Download Standard Normal Distribution.pptx

Normal Distribution
1
f (x) =
e
σ 2π
1 % x− µ (
− '
*
2& σ )
For Standard Normal Distribution
µ = 0, σ = 1
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1
f (x) =
e
2π
1 2
− x
2
2
Standard Normal Distribution
Area Under the Curve
=
1
2π
∫
b
a
e
−x 2
2
dx
Numerical Integration
= 1 for a = -∞, b = ∞
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= 0.67 for a = -1, b = 1
= 0.95 for a = -2, b = 2
= 0.025 for a = -∞, b = -2 (left tail probability)
= 0.025 for a = 2, b = ∞ (right tail probability)
Standard Normal Distribution
Maximum Frequency at Mean
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f (x) =
1
e
2π
1
− x2
2
Take first derivative, set equal to 0, solve for x:
Derivative of e u = due u
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f "(x) =
€
1
2π
1
1
%
− x2 (
− x2
−x
e 2
'−xe 2 * =
2π
&
)
= 0 only if x = 0 = mean of distribution
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Standard Normal Distribution
Inflection Points at Mean ± Standard Deviation
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Take second derivative, set equal to 0, solve for x:
−x
f "(x) =
e
2π
−1
f ""(x) =
e
2π
€
−1 2
x
2
1
− x2
2
from above
1 2(
−1 2
%
− x
x
−x
−1
2
2
2
+
e (1− x )
'−xe * =
2π &
) 2π
= 0 only if x = −1 or 1