Problem Analysis #38

[THOUGHT] To solve this problem, I'll follow these steps:

1. Determine the probability of recording 3 heads:

   • If the die roll is ≤ $a$, the probability of getting 3 heads from 5 coin tosses is calculated using the binomial distribution.
   • If the die roll is > $a$, the probability of getting 3 heads from 3 coin tosses is also calculated using the binomial distribution.

2. Calculate the expected number of times the recorded value is 3:

   • The expected value $\text{E}(X)$ is given as 4800. This will help us find $a$.

3. Find the variance of $X$:

   • The variance of the number of times the recorded value is 3 can be calculated using the law of total variance.

4. Standardize the value $4800 + 30a$ and use the standard normal table to find the probability:

   • Use the Z-score formula to standardize $4800 + 30a$.
   • Find the corresponding probability from the given standard normal table.

5. Calculate $1000 \times k$ where $k$ is the probability found.

Let's start by calculating the probabilities for the number of heads:

Probability Calculations:

• For the die roll ≤ $a$, we toss the coin 5 times. The probability of getting exactly 3 heads:

$$P(H = 3 \mid \text{Die} \leq a) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \cdot \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$$

• For the die roll > $a$, we toss the coin 3 times. The probability of getting exactly 3 heads:

$$P(H = 3 \mid \text{Die} > a) = \binom{3}{3} \left(\frac{1}{2}\right)^3 = 1 \cdot \frac{1}{8} = \frac{1}{8}$$

Expected Value Calculation:

Given $\text{E}(X) = 4800$ and the experiment is repeated 19200 times, we have:

$$\text{E}(X) = 19200 \cdot \left( \frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8} \right) = 4800$$

Solve for $a$:

$$19200 \cdot \left( \frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8} \right) = 4800$$ $$19200 \cdot \left( \frac{5a}{96} + \frac{6-a}{48} \right) = 4800$$ $$19200 \cdot \left( \frac{5a}{96} + \frac{12-2a}{96} \right) = 4800$$ $$19200 \cdot \frac{5a + 12 - 2a}{96} = 4800$$ $$19200 \cdot \frac{3a + 12}{96} = 4800$$ $$3a + 12 = \frac{4800 \cdot 96}{19200}$$ $$3a + 12 = 24$$ $$3a = 12$$ $$a = 4$$

Variance Calculation:

Let $X_i$ be the indicator random variable for the $i$-th experiment where the recorded value is 3. Then $\text{Var}(X) = 19200 \cdot \text{Var}(X_i)$.

$$\text{Var}(X_i) = \text{E}(X_i^2) - (\text{E}(X_i))^2$$ $$\text{E}(X_i) = \frac{4}{6} \cdot \frac{5}{16} + \frac{2}{6} \cdot \frac{1}{8} = \frac{1}{3} \cdot \frac{5}{16} + \frac{1}{3} \cdot \frac{1}{8} = \frac{5}{48} + \frac{6}{48} = \frac{11}{48}$$

For $\text{E}(X_i^2)$, note $X_i$ is 0 or 1:

$$\text{E}(X_i^2) = \text{E}(X_i) = \frac{11}{48}$$

Thus,

$$\text{Var}(X_i) = \frac{11}{48} - \left(\frac{11}{48}\right)^2 = \frac{11}{48} - \frac{121}{2304} = \frac{528 - 121}{2304} = \frac{407}{2304}$$ $$\text{Var}(X) = 19200 \cdot \frac{407}{2304} = 19200 \cdot \frac{407}{2304} \approx 3325$$

Standardizing $4800 + 30a$:

For $a = 4$, we have $4800 + 30 \times 4 = 4800 + 120 = 4920$.

The Z-score is:

$$Z = \frac{4920 - 4800}{\sqrt{3325}} = \frac{120}{\sqrt{3325}} \approx \frac{120}{57.67} \approx 2.08$$

Using the standard normal table, $P(0 \leq Z \leq 2.08)$ is closest to $P(0 \leq Z \leq 2.0) = 0.477$.

The cumulative probability up to $Z = 2.08$ is:

$$P(Z \leq 2.08) \approx 0.5 + 0.477 = 0.977$$

Thus, $k = 0.977$.

Final Calculation:

$$1000 \times k = 1000 \times 0.977 = 977$$

[/THOUGHT] $\boxed{977}$