Problem Analysis #38

[THOUGHT] First, determine a such that E(X)=4800. n=19200. p = (a+4)/32 from derivation. np=19200*(a+4)/32=600(a+4)=4800 ⇒ a+4=8 ⇒ a=4. Verify with code by computing for a=1 to 6.

Then μ=4800, p=0.25, σ=sqrt(192000.250.75)=60.

Then 4800+30*4=4920. For normal approx P(X≤4920) ≈ Φ( (4920.5-4800)/60 ) or without cc.

But since table and exact 120/60=2, likely z=2, Φ(2)=0.5+0.477=0.977, 1000k=977.

Use Python to confirm a, μ, σ, and perhaps compute exact cdf with scipy.stats.norm.cdf for reference, but final from table. [/THOUGHT] [PYTHON] import numpy as np from scipy.stats import binom, norm

n = 19200 for a in range(1,7): p = (a + 4) / 32 mu = n * p print(f'a={a}, p={p}, mu={mu}')

p = 8/32 var = n * p * (1-p) sigma = np.sqrt(var) print(f'sigma={sigma}')

threshold = 4800 + 30*4 print(f'threshold={threshold}')

z_no_cc = (threshold - 4800) / sigma z_cc = (threshold + 0.5 - 4800) / sigma print(f'z no cc={z_no_cc}, z cc={z_cc}')

phi2 = 0.5 + 0.477 print(f'table approx={phi2}')

k_scipy_cc = norm.cdf(z_cc) k_scipy_no = norm.cdf(z_no_cc) print(f'scipy cc={k_scipy_cc}, 1000k={1000k_scipy_cc}') print(f'scipy no cc={k_scipy_no}, 1000k={1000k_scipy_no}') [/PYTHON]

[THOUGHT] From output, a=4, sigma=60, threshold=4920, z=2.0. Table: P(Z≤2)=0.5+0.477=0.977, so 1000k=977. [/THOUGHT] $\boxed{977}$

Explanation

The reasoning follows a step-by-step process to solve the problem using the provided Python output as ground truth.

1. Identify $a$: The output shows that for $a=4$, $\mu = 19200 \times p = 4800$, where $p = (a + 4)/32 = 8/32 = 0.25$, matching $\text{E}(X) = 4800$.

2. Compute variance and standard deviation: With $n=19200$ and $p=0.25$, variance = $n \times p \times (1-p) = 19200 \times 0.25 \times 0.75 = 3600$, so $\sigma = \sqrt{3600} = 60$.

3. Determine the threshold: $4800 + 30a = 4800 + 30 \times 4 = 4920$.

4. Calculate z-score for normal approximation: $z = \frac{4920 - 4800}{60} = \frac{120}{60} = 2.0$.

5. Use standard normal table: The table gives $\text{P}(0 \leq Z \leq 2.0) = 0.477$, so $\text{P}(Z \leq 2.0) = 0.5 + 0.477 = 0.977$.

6. Compute final value: $k = 0.977$, thus $1000 \times k = 977$.