Question

A random variable \( X \) is said to be distributed as \( \text{Bernoulli}(\theta) \), denoted by \( X \sim \text{Bernoulli}(\theta) \), if \[ P(X = 1) = \theta, \quad P(X = 0) = 1 - \theta \] for \( 0<\theta<1 \). Let \( Y = \sum_{i=1}^{300} X_i \), where \( X_i \sim \text{Bernoulli}(\theta) \), \( i = 1, 2, \dots, 300 \) be independent and identically distributed random variables with \( \theta = 0.25 \). The value of \( P(60 \leq Y \leq 90) \), after approximation through the Central Limit Theorem, is given by

• A. \( \phi(2) - \phi(-2) \)
• B. \( \phi(1) - \phi(-1) \)
• C. \( \phi(3) - \phi(-3) \)
• D. \( \phi(90) - \phi(60) \)

Hint:

When applying the Central Limit Theorem to a sum of random variables, standardize the variable by subtracting the mean and dividing by the standard deviation. This transforms the sum into a normal distribution, which can be used to calculate probabilities.

Answer 1

The Correct Option is A

Given that \( Y = \sum_{i=1}^{300} X_i \), where \( X_i \sim \text{Bernoulli}(0.25) \), we approximate the distribution of \( Y \) using the Central Limit Theorem. The mean and variance of \( Y \) are: \[ \mu_Y = 300 \times 0.25 = 75, \quad \sigma_Y^2 = 300 \times 0.25 \times 0.75 = 56.25, \quad \sigma_Y = 7.5 \] The probability \( P(60 \leq Y \leq 90) \) is approximated by: \[ P\left( \frac{60 - 75}{7.5} \leq Z \leq \frac{90 - 75}{7.5} \right) = P(-2 \leq Z \leq 2) \] Using the cumulative distribution function \( \phi(x) \) of the standard normal distribution: \[ P(-2 \leq Z \leq 2) = \phi(2) - \phi(-2) \] Thus, the value of \( P(60 \leq Y \leq 90) \) is \( \phi(2) - \phi(-2) \). 
Correct Answer: (A) \( \phi(2) - \phi(-2) \).