A random variable \(X\) is said to be distributed as \(Bernoulli(\theta)\), denoted by \(X \sim Bernoulli(\theta)\), if
\[P(X = 1) = \theta, \quad P(X = 0) = 1 - \theta\]
for \(0 \lt \theta \lt 1\). Let \(Y = \sum_{i=1}^{300} X_i\), where \(X_i \sim Bernoulli(\theta)\), \(i = 1, 2, \ldots, 300\) be independent and identically distributed random variables with \(\theta = 0.25\). The value of \(P(60 \le Y \le 90)\), after approximation through Central Limit Theorem, is given by
(Recall that \(\phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{x} e^{-\frac{t^{2}}{2}}\, dt\))
Answer
(A)
Explanation
\(Y\) is a sum of 300 independent Bernoulli variables with \(\theta = 0.25\), so
- \(E[Y] = 300 \times 0.25 = 75\);
- \(\operatorname{Var}(Y) = 300 \times 0.25 \times 0.75 = 56.25\), a standard deviation of \(7.5\).
By the Central Limit Theorem, \(\frac{Y - 75}{7.5}\) is approximately standard normal. The limits standardise to \(\frac{60 - 75}{7.5} = -2\) and \(\frac{90 - 75}{7.5} = 2\), so
\[P(60 \le Y \le 90) \approx \phi(2) - \phi(-2).\]
Option D puts the raw values of \(Y\) into \(\phi\) without standardising.
Answer: A