Question

The probability density function \(f(x)\) of a random variable \(X\) which takes real values 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:

Compare the given density with the standard normal form 1/(sigma*sqrt(2*pi)) exp(-x^2/(2*sigma^2)); it matches with mean 0 and sigma = 3.

Answer 1

The Correct Option is B

Step 1: Rather than matching constants against the general normal formula directly, check the distinguishing fingerprints of each candidate distribution against $f(x) = \dfrac{1}{3\sqrt{2\pi}} \exp\left(-\dfrac{x^2}{18}\right)$ one property at a time.

Step 2: Fingerprint of symmetry: replacing $x$ with $-x$ in $f(x)$ leaves it unchanged, since only $x^2$ appears in the exponent, so $f(x)$ is an even function, symmetric about $x = 0$. An exponential density is not symmetric, it is zero for $x<0$ and strictly decreasing for $x \geq 0$, so this already rules out the exponential option.

Step 3: Fingerprint of domain type: $f(x)$ is a smooth function defined for every real number, giving a genuine probability density over a continuum of values. A Poisson random variable only assigns probabilities to the discrete values $0, 1, 2, 3, \dots$ using a formula with factorials, it has no density function over a continuum at all, so a continuous bell curve like $f(x)$ cannot represent a Poisson variable.

Step 4: Fingerprint of shape: $f(x)$ strictly decreases as $|x|$ grows away from $0$, since $\exp(-x^2/18)$ shrinks as $x^2$ grows, giving a peak at the center and tapering tails on both sides. A uniform density is flat, constant on some interval and exactly zero outside it, it never tapers smoothly like this, so uniform is ruled out too.

Step 5: The one distribution whose density is a smooth, symmetric, real-line-supported, exponentially tapering bell curve $\frac{1}{\sigma\sqrt{2\pi}}\exp(-x^2/2\sigma^2)$ is exactly the normal distribution, here with $\sigma = 3$ since $2\sigma^2 = 18$. Every fingerprint checked, symmetry, continuous real-line support, smooth tapering peak, matches normal and mismatches the other three.

\[ \boxed{X \text{ is a normal random variable}} \]