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 \( \Phi(0.5) = 0.6915 \), \( \Phi(1) = 0.8413 \), \( \Phi(1.5) = 0.9332 \), \( \Phi(2) = 0.9772 \)).

• A. 1.6915
• B. 1.3085
• C. 1.1587
• D. 0.6915

Answer 1

The Correct Option is B

Step 1: Apply the Central Limit Theorem (CLT).
For a binomial random variable \( X_n \sim \text{Bin}\left(n, \frac{1}{4}\right) \), we have: \[ \mathbb{E}(X_n) = n \cdot \frac{1}{4} = \frac{n}{4}, \quad \text{Var}(X_n) = n \cdot \frac{1}{4} \cdot \left(1 - \frac{1}{4}\right) = \frac{3n}{16}. \] Using the CLT, the standardized random variable: \[ Z_n = \frac{X_n - \mathbb{E}(X_n)}{\sqrt{\text{Var}(X_n)}} = \frac{X_n - \frac{n}{4}}{\sqrt{\frac{3n}{16}}} = \frac{4X_n - n}{\sqrt{3n}} \] approximately follows a standard normal distribution \( \mathcal{N}(0, 1) \) for large \( n \).\noindent\textbf{Step 2: Analyze the first probability term.} \[ P\left( X_n \leq \frac{2n - \sqrt{3n}}{8} \right) \] Rewriting the inequality in terms of \( Z_n \): \[ \frac{4X_n - n}{\sqrt{3n}} \leq \frac{4 \cdot \frac{2n - \sqrt{3n}}{8} - n}{\sqrt{3n}} = \frac{n - \sqrt{3n} - n}{\sqrt{3n}} = -1. \] Thus, \[ P\left( X_n \leq \frac{2n - \sqrt{3n}}{8} \right) \approx \Phi(-1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587. \] \noindent\textbf{Step 3: Analyze the second probability term.} \[ P\left( \frac{n}{6} \leq X_n \leq \frac{n}{3} \right) \] For \( X_n \geq \frac{n}{6} \): \[ \frac{4X_n - n}{\sqrt{3n}} \geq \frac{4 \cdot \frac{n}{6} - n}{\sqrt{3n}} = \frac{\frac{2n}{3} - n}{\sqrt{3n}} = -\frac{n}{3\sqrt{3n}} = -\frac{\sqrt{n}}{3\sqrt{3}}. \] For large \( n \), \( -\frac{\sqrt{n}}{3\sqrt{3}} \to 0 \), so: \[ P\left( \frac{n}{6} \leq X_n \right) \approx \Phi(0) = 0.5. \] For \( X_n \leq \frac{n}{3} \): \[ \frac{4X_n - n}{\sqrt{3n}} \leq \frac{4 \cdot \frac{n}{3} - n}{\sqrt{3n}} = \frac{\frac{4n}{3} - n}{\sqrt{3n}} = \frac{n}{3\sqrt{3n}} = \frac{\sqrt{n}}{3\sqrt{3}}. \] For large \( n \), \( \frac{\sqrt{n}}{3\sqrt{3}} \to 1 \), so: \[ P\left( X_n \leq \frac{n}{3} \right) \approx \Phi(1) = 0.8413. \] Thus: \[ P\left( \frac{n}{6} \leq X_n \leq \frac{n}{3} \right) \approx \Phi(1) - \Phi(0) = 0.8413 - 0.5 = 0.3413. \] \bigskip \noindent\textbf{Step 4: Combine the probabilities.} \[ \lim_{n \to \infty} \left[ P\left( X_n \leq \frac{2n - \sqrt{3n}}{8} \right) + P\left( \frac{n}{6} \leq X_n \leq \frac{n}{3} \right) \right] \approx 0.1587 + 0.3413 = 1.3085. \]