The normal distribution pdf is
![f(x|\mu,\sigma ^{2} ) = \frac{1}{\sigma \sqrt{2 \pi } } e^{- \frac{(x-\mu)^{2}}{2\sigma^{2}}](https://tex.z-dn.net/?f=f%28x%7C%5Cmu%2C%5Csigma%20%5E%7B2%7D%20%29%20%3D%20%5Cfrac%7B1%7D%7B%5Csigma%20%5Csqrt%7B2%20%5Cpi%20%7D%20%7D%20e%5E%7B-%20%5Cfrac%7B%28x-%5Cmu%29%5E%7B2%7D%7D%7B2%5Csigma%5E%7B2%7D%7D%20)
where
μ = population mean,
σ = population standard deviation
For a random variable x, the probability (area under the normal curve) is
![\int _{-\infty} ^{x} f(x | \mu. \sigma ^{2})} dx](https://tex.z-dn.net/?f=%5Cint%20_%7B-%5Cinfty%7D%20%5E%7Bx%7D%20f%28x%20%7C%20%5Cmu.%20%5Csigma%20%5E%7B2%7D%29%7D%20dx)
Define z = (x - μ)/σ.
Then
dx = σ (dz)
and the area under the curve transforms to
![\int _{-\infty} ^{z} \frac{1}{ \sqrt{2 \pi } } e^{ - \frac{z^{2}}{2} } dz](https://tex.z-dn.net/?f=%5Cint%20_%7B-%5Cinfty%7D%20%5E%7Bz%7D%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2%20%5Cpi%20%7D%20%7D%20e%5E%7B%20-%20%5Cfrac%7Bz%5E%7B2%7D%7D%7B2%7D%20%7D%20dz)
This integral is evaluated numerically (trapezoidal or Simpson's rule). For the lower limit, a value of z = -4 instead of ∞ is sufficient for good accuracy.
For example, with the trapezoidal rule, use a step size of h = 0.05.
Between z = -4 to z = 0.3, there are 87 values.
Calculate the function values as y₀, y₁, y₂, ..., y₈₇.
y₀ + y₈₇ = 0.3815
y₁+y₂+ ... +y₈₆ = 12.1664
The area is
A = (0.05/2)*(0.3815 + 2*12.1664) = 0.025*24.7142 = 0.6179.
Results for specified values of z are
(a) z = 0.30, Area = 0.6179
(b) z = 1.75, Area = 0.9599
(c) z = 2.42, Area = 0.9922
(d) z = -0.68, Area = 0.2483
(e) z = -1.11, Area = 0.1335