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] Find the per-trial probability p that the recorded value equals 3 as a function of a, use E(X)=19200p=4800 to determine a. Then approximate X~Normal(mean=4800, sd=sqrt(19200p(1-p))) and compute P(X<=4800+30a) via the provided standard normal table. [/THOUGHT] [PYTHON] import math from fractions import Fraction
solve for a given E(X)=4800 with n=19200
n = 19200 target_mean = 4800 p_target = Fraction(target_mean, n)
def p_of_a(a): # probability recorded value is 3 p5 = Fraction(math.comb(5,3), 25) # 3 heads in 5 tosses p3 = Fraction(1, 23) # 3 heads in 3 tosses return Fraction(a,6)*p5 + Fraction(6-a,6)*p3
vals = {a: p_of_a(a) for a in range(1,7)} a_sol = [a for a,p in vals.items() if p == p_target] print("p_target =", p_target) print("p(a) values:", vals) print("a solving:", a_sol)
a = a_sol[0] p = float(p_of_a(a)) mu = np sigma = math.sqrt(np*(1-p)) threshold = 4800 + 30*a z = (threshold - mu)/sigma print("a =", a, "mu =", mu, "sigma =", sigma, "threshold =", threshold, "z =", z)
Using the provided table: for z=2.0, P(0<=Z<=z)=0.477
k = 0.5 + 0.477 # P(Z<=2) print("k =", k, "1000k =", 1000k) [/PYTHON]
[THOUGHT] From the output, (a=4), (z=2.0), so (k=0.977) and (1000k=977). [ \boxed{977} ]