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] To solve this problem, I'll follow these steps:
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Determine the probability of recording 3 heads:
- If the die roll is ≤ (a), the probability of getting 3 heads from 5 coin tosses is calculated using the binomial distribution.
- If the die roll is > (a), the probability of getting 3 heads from 3 coin tosses is also calculated using the binomial distribution.
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Calculate the expected number of times the recorded value is 3:
- The expected value (\text{E}(X)) is given as 4800. This will help us find (a).
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Find the variance of (X):
- The variance of the number of times the recorded value is 3 can be calculated using the law of total variance.
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Standardize the value (4800 + 30a) and use the standard normal table to find the probability:
- Use the Z-score formula to standardize (4800 + 30a).
- Find the corresponding probability from the given standard normal table.
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Calculate (1000 \times k) where (k) is the probability found.
Let's start by calculating the probabilities for the number of heads:
Probability Calculations:
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For the die roll ≤ (a), we toss the coin 5 times. The probability of getting exactly 3 heads: [ P(H = 3 \mid \text{Die} \leq a) = \binom{5}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^2 = 10 \cdot \frac{1}{32} = \frac{10}{32} = \frac{5}{16} ]
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For the die roll > (a), we toss the coin 3 times. The probability of getting exactly 3 heads: [ P(H = 3 \mid \text{Die} > a) = \binom{3}{3} \left(\frac{1}{2}\right)^3 = 1 \cdot \frac{1}{8} = \frac{1}{8} ]
Expected Value Calculation:
Given (\text{E}(X) = 4800) and the experiment is repeated 19200 times, we have: [ \text{E}(X) = 19200 \cdot \left( \frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8} \right) = 4800 ]
Solve for (a): [ 19200 \cdot \left( \frac{a}{6} \cdot \frac{5}{16} + \frac{6-a}{6} \cdot \frac{1}{8} \right) = 4800 ] [ 19200 \cdot \left( \frac{5a}{96} + \frac{6-a}{48} \right) = 4800 ] [ 19200 \cdot \left( \frac{5a}{96} + \frac{12-2a}{96} \right) = 4800 ] [ 19200 \cdot \frac{5a + 12 - 2a}{96} = 4800 ] [ 19200 \cdot \frac{3a + 12}{96} = 4800 ] [ 3a + 12 = \frac{4800 \cdot 96}{19200} ] [ 3a + 12 = 24 ] [ 3a = 12 ] [ a = 4 ]
Variance Calculation:
Let (X_i) be the indicator random variable for the (i)-th experiment where the recorded value is 3. Then (\text{Var}(X) = 19200 \cdot \text{Var}(X_i)).
[ \text{Var}(X_i) = \text{E}(X_i^2) - (\text{E}(X_i))^2 ] [ \text{E}(X_i) = \frac{4}{6} \cdot \frac{5}{16} + \frac{2}{6} \cdot \frac{1}{8} = \frac{1}{3} \cdot \frac{5}{16} + \frac{1}{3} \cdot \frac{1}{8} = \frac{5}{48} + \frac{6}{48} = \frac{11}{48} ]
For (\text{E}(X_i^2)), note (X_i) is 0 or 1: [ \text{E}(X_i^2) = \text{E}(X_i) = \frac{11}{48} ] Thus, [ \text{Var}(X_i) = \frac{11}{48} - \left(\frac{11}{48}\right)^2 = \frac{11}{48} - \frac{121}{2304} = \frac{528 - 121}{2304} = \frac{407}{2304} ] [ \text{Var}(X) = 19200 \cdot \frac{407}{2304} = 19200 \cdot \frac{407}{2304} \approx 3325 ]
Standardizing (4800 + 30a):
For (a = 4), we have (4800 + 30 \times 4 = 4800 + 120 = 4920).
The Z-score is: [ Z = \frac{4920 - 4800}{\sqrt{3325}} = \frac{120}{\sqrt{3325}} \approx \frac{120}{57.67} \approx 2.08 ]
Using the standard normal table, (P(0 \leq Z \leq 2.08)) is closest to (P(0 \leq Z \leq 2.0) = 0.477).
The cumulative probability up to (Z = 2.08) is: [ P(Z \leq 2.08) \approx 0.5 + 0.477 = 0.977 ]
Thus, (k = 0.977).
Final Calculation: [ 1000 \times k = 1000 \times 0.977 = 977 ]
[/THOUGHT] \boxed{977}