The probability distribution function of a normal variate with mean $\mu$ and variance $\sigma^2$ is given by:

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

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 $x = \mu$.

(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:

The given probability distribution function is:

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$

where $\mu$ is the mean, $\sigma$ is the standard deviation, $-\infty < x < \infty$, $-\infty < \mu < \infty$, and $\sigma > 0$.

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

The term $(x-\mu)$ appears squared in the exponential function.

For any value $a$:

$((\mu + a) - \mu)^2 = a^2$

$((\mu - a) - \mu)^2 = (-a)^2 = a^2$

Points equidistant from $\mu$ yield the same function value, creating perfect symmetry about $x = \mu$.

Statement (A) is correct.

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Statement (B): Mean, median and mode of the distribution coincide.

The mean equals $\mu$ by definition.

Due to symmetry about $x = \mu$, exactly half the area under the curve lies on each side, placing the median at $x = \mu$.

The maximum value of $f(x)$ occurs when the exponent equals zero, which happens at $x = \mu$, making this the mode.

All three measures equal $\mu$.

Statement (B) is correct.

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Statement (C): Y-axis is an asymptote to the normal curve.

The Y-axis is the vertical line $x = 0$.

At $x = 0$, the function has the value:

$f(0) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{-\mu}{\sigma})^2}$

The curve crosses the Y-axis at a defined point rather than approaching it asymptotically.

The horizontal asymptote is the X-axis ($y = 0$), which the curve approaches as $x \to \pm\infty$.

Statement (C) is incorrect.

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Statement (D): If x increases numerically, $f(x)$ decreases rapidly.

"Increases numerically" refers to increasing $|x - \mu|$, the distance from the mean.

The exponential term is $e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$.

As $|x - \mu|$ increases:

$\frac{(x-\mu)^2}{\sigma^2}$ increases

$-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}$ becomes increasingly negative

$e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$ approaches zero rapidly

Statement (D) is correct.

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Statements (A), (B), and (D) are correct.

Statement (C) is incorrect.