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GATE 2025 DA – Question 40

Probability and Statistics · Central limit theorem, confidence intervals and hypothesis tests · 2 marks · Multiple choice

A random variable $X$ is said to be distributed as $Bernoulli(\theta)$, denoted by $X \sim Bernoulli(\theta)$, if

$P(X = 1) = \theta$, $P(X = 0) = 1 - \theta$

for $0 < \theta < 1$. 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 $\phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x} e^{-\frac{t^2}{2}}dt$)

  1. $\phi(2) - \phi(-2)$
  2. $\phi(1) - \phi(-1)$
  3. $\phi(3) - \phi(-3)$
  4. $\phi(90) - \phi(60)$

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Correct answer: (A) $\phi(2) - \phi(-2)$

Explanation

$Y$ is binomial with mean $300 \times 0.25 = 75$ and variance $300 \times 0.25 \times 0.75 = 56.25$, so the standard deviation is 7.5. By the central limit theorem, $P(60 \le Y \le 90) \approx \phi\left(\frac{90 - 75}{7.5}\right) - \phi\left(\frac{60 - 75}{7.5}\right) = \phi(2) - \phi(-2)$.