How to prove that Mean is a extreme Point in normal distribution Curve? - extreme curves video
For the first time you open your guide, after reading all the curves more or less normal distribution, and finally, he quit the answer.
Friday, December 11, 2009
Extreme Curves Video How To Prove That Mean Is A Extreme Point In Normal Distribution Curve?
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Density function of the probability of a normal distribution curve:
y = 1 / σ √ (2π) * exp (- (x - μ) ² / 2σ ²)
where σ, the standard deviation and μ is the mean.
Derivatives:
y '= 1 / σ √ (2π) -2 * (x - μ) / 2σ ² * exp (- (x - μ) ² / 2σ ²)
Setting the derivative equal to zero:
1 / σ √ (2π) -2 * (x - μ) / 2σ ² * exp (- (x - μ) ² / 2σ ²) = 0
Divide both sides by 1 / σ √ (2π)
-2 (X - μ) / 2σ ² * exp (- (x - μ) ² / 2σ ²) = 0
Divide both sides by exp (- (x - μ) ² / 2σ ²), because it can not be zero:
-2 (X - μ) / 2σ ² = 0
Divide the numerator and denominator by 2:
- (X - μ) / σ ² = 0
Multiply both sides by σ ²:
- (X - μ) = 0
Divide both sides by -1:
x - μ = 0
Add μ for both parties:
x = μ
So x = μ is a critical point.
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If you want to show that just used, you should test the second derivative.
Take the second derivative:
and''= 1 / σ √ (2π) ([-2 (x - μ) / 2σ ²] ² exp (- (x - μ) ² / 2σ ²) + exp (- (x --²) / 2σ ²) / σ ²)
Close the critical point:
Y''(μ) = 1 / σ √ (2π) ([-2 (μ - μ) / 2σ ²] ² exp (- (μ - μ) ² / 2σ ²) + exp (- (μ - μ) ² / 2σ ²) / σ ²)
and''(μ) = 1 / σ √ (2π) -1 * / σ ²
Y''(μ) = -1 / σ ³ √ (2π)
U0026lt For this reason, and''(μ) \\ \\ \\ \\ \\ \\ \\ \\, 0
→ x = μ is a local maximum.
f (x) = (1/sqrt 2) σ exp [- (x-μ) ^ 2 / 2σ ^ 2]
Differentiation with respect to x
f '(x) = (1/sqrt 2) σ exp [- (x-μ) ^ 2 / 2σ ^ 2] [-2 (x-μ) / 2σ ^ 2] = 0
the solution x = μ
verify u0026lt f''(x) \\ \\ \\ \\ \\ \\ \\ \\, 0, is the end point, the curve is maximum when x = μ.
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