The probability distribution function of a normal variate with mean $μ$ and variance $σ^2$ is given by:
$f(x) =\frac{1}{σ\sqrt{2}π}e^{-\frac{1}{2}(\frac{x-μ}{σ})^2},-∞<x<∞,-∞<μ<∞, σ> 0$
If $y = f(x)$ be the normal probability curve, then which of the following is correct?

(A) The normal curve is symmetrical about the line $x = μ$.
(B) Mean, median and mode of the distribution coincide.
(C) Y-axis is an asymptote to the normal curve.
(D) If $x$ increases numerically, $f(x)$ decreases rapidly.

Choose the correct answer from the options given below:

(A), (B) and (D) only

The correct answer is Option (2) → (A), (B) and (D) only

For the normal probability distribution:

$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}, \quad -\infty < x < \infty, \; \sigma > 0$

(A) The normal curve is symmetrical about the line $x=\mu$. ✅ Correct.

(B) Mean, median and mode of the distribution coincide. ✅ Correct.

(C) Y-axis is not an asymptote; as $x \to \pm \infty$, $f(x)\to 0$, so $x$-axis is the asymptote. ❌ Incorrect.

(D) If $x$ increases numerically (i.e., $|x|$ increases), $f(x)$ decreases rapidly. ✅ Correct.

Therefore, the correct statements are (A), (B), and (D).