Question

The point of inflexion of Normal ($\mu,\sigma^2$) curve are :

• A. $\mu \pm \sigma$
• B. $\mu \pm 2\sigma$
• C. $\mu \pm 3\sigma$
• D. $\mu \pm 0.3\sigma$

Hint:

The inflection point of a normal curve is always located exactly $1$ standard deviation away from the mean ($\mu \pm \sigma$). This is a standard property often queried in theory exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A point of inflection is a point on a curve at which the concavity changes (from concave up to concave down, or vice versa).
Mathematically, this occurs where the second derivative of the probability density function with respect to $x$ is equal to zero: \[ \frac{d^2 f(x)}{dx^2} = 0 \] Detailed Explanation:
The probability density function of a normal distribution is: \[ f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \] Taking the first derivative with respect to $x$: \[ f'(x) = -\frac{(x-\mu)}{\sigma^2} f(x) \] Taking the second derivative: \[ f''(x) = \left[ -\frac{1}{\sigma^2} + \frac{(x-\mu)^2}{\sigma^4} \right] f(x) \] Set $f''(x) = 0$. Since $f(x) \neq 0$ for all real $x$: \[ \frac{(x-\mu)^2}{\sigma^4} = \frac{1}{\sigma^2} \] \[ (x-\mu)^2 = \sigma^2 \] \[ x-\mu = \pm \sigma \] \[ x = \mu \pm \sigma \] Therefore, the points of inflection of the normal curve are at a distance of one standard deviation from the mean.

Step 2: Final Answer:
The points of inflection are $\mu \pm \sigma$, which matches Option (A).