Question

The probability density function \(f(x)\) of a real-valued random variable \(X\) is
\[f(x)=\frac{1}{3\sqrt{2\pi}}\exp\!\left(-\frac{x^2}{18}\right),\quad x\in(-\infty,+\infty).\]
Which one of the following statements is correct about the random variable X?

• A. 𝑋 is an exponential random variable
• B. 𝑋 is a normal random variable
• C. 𝑋 is a Poisson random variable
• D. 𝑋 is a uniform random variable

Hint:

Match the given density with the standard normal form (1 over sigma times sqrt(2 pi)) times exp(-(x minus mu) squared over 2 sigma squared) to read off mu = 0 and sigma = 3.

Answer 1

The Correct Option is B

Step 1: Rather than matching formulas directly, eliminate the wrong options first. An exponential random variable has density \(\lambda e^{-\lambda x}\) defined only for \(x \geq 0\), but the given \(f(x)\) is defined and positive for all real x, so X cannot be exponential.
Step 2: A Poisson random variable is discrete, described by a probability mass function over non-negative integers, not a continuous density like the \(f(x)\) given here, so X cannot be Poisson.
Step 3: A uniform random variable has a constant flat density over a finite interval and zero elsewhere, but \(f(x)\) here varies smoothly with x through the exponential term and never becomes exactly zero, so X cannot be uniform.
Step 4: By process of elimination the only remaining choice is that X is normal, and indeed \(f(x) = \frac{1}{3\sqrt{2\pi}}\exp(-x^2/18)\) has exactly the bell-shaped Gaussian form with mean 0 and standard deviation 3.
Final Answer: Option (B)