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Official PYQ · 2026Set 2 · Q4Probability: Random Variables and DistributionsMCQ1 mark

GATE CS 2026 Set 2 Q4 — Probability: Random Variables and Distributions

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The probability density function f(x)f(x) of a random variable XX which takes real values is
f(x)=132πexp⁡(−x218),x∈(−∞,+∞)f(x)=\frac{1}{3 \sqrt{2 \pi}} \exp \left(-\frac{x^{2}}{18}\right), \quad x \in(-\infty,+\infty)
Which one of the following statements is correct about the random variable XX ?

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Answer: B

The probability density function (PDF) of a normal random variable XX with mean μ\mu and variance σ2\sigma^2 is given by:

f(x)=1σ2πexp⁡(−(x−μ)22σ2),x∈(−∞,∞)f(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp\left( -\frac{(x-\mu)^2}{2\sigma^2} \right), \quad x \in (-\infty, \infty)

Comparing this standard form with the given PDF:

f(x)=132πexp⁡(−x218)f(x) = \frac{1}{3 \sqrt{2 \pi}} \exp \left(-\frac{x^{2}}{18}\right)

We can identify the parameters by matching the terms:

  • The coefficient in front of the exponential gives σ=3\sigma = 3.
  • The exponent term −x218-\frac{x^2}{18} can be written as −(x−0)22(32)-\frac{(x-0)^2}{2(3^2)}, which gives μ=0\mu = 0 and 2σ2=182\sigma^2 = 18.

Since the given PDF perfectly matches the normal distribution with mean μ=0\mu = 0 and standard deviation σ=3\sigma = 3, XX is a normal random variable.

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