Let be a natural number with . Using one die and one coin, perform the following experiment.\begin{center}\fbox{\begin{minipage}{0.9\textwidth}Roll the die once.\quad-- If the outcome is less than or equal to , toss the coin times and record the number of heads.\quad-- If the outcome is greater than , toss the coin times and record the number of heads.\end{minipage}}\end{center}Repeat this experiment times, and let be the number of times the recorded value is . Suppose that . Using the standard normal table on the right, computeand let this value be . Compute 1000 $$\times$$ k. [4 points]\begin{center}\begin{tabular}{|c|c|}\hline & \\hline0.5 & 0.191 \\hline1.0 & 0.341 \\hline1.5 & 0.433 \\hline2.0 & 0.477 \\hline2.5 & 0.494 \\hline3.0 & 0.499 \\hline\end{tabular}\end{center}\
[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.
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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).
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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).
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Determine the threshold: (4800 + 30a = 4800 + 30 \times 4 = 4920).
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Calculate z-score for normal approximation: (z = \frac{4920 - 4800}{60} = \frac{120}{60} = 2.0).
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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).
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Compute final value: (k = 0.977), thus (1000 \times k = 977).