A random variable X is said to be distributed as Bernoulli(theta), denoted by X ~ Bernoulli(theta), if P(X = 1) = theta ...

Question

A random variable X is said to be distributed as Bernoulli(theta), denoted by X ~ Bernoulli(theta), if P(X = 1) = theta and P(X = 0) = 1 - theta for 0 < theta < 1. Let Y = (1/300) * sum_{i=1}^{300} X_i, where X_1, X_2, ..., X_300 are independent and identically distributed Bernoulli(theta) random variables with theta = 0.25. The value of P(Y <= 0.3), after approximation through the Central Limit Theorem, is given by (Recall that phi(z) = (1 / sqrt(2*pi)) * integral from -inf to z of e^{-t^2/2} dt):

A. phi(2)
B. phi(1) - phi(-2)
C. phi(-2)
D. phi(0)

Accepted Answer

P(Y <= 0.3) = phi(2), which is approximately 0.9772. Y is the sample mean of 300 independent Bernoulli(0.25) random variables. By the Central Limit Theorem, Y is approximately normally distributed with mean theta = 0.25 and variance theta*(1-theta)/n = 0.25*0.75/300. To find P(Y <= 0.3), we standardise by subtracting the mean and dividing by the standard deviation, which converts the probability into a standard normal CDF value phi(z). The standardised value z = (0.3 - 0.25) / sqrt(0.25*0.75/300) equals 2, so the answer is phi(2).