GATE 2025 DA Q.40: solution and answer

GATE 2025 · DAQ.40 · 2 marks · MCQ

A random variable $X$ is said to be distributed as Bernoulli( $θ$ ), denoted by $X \sim$ Bernoulli( $θ$ ), if $P (X = 1) = θ$ , $P (X = 0) = 1 - θ$ for $0 < θ < 1$ . Let $Y = \sum_{i = 1}^{300} X_{i}$ , where $X_{i} \sim$ Bernoulli( $θ$ ), $i = 1, 2, \dots, 300$ be independent and identically distributed random variables with $θ = 0.25$ . The value of $P (60 \leq Y \leq 90)$ , after approximation through Central Limit Theorem, is given by (Recall that $ϕ (x) = \frac{1}{2 π} \int_{- \infty}^{x} e^{- t^{2} /2} d t$ )

(A)
$ϕ (2) - ϕ (- 2)$
(B)
$ϕ (1) - ϕ (- 1)$
(C)
$ϕ (3) - ϕ (- 3)$
(D)
$ϕ (90) - ϕ (60)$

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Answer and solution

ANSWER

$P (60 \leq Y \leq 90) \approx ϕ (2) - ϕ (- 2)$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 DA (IIT Roorkee)

Each $X_{i} \sim Bernoulli (0.25)$ has $E [X_{i}] = θ = 0.25$ and $Var (X_{i}) = θ (1 - θ) = 0.1875$ . Since the $X_{i}$ are independent, $Y = \sum_{i = 1}^{300} X_{i} \sim Binomial (300, 0.25)$ .
Mean: $E [Y] = 300 \times 0.25 = 75$ .
Variance: $Var (Y) = 300 \times 0.25 \times 0.75 = 56.25$ , so the standard deviation is $56.25 = 7.5$ .
By the Central Limit Theorem, $Z = \frac{Y - 75}{7.5}$ is approximately standard normal, so $P (a \leq Y \leq b) \approx ϕ (\frac{b - 75}{7.5}) - ϕ (\frac{a - 75}{7.5})$ , where $ϕ$ is the standard normal CDF as defined in the question.
Standardise the limits: $\frac{60 - 75}{7.5} = - 2$ and $\frac{90 - 75}{7.5} = 2$ .
Therefore $P (60 \leq Y \leq 90) \approx ϕ (2) - ϕ (- 2)$ , which is option (A).
Option (B) would need limits at $75 \pm 7.5$ and option (C) at $75 \pm 22.5$ , so neither matches. Option (D) plugs the raw values 60 and 90 into $ϕ$ without standardising.

KEY IDEA

A sum of $n$ i.i.d. Bernoulli( $θ$ ) variables has mean $n θ$ and variance $n θ (1 - θ)$ . The CLT lets you standardise with these and read the probability as a difference of standard normal CDF values.

WATCH OUT FOR

Using the Bernoulli variance $θ (1 - θ)$ instead of the variance of the sum $n θ (1 - θ)$ . Another slip is forgetting to standardise, which gives option (D), or dividing by the variance 56.25 instead of the standard deviation 7.5.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 2 CLT & tests questions, with notes