To solve these problems, let's first clarify the scenario. We're rolling a fair six-sided die 360 times, and we're interested in the number of times a 4 is rolled. This can be modeled with a binomial distribution, where:
- \( n = 360 \) (the number of trials)
- \( p = \frac{1}{6} \) (the probability of rolling a 4)
Mean and Standard Deviation
For a binomial distribution, the mean (\( \mu \)) and standard deviation (\( \sigma \)) can be calculated as follows:
- \( \mu = n \cdot p = 360 \cdot \frac{1}{6} = 60 \)
- \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{360 \cdot \frac{1}{6} \cdot \frac{5}{6}} = \sqrt{50} \approx 7.071 \)
Part 1: Normal Approximation for \( X < 100 \)
We want to find \( P(X < 100) \).
With the continuity correction for the inequality, we adjust our value:
\[ P(X < 100) \approx P(X < 99.5) \]
Now, we standardize using the normal approximation:
\[ Z = \frac{X - \mu}{\sigma} = \frac{99.5 - 60}{\sqrt{50}} \approx \frac{39.5}{7.071} \approx 5.58 \]
Now, we look up \( Z = 5.58 \) in the standard normal table or use a calculator. The probability for \( Z \) values greater than 3 is very close to 1, hence:
\[ P(X < 100) \approx 1 \]
Part 2: Normal Approximation for \( X = 100 \)
We want to find \( P(X = 100) \).
With continuity correction for this case:
\[ P(X = 100) \approx P(99.5 < X < 100.5) \]
We can compute this using the normal approximation:
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For \( X = 99.5 \): \[ Z = \frac{99.5 - 60}{\sqrt{50}} \approx 5.58 \text{ (already calculated)} \]
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For \( X = 100.5 \): \[ Z = \frac{100.5 - 60}{\sqrt{50}} \approx \frac{40.5}{7.071} \approx 5.73 \]
Now we find the probabilities corresponding to these Z-values. As noted before, both Z-scores are extremely high, which means:
\[ P(Z < 5.58) \approx 1 \] and \[ P(Z < 5.73) \approx 1 \]
Thus:
\[ P(99.5 < X < 100.5) \approx 1 - 1 = 0 \]
Summary
- For \( P(X < 100) \), the probability is approximately 1.
- For \( P(X = 100) \), the probability is approximately 0.
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