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: The general form of a normal (Gaussian) probability density function with mean \(\mu\) and standard deviation \(\sigma\) is \[f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), \quad x \in (-\infty, +\infty)\]
Step 2: Compare this general form with the given density \(f(x) = \dfrac{1}{3\sqrt{2\pi}} \exp\left(-\dfrac{x^2}{18}\right)\). Matching the coefficient \(\dfrac{1}{3\sqrt{2\pi}}\) with \(\dfrac{1}{\sigma\sqrt{2\pi}}\) gives \(\sigma = 3\).
Step 3: Check the exponent: \(2\sigma^2 = 2(3)^2 = 18\), which exactly matches the denominator 18 appearing in the given exponent, and since there is no shift term inside the square, the mean is \(\mu = 0\).
Step 4: Since the density has precisely the Gaussian bell-curve shape over the entire real line, X must be a normal random variable. It cannot be exponential (only defined for x greater than or equal to 0), Poisson (a discrete distribution), or uniform (a flat density over a bounded interval).
Final Answer: Option (B): X is a normal random variable