[THOUGHT] The problem asks us to find the value of $1000 \times k$, where $k = P(X \le 4800 + 30a)$. First, I need to determine the value of $a$. Let $Y$ be the random variable representing the number of heads recorded in a single experiment. The experiment has two cases based on the die roll $D$:

1. If $D \le a$ (probability $a/6$), toss coin 5 times. Let $H_5$ be heads in 5 tosses. We record $H_5$.
2. If $D > a$ (probability $(6-a)/6$), toss coin 3 times. Let $H_3$ be heads in 3 tosses. We record $H_3$.

We are interested in the event that the recorded value is 3. Let $p$ be the probability of this event in a single experiment. $p = P(\text{recorded value} = 3) = P(D \le a) P(H_5 = 3) + P(D > a) P(H_3 = 3)$.

For a fair coin: $P(H_5 = 3) = \binom{5}{3} (0.5)^5 = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$. $P(H_3 = 3) = \binom{3}{3} (0.5)^3 = 1 \times \frac{1}{8} = \frac{1}{8}$.

So, $p = \frac{a}{6} \times \frac{5}{16} + \frac{6-a}{6} \times \frac{1}{8}$.

The experiment is repeated $n = 19200$ times. Let $X$ be the number of times the recorded value is 3. $X$ follows a Binomial distribution $B(n, p)$. The expected value is given as $E(X) = 4800$. Since $E(X) = n \times p$, we have $19200 \times p = 4800$. This allows us to solve for $p$, and subsequently for $a$.

Once $a$ is found, we need to compute $P(X \le 4800 + 30a)$. Since $n$ is large, we can approximate the Binomial distribution with a Normal distribution $N(\mu, \sigma^2)$, where $\mu = np$ and $\sigma = \sqrt{np(1-p)}$. We are given $\mu = 4800$. We need to find $P(X \le 4800 + 30a) \approx P(Z \le \frac{(4800 + 30a) - \mu}{\sigma})$. Note: The problem provides a standard normal table, suggesting this approximation is intended. Continuity correction might or might not be expected, but usually with such large numbers and the phrasing "Using the standard normal table", standard approximation is key. However, I should check if continuity correction ($X \le K \to X < K + 0.5$) significantly changes the Z-score given the table values. Usually, textbook problems align with the table values (0.5, 1.0, etc.).

Let's calculate $a$ first using Python.

[/THOUGHT] [PYTHON] from sympy import symbols, solve

Define variables

a = symbols('a')

Probabilities

p_H5_3 = 10 / 32 # 5C3 * (1/2)^5 p_H3_3 = 1 / 8 # 3C3 * (1/2)^3

Probability p of recording a 3 in one experiment

p = (a/6) * p_H5_3 + ((6-a)/6) * p_H3_3

p = (a/6) * p_H5_3 + ((6-a)/6) * p_H3_3

Expected value E(X) = n * p

n = 19200 E_X = n * p

Given E(X) = 4800

equation = E_X - 4800

Solve for a

sol_a = solve(equation, a) print(f"{sol_a=}") [/PYTHON] [THOUGHT] The code output shows sol_a=[4]. So $a = 4$. Now I have $a=4$. The probability $p$ is $4800 / 19200 = 1/4 = 0.25$. Let's verify $p$ with $a=4$: $p = \frac{4}{6} \frac{5}{16} + \frac{2}{6} \frac{1}{8} = \frac{2}{3} \frac{5}{16} + \frac{1}{3} \frac{1}{8} = \frac{5}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}$. Correct.

Now we need to compute $P(X \le 4800 + 30a)$. Substitute $a=4$: target value is $4800 + 30(4) = 4800 + 120 = 4920$. So we want $P(X \le 4920)$. $X \sim B(n, p)$ with $n=19200, p=0.25$. Mean $\mu = 4800$. Variance $\sigma^2 = np(1-p) = 4800 \times 0.75 = 3600$. Standard deviation $\sigma = \sqrt{3600} = 60$.

We want $P(X \le 4920)$. Using Normal approximation: $Z = \frac{X - \mu}{\sigma} = \frac{4920 - 4800}{60} = \frac{120}{60} = 2.0$.

We need to check if continuity correction is needed. With continuity correction: $P(X \le 4920) \approx P(Z \le \frac{4920.5 - 4800}{60}) = P(Z \le 2.0083)$. Without continuity correction: $P(Z \le 2.0)$. Looking at the provided table: $z=2.0 \implies P(0 \le Z \le 2.0) = 0.477$. The table gives $P(0 \le Z \le z)$. We want $P(Z \le 2.0) = P(Z \le 0) + P(0 \le Z \le 2.0) = 0.5 + 0.477 = 0.977$.

If I used continuity correction, $z \approx 2.01$, which is not in the table. The table has steps of 0.5. This strongly suggests we should use the integer value directly resulting in a clean Z-score found in the table. So $z = 2.0$. $k = 0.977$. We need $1000 \times k$. $1000 \times 0.977 = 977$.

Let me double check the calculations in Python to be absolutely sure. [/THOUGHT] [PYTHON] import math

n = 19200 a = 4 p = 0.25 mu = n * p sigma = math.sqrt(n * p * (1 - p))

target_X = 4800 + 30 * a z_score = (target_X - mu) / sigma

print(f"{mu=}") print(f"{sigma=}") print(f"{target_X=}") print(f"{z_score=}") [/PYTHON] [THOUGHT] The Python output confirms: $\mu = 4800.0$ $\sigma = 60.0$ Target $X = 4920$ $z = \frac{4920 - 4800}{60} = 2.0$

The table provided is: $z=2.0 \rightarrow P(0 \le Z \le 2.0) = 0.477$ We want $P(X \le 4920) \approx P(Z \le 2.0)$. Since the normal distribution is symmetric, $P(Z \le 0) = 0.5$. So $P(Z \le 2.0) = 0.5 + 0.477 = 0.977$. $k = 0.977$. The question asks for $1000 \times k$. $1000 \times 0.977 = 977$.

Is there any ambiguity? "Let $X$ be the number of times the recorded value is 3." -> Binomial variable. "Compute $P(X \le 4800 + 30a)$". "Using the standard normal table". The table has specific values. $z=2.0$ is exactly there. If I used continuity correction, $z = 120.5 / 60 = 2.00833$. The table doesn't support 2.008. It supports 2.0. So the intended solution is without continuity correction or the problem is designed such that the difference is negligible for the "lookup" nature of the task. Given the discrete table steps, exact $z=2.0$ is the intended path.

Final check of the question text: "Compute $1000 \times k$." $k$ is the probability. Result is 977.

Let's re-read to ensure no misinterpretation of "recorded value".

• Roll die.
• If $\le a$: toss 5 times. Record heads.
• If $> a$: toss 3 times. Record heads.
• $X$ is count of experiments where recorded value is 3. This matches my derivation of $p$.

Let's re-verify the calculation of $p$ for $a=4$. $P(D \le 4) = 4/6 = 2/3$. $P(D > 4) = 2/6 = 1/3$. If $D \le 4$, toss 5 coins. $P(3H) = \binom{5}{3} (1/2)^5 = 10/32 = 5/16$. If $D > 4$, toss 3 coins. $P(3H) = \binom{3}{3} (1/2)^3 = 1/8$. $p = (2/3)(5/16) + (1/3)(1/8) = 10/48 + 1/24 = 5/24 + 1/24 = 6/24 = 1/4$. Correct.

$E(X) = 19200 \times (1/4) = 4800$. Matches given $E(X)$. So $a=4$ is correct.

Target $X \le 4800 + 30(4) = 4920$. $Z = (4920 - 4800) / \sqrt{19200 \times 0.25 \times 0.75}$. $\sigma = \sqrt{4800 \times 0.75} = \sqrt{3600} = 60$. $Z = 120 / 60 = 2$.

Table value for $Z=2.0$ is $0.477$ (for $0 \le Z \le z$). Total probability $P(Z \le 2.0) = 0.5 + 0.477 = 0.977$. $1000 k = 977$.

The solution seems robust.

One minor detail: "natural number $a \le 6$". If $a$ was different, say $a=1$. $p = (1/6)(5/16) + (5/6)(1/8) = 5/96 + 5/48 = 15/96 = 5/32 \approx 0.156$. $E(X) = 19200 \times 5/32 = 600 \times 5 = 3000 \ne 4800$. So $a=4$ is unique.

The steps are clear.

1. Formulate $p$ in terms of $a$.
2. Equate $np$ to 4800 to find $a$.
3. Calculate $\mu$ and $\sigma$.
4. Calculate $z$-score for the bound.
5. Look up probability in the table.
6. Multiply by 1000.

The value $k$ is the probability. $k = 0.977$. $1000k = 977$.

I will output the final answer. [/THOUGHT] [THOUGHT] The problem asks for $1000 \times k$ where $k = P(X \le 4800 + 30a)$. From the Python code, $a=4$, $\mu=4800$, $\sigma=60$. The bound is $4800 + 30(4) = 4920$. The Z-score is $(4920 - 4800) / 60 = 2.0$. From the table, $P(0 \le Z \le 2.0) = 0.477$. Thus, $P(Z \le 2.0) = 0.5 + 0.477 = 0.977$. $k = 0.977$. $1000 \times k = 977$. [/THOUGHT]

\boxed{977}

The random variable $X$ represents the number of times the recorded value is 3 in $n=19200$ trials. Let $p$ be the probability of recording a 3 in a single trial. The experiment involves rolling a die (outcome $D$) and then tossing a coin.

• If $D \leq a$, toss 5 times. Probability of 3 heads is $\binom{5}{3}(0.5)^5 = \frac{10}{32} = \frac{5}{16}$.
• If $D > a$, toss 3 times. Probability of 3 heads is $\binom{3}{3}(0.5)^3 = \frac{1}{8}$. The probability of rolling $D \leq a$ is $a/6$, and $D > a$ is $(6-a)/6$. Thus, $p = \frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8}$.

Given $E(X) = 4800$ and $E(X) = np = 19200p$, we have: $19200p = 4800 \implies p = \frac{4800}{19200} = \frac{1}{4}$.

We solve for $a$: $\frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8} = \frac{1}{4}$ Multiply by 48 to clear denominators: $8a \cdot \frac{5}{16} + 8(6-a) \cdot \frac{1}{8} = 12$ $\frac{5a}{2} + (6-a) = 12$ $2.5a + 6 - a = 12$ $1.5a = 6$ $a = 4$.

We need to compute $k = P(X \leq 4800 + 30a)$. Substitute $a=4$: $k = P(X \leq 4800 + 120) = P(X \leq 4920)$.

We approximate $X$ using a normal distribution $N(\mu, \sigma^2)$. $\mu = 4800$. $\sigma = \sqrt{np(1-p)} = \sqrt{4800(1 - 0.25)} = \sqrt{4800 \cdot 0.75} = \sqrt{3600} = 60$.

Standardize the value $X = 4920$: $z = \frac{4920 - 4800}{60} = \frac{120}{60} = 2.0$.

Using the provided table for $z=2.0$: $P(0 \leq Z \leq 2.0) = 0.477$. Since we want $P(Z \leq 2.0)$, we add the probability of the lower half: $P(Z \leq 2.0) = 0.5 + 0.477 = 0.977$.

Thus, $k = 0.977$. The value to compute is $1000 \times k = 977$.

\boxed{977}