Problem Analysis #38

The value of $a$ is determined by the given expected value $\text{E}(X) = 4800$. For a binomial random variable $X$ with $n = 19200$ trials, $\text{E}(X) = n \cdot p$, so $19200 \cdot p = 4800$ gives $p = 0.25$. The probability $p$ depends on $a$ as follows:

[ p = \left(\frac{a}{6}\right) \cdot \frac{5}{16} + \left(\frac{6 - a}{6}\right) \cdot \frac{1}{8} = 0.25 ]

Solving for $a$:

[ \frac{5a}{96} + \frac{6 - a}{48} = 0.25 ]

Multiplying through by 96:

[ 5a + 2(6 - a) = 24 \implies 5a + 12 - 2a = 24 \implies 3a = 12 \implies a = 4 ]

With $a = 4$, compute $4800 + 30a = 4800 + 120 = 4920$. Thus, $P(X \leq 4920)$ is needed.

$X$ is binomial with $n = 19200$ and $p = 0.25$. Using the normal approximation, the mean $\mu = n \cdot p = 4800$ and variance $\sigma^2 = n \cdot p \cdot (1 - p) = 3600$, so $\sigma = 60$. Without continuity correction, $z = \frac{4920 - 4800}{60} = 2.0$.

From the standard normal table, $P(0 \leq Z \leq 2.0) = 0.477$, so $P(Z \leq 2.0) = 0.5 + 0.477 = 0.977$. Thus, $k = 0.977$.

Finally, $1000 \times k = 1000 \times 0.977 = 977$.

\boxed{977}

[THOUGHT] We are to find the value of (a) that satisfies the given expected value condition. Since (a) is a natural number with (a \leq 6), we will iterate through possible values of (a) from 1 to 6 and compute the corresponding probability (p) defined by the experiment. This will help verify which (a) yields (p = 0.25).

[PYTHON] a_values = [1, 2, 3, 4, 5, 6] results = [] for a in a_values: p = (a/6) * (5/16) + ((6 - a)/6) * (1/8) results.append((a, p))

for a, p in results: print(f"a={a}, p={p:.6f}") [/PYTHON]

\boxed{977}