Question:
Let \( \{X_n\}_{n \geq 1} \) be a sequence of i.i.d. random variables with the probability mass function
Let \( \bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i, n = 1, 2, \dots \). If \( \lim_{n \to \infty} P(m \leq \bar{X}_n \leq M) = 1 \), then possible values of \( m \) and \( M \) are
Let \( \{X_n\}_{n \geq 1} \) be a sequence of i.i.d. random variables with the probability mass function
Let \( \bar{X}_n = \frac{1}{n} \sum_{i=1}^n X_i, n = 1, 2, \dots \). If \( \lim_{n \to \infty} P(m \leq \bar{X}_n \leq M) = 1 \), then possible values of \( m \) and \( M \) are
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The Law of Large Numbers ensures that the sample mean converges to the expected value as the sample size increases.
Updated On: Dec 15, 2025
- \( m = 2.1, M = 3.1 \)
- \( m = 3.2, M = 4.1 \)
- \( m = 4.2, M = 5.7 \)
- \( m = 6.1, M = 7.1 \)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the distribution.
The random variable \( X_n \) takes values 4 and 8 with probabilities \( \frac{1}{4} \) and \( \frac{3}{4} \), respectively.
Step 2: Finding the mean and variance.
The expected value \( \mathbb{E}[X_n] \) is: \[ \mathbb{E}[X_n] = 4 \times \frac{1}{4} + 8 \times \frac{3}{4} = 7 \] The variance \( \text{Var}(X_n) \) is: \[ \text{Var}(X_n) = \mathbb{E}[X_n^2] - (\mathbb{E}[X_n])^2 = \left( 4^2 \times \frac{1}{4} + 8^2 \times \frac{3}{4} \right) - 7^2 = 1.75 \]
Step 3: Using the Law of Large Numbers.
As \( n \to \infty \), by the law of large numbers, \( \bar{X}_n \) converges to \( \mathbb{E}[X_n] = 7 \). Hence, the values of \( m \) and \( M \) should be chosen so that the probability of \( \bar{X}_n \) falling between \( m \) and \( M \) converges to 1. The correct answer is \( m = 6.1 \) and \( M = 7.1 \).
The random variable \( X_n \) takes values 4 and 8 with probabilities \( \frac{1}{4} \) and \( \frac{3}{4} \), respectively.
Step 2: Finding the mean and variance.
The expected value \( \mathbb{E}[X_n] \) is: \[ \mathbb{E}[X_n] = 4 \times \frac{1}{4} + 8 \times \frac{3}{4} = 7 \] The variance \( \text{Var}(X_n) \) is: \[ \text{Var}(X_n) = \mathbb{E}[X_n^2] - (\mathbb{E}[X_n])^2 = \left( 4^2 \times \frac{1}{4} + 8^2 \times \frac{3}{4} \right) - 7^2 = 1.75 \]
Step 3: Using the Law of Large Numbers.
As \( n \to \infty \), by the law of large numbers, \( \bar{X}_n \) converges to \( \mathbb{E}[X_n] = 7 \). Hence, the values of \( m \) and \( M \) should be chosen so that the probability of \( \bar{X}_n \) falling between \( m \) and \( M \) converges to 1. The correct answer is \( m = 6.1 \) and \( M = 7.1 \).
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