earlier we derived:

Equating the two expressions for $p$:

Cross-multiplying:

So let's recheck the derivation.

Actually, from earlier:

Given \text{E}(X) = 19200 $$\cdot$$ p = 4800, we get:

So:

✅ Confirmed: $a = 4$

Step 3: Use Normal Approximation for $X$

Now, X $$\sim$$ \text{Binomial}(n = 19200, p = $$\frac{1}{4}$$)

• Mean: $$\mu$$ = np = 19200 $$\cdot$$ $$\frac{1}{4}$$ = 4800
• Variance: $$\sigma$$^2 = np(1 - p) = 19200 $$\cdot$$ $$\frac{1}{4}$$ $$\cdot$$ $$\frac{3}{4}$$ = 3600
• Standard deviation: $$\sigma$$ = $$\sqrt{3600}$$ = 60

We are asked to compute:

Use the normal approximation:

So,

From the standard normal table:

Thus, $k = 0.977$

Step 4: Final Calculation

✅ Final Answer:

[THOUGHT] We need to solve for $a$, confirm it's 4, then compute the z-score for X $$\leq$$ 4800 + 30a, use the normal approximation, and find $1000k$.

First, calculate $p$, the probability of success (recording 3 heads) in one trial. The expectation \text{E}(X) = 19200 $$\cdot$$ p = 4800 \implies p = $$\frac{4800}{19200}$$ = 0.25.

The probability $p$ is also given by:

Wait, earlier calculation was wrong. Let's recalculate $p$ correctly:

For $a$ (die ≤ $a$): probability of 3 heads in 5 tosses is $\binom{5}{3}/2^5 = 10/32 = 5/16 = 0.3125$.

For $6 - a$ (die > $a$): probability of 3 heads in 3 tosses is $1/8 = 0.125$.

Thus,

Convert to common denominator 96:

Wait, but earlier we had $p = (a + 4)/32$. Let's check with $a = 4$:

$(2*4 + 3)/24 = 11/24 ≈ 0.458$, which is wrong. I must have made a mistake in the algebra.

Let's recalculate $p$ correctly:

$$\frac{a}{6}$$ $$\cdot$$ $$\frac{10}{32}$$ + $$\frac{6 - a}{6}$$ $$\cdot$$ $$\frac{1}{8}$$ = $$\frac{10a}{192}$$ + $$\frac{6 - a}{48}$$ = $$\frac{10a + 4(6 - a)}{96}$$ = $$\frac{10a + 24 - 4a}{96}$$ = $$\frac{6a + 24}{96}$$ = $$\frac{a + 4}{16}$$.

Ah, there we go. I missed a factor earlier. So $p = (a + 4)/16$? Wait no:

Wait $10a/192 = 5a/96$, and $ (6 - a)/48 = 12/96? No, 482=96, so (6 - a)/48 = 2(6 - a)/96 = (12 - 2a)/96. So total p = (5a + 12 - 2a)/96 = (3a + 12)/96 = (a + 4)/32. Yes, that's correct. So 3a + 12 = 3(a + 4), so 3(a + 4)/96 = (a + 4)/32. So p = (a + 4)/32.

Given p = 0.25, then (a + 4)/32 = 0.25 → a + 4 = 8 → a = 4. Correct.

Now, 30a = 120, so the upper limit is 4800 + 120 = 4920.

Mean μ = 4800, variance σ² = 19200 * 0.25 * 0.75 = 3600, so σ = 60.

Z = (4920 - 4800)/60 = 120/60 = 2.0.

From the table, P(0 ≤ Z ≤ 2.0) = 0.477, so P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977.

Thus, k = 0.977, so 1000k = 977.

Let's use Python to confirm the z-score and probability.

[PYTHON]

Confirm a

n = 19200 E_X = 4800 p = E_X / n # p = 0.25 a = 32 * p - 4 print("a =", a) # Should be 4.0

Calculate upper limit

upper_limit = E_X + 30 * a print("Upper limit (4800 + 30a) =", upper_limit) # 4920.0

Calculate z-score

import math mu = E_X sigma = math.sqrt(n * p * (1 - p)) z = (upper_limit - mu) / sigma print("z-score =", z) # Should be 2.0

From table, P(0 ≤ Z ≤ 2.0) = 0.477, so P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977

k = 0.5 + 0.477 print("k =", k) # 0.977 print("1000k =", 1000 * k) # 977.0 [/PYTHON]

[THOUGHT] The problem requires finding $a$, confirming it's 4, then computing the z-score for X $$\leq$$ 4800 + 30a, using the standard normal table to find the cumulative probability, and then scaling by 1000.

First, calculate $a$:

• \text{E}(X) = 19200 $$\cdot$$ p = 4800 \implies p = 0.25
• p = $$\frac{a + 4}{32}$$ \implies 0.25 = $$\frac{a + 4}{32}$$ \implies a + 4 = 8 \implies a = 4

Now, compute $4800 + 30a = 4800 + 120 = 4920$.

For the normal approximation:

• Mean $$\mu$$ = 4800
• Variance $$\sigma$$^2 = 19200 $$\cdot$$ 0.25 $$\cdot$$ 0.75 = 3600 \implies $$\sigma$$ = 60
• Z-score: z = $$\frac{4920 - 4800}{60}$$ = 2.0

From the table, \text{P}(0 $$\leq$$ Z $$\leq$$ 2.0) = 0.477 \implies \text{P}(Z $$\leq$$ 2.0) = 0.5 + 0.477 = 0.977

Thus, $k = 0.977 \implies 1000k = 977$.

[PYTHON] import math

Given values

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

Calculate a

a = 32 * p - 4 # From p = (a + 4)/32 print("a =", a) # Output: 4.0

Upper limit

upper_limit = E_X + 3 </think>

[THOUGHT]

We need to solve for $a$, confirm it's 4, then compute the z-score for X $$\leq$$ 4800 + 30a, use the normal approximation, and find $ 1000.

First, calculate $a$:

• \text{E}(X) = 19200 $$\cdot$$ p = 4800 \implies p = $$\frac{4800}{19200}$$ = 0.25
• p = $$\frac{a}{6}$$ $$\cdot$$ \binom{5}{3} \left($$\frac{1}{2}$$\right)^5 + $$\frac{6 - a}{6}$$ $$\cdot$$ \binom{3}{3} \left($$\frac{1}{2}$$\right)^3
• p = $$\frac{a}{6}$$ $$\cdot$$ $$\frac{10}{32}$$ + $$\frac{6 - a}{6}$$ $$\cdot$$ $$\frac{1}{8}$$ = $$\frac{5a}{96}$$ + $$\frac{6 - a}{48}$$ = $$\frac{5a + 2(6 - a)}{96}$$ = $$\frac{3a + 12}{96}$$
• Wait, earlier correct expression: p = $$\frac{a + 4}{32}$$
• So $$\frac{a + 4}{32}$$ = 0.25 \implies a + 4 = 8 \implies a = 4

Now, $30a = 120$, so the upper limit is $4800 + 120 = 4920$.

For normal approximation:

• Mean $$\mu$$ = 4800
• Variance $$\sigma$$^2 = 19200 $$\cdot$$ 0.25 $$\cdot$$ 0.75 = 3600 \implies $$\sigma$$ = 60
• Z-score: z = $$\frac{4920 - 4800}{60}$$ = 2.0

From the table, \text{P}(0 $$\leq$$ Z $$\leq$$ 2.0) = 0.477 \implies \text{P}(Z $$\leq$$ 2.0) = 0.5 + 0.477 = 0.977

Thus, $k = 0.977 \implies 1000k = 977$

Let's verify with Python: [PYTHON] import math

Given values

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

Calculate a from p = (a + 4)/32

a = 32 * p - 4 print("a =", a) # Output: 4.0

Upper limit

upper_limit = E_X + 30 * a print("Upper limit (4800 + 30a) =", upper_limit) # Output: 4920.0

Calculate z-score

mu = E_X sigma = math.sqrt(n * p * (1 - p)) z = (upper_limit - mu) / sigma print("z-score =", z) # Output: 2.0

From table, P(0 ≤ Z ≤ 2.0) = 0.477

k = 0.5 + 0.477 print("k =", k) # Output: 0.977

Final answer

result = 1000 * k print("1000k =", result) # Output: 977.0 [/PYTHON] [THOUGHT] All calculations confirm $a = 4$, z-score $= 2.0$, $k = 0.977$, and $1000k = 977$.

$\boxed{977}$

[THOUGHT] First, I need to find the value of $a$ using the given expectation $\text{E}(X) = 4800$. Then, I'll compute the required probability using the normal approximation.

[PYTHON] import math

Given values

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

Calculate a from p = (a + 4)/32

a = 32 * p - 4 print(f"a = {a}")

Upper limit for X

upper_limit = E_X + 30 * a print(f"Upper limit (4800 + 30a) = {upper_limit}")

Calculate z-score

mu = E_X sigma = math.sqrt(n * p * (1 - p)) z = (upper_limit - mu) / sigma print(f"z-score = {z}")

From the table, P(0 ≤ Z ≤ 2.0) = 0.477

So P(Z ≤ 2.0) = 0.5 + 0.477 = 0.977

k = 0.5 + 0.477 print(f"k = {k}") print(f"1000k = {1000 * k}") [/PYTHON]

The value of $a$ is determined by solving $$\frac{a + 4}{32}$$ = 0.25, yielding $a = 4$. The upper limit for $X$ is 4800 + 30 $$\times$$ 4 = 4920. Using the normal approximation with mean $4800$ and standard deviation $60$, the z-score is $$\frac{4920 - 4800}{60}$$ = 2.0. From the standard normal table, \text{P}(0 $$\leq$$ Z $$\leq$$ 2.0) = 0.477, so \text{P}(Z $$\leq$$ 2.0) = 0.5 + 0.477 = 0.977. Thus, $k = 0.977$ and $1000k = 977$.

$\boxed{977}$