Question:
For \( n \in \mathbb{N} \), let \( X_n \) be a random variable having the \( \text{Bin}(n, \frac{1}{4}) \) distribution. Then
\[\lim_{n \to \infty} \left[ P(X_n \leq 2n - \frac{\sqrt{3n}}{8}) + P\left(\frac{n}{6} \leq X_n \leq \frac{n}{3}\right) \right]\]
is equal to (You may use \( \Phi(0.5) = 0.6915 \), \( \Phi(1) = 0.8413 \), \( \Phi(1.5) = 0.9332 \), \( \Phi(2) = 0.9772 \)).
For \( n \in \mathbb{N} \), let \( X_n \) be a random variable having the \( \text{Bin}(n, \frac{1}{4}) \) distribution. Then
\[\lim_{n \to \infty} \left[ P(X_n \leq 2n - \frac{\sqrt{3n}}{8}) + P\left(\frac{n}{6} \leq X_n \leq \frac{n}{3}\right) \right]\]
is equal to (You may use \( \Phi(0.5) = 0.6915 \), \( \Phi(1) = 0.8413 \), \( \Phi(1.5) = 0.9332 \), \( \Phi(2) = 0.9772 \)).
\[\lim_{n \to \infty} \left[ P(X_n \leq 2n - \frac{\sqrt{3n}}{8}) + P\left(\frac{n}{6} \leq X_n \leq \frac{n}{3}\right) \right]\]
is equal to (You may use \( \Phi(0.5) = 0.6915 \), \( \Phi(1) = 0.8413 \), \( \Phi(1.5) = 0.9332 \), \( \Phi(2) = 0.9772 \)).
Updated On: Oct 1, 2024
- 1.6915
- 1.3085
- 1.1587
- 0.6915
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The Correct Option is B
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The correct option is (B): 1.3085
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