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] I’ll use Python to:
- Express the single-trial probability of recording value 3 in terms of (a).
- Solve for (a) using the condition (E(X)=4800).
- Use the normal approximation to compute (P(X \le 4800 + 30a)) and then (1000k).
[/THOUGHT] [PYTHON] import sympy as sp import math
Step 1: Define p(a)
a = sp.symbols('a', integer=True) p = (a/6)(5/sp.Integer(16)) + ((6-a)/6)(1/sp.Integer(8)) # probability of recording 3
Step 2: Solve for a given E(X)=4800 with n=19200
n = 19200 solution = sp.solve(sp.Eq(n*p, 4800), a) solution [/PYTHON]
[THOUGHT] From the Python result, (a = 4). Then (p = \frac{4800}{19200} = 0.25), so (X \sim \text{Bin}(19200, 0.25)) with mean (4800) and variance (19200 \cdot 0.25 \cdot 0.75 = 3600), hence (\sigma = 60).
We need (P(X \le 4800 + 30a) = P(X \le 4920)). Using the normal approximation, [ Z = \frac{4920 - 4800}{60} = 2. ] The table gives (P(0 \le Z \le 2) = 0.477), so (P(Z \le 2) = 0.5 + 0.477 = 0.977). Thus (k = 0.977), and (1000k = 977). [/THOUGHT] \boxed{977}