The probability distribution function of a normal variate with mean \(\mu\) and variance \(\sigma^2\) is given by:
\(f(x) = \displaystyle\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}, -\infty \lt x \lt \infty, -\infty \lt \mu \lt \infty, \sigma \gt 0\)
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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
In f(x), the variable x appears only through \(\left(\frac{x-\mu}{\sigma}\right)^2\), and the exponent is never positive. Its largest value, 0, is at \(x = \mu\).
(A) The normal curve is symmetrical about the line \(x = \mu\). — Correct. Points at the same distance h on either side of \(\mu\) give the same square \(h^2\), so \(f(\mu + h) = f(\mu - h)\): the curve is symmetrical about x = μ.
(B) Mean, median and mode of the distribution coincide. — Correct. The mean is \(\mu\); by the symmetry, half the area lies on each side of \(\mu\), so the median is \(\mu\); and f is largest at \(x = \mu\), so the mode is \(\mu\). All three coincide.
(C) Y-axis is an asymptote to the normal curve. — Incorrect. f(x) is positive for every x and tends to 0 as \(x \to \pm\infty\), so the x-axis is the asymptote. The y-axis meets the curve at the finite height f(0).
(D) If \(x\) increases numerically, \(f(x)\) decreases rapidly. — Correct. As x moves away from \(\mu\) on either side (for \(\mu = 0\), as |x| increases), the square \(\left(\frac{x-\mu}{\sigma}\right)^2\) grows and the exponential factor falls off very fast, so f(x) decreases rapidly.