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GATE DA 2025 Q40: CLT approximation for a sum of 300 Bernoulli(0.25)

Question

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

P(X=1)=θ,P(X=0)=1−θP(X = 1) = \theta, \quad P(X = 0) = 1 - \theta

for 0<θ<10 < \theta < 1. Let Y=∑i=1300XiY = \sum_{i=1}^{300} X_i, where Xi∼Bernoulli(θ)X_i \sim Bernoulli(\theta), i=1,2,…,300i = 1, 2, \ldots, 300 be independent and identically distributed random variables with θ=0.25\theta = 0.25. The value of P(60≤Y≤90)P(60 \le Y \le 90), after approximation through Central Limit Theorem, is given by

(Recall that ϕ(x)=12π∫−∞xe−t22 dt\phi(x) = \dfrac{1}{\sqrt{2\pi}} \displaystyle\int_{-\infty}^{x} e^{-\frac{t^2}{2}}\, dt)

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Answer: A (official GATE 2025 final answer key)

Concept used: Central limit theorem: standardise the sum using its mean nθn\theta and variance nθ(1−θ)n\theta(1 - \theta).

First step: Compute E[Y]=300×0.25=75E[Y] = 300 \times 0.25 = 75 and Var⁡(Y)=300×0.25×0.75=56.25\operatorname{Var}(Y) = 300 \times 0.25 \times 0.75 = 56.25, so the standard deviation is 7.5. Standardise 60 and 90.

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Source: official GATE 2025 DA question paper and final answer key, organised by IIT Roorkee. Question paper · Answer key