We know that the probability density function of a variable that is normally distributed is f(x) = 1/(σ√2π) * exp[1/2 (x – µ). Its inflection point is the point where f"(x) = 0. Taking the first derivative, we get f'(x) = –(x–µ)/(σ³/√2π) exp[–(x–µ)²/(2σ²)] = –(x–µ) f(x)/σ². The second derivative would be f"(x) = [ –(x–µ) f(x)/σ]' = –f(x)/σ² – (x–µ) f'(x)/σ² = –f(x)/σ² + (x-µ)² f(x)/σ⁴. Setting this expression equal to zero, we get –f(x)/σ² + (x-µ)² f(x)/σ⁴ = 0 Multiply both sides by σ⁴/f(x): –σ² + (x-µ)² = 0 (x-µ)² = σ² x-µ= + – σ x = µ +– σ So the answers are x = µ – σ and x = µ + σ.
