Figure 11-9.
Standard normal probability curve and distribution function as area.
Note that with a change of variable F(x) = Φ
(
x−μ
σ
)
so that in terms of proba-
bilities
P(a < X ≤b) = F(b)−F(a) = Φ
(
b−μ
σ
)
−Φ
(
a−μ
σ
)
(11 .59)
In figure 11-9, the normal probability curve is symmetric about z= 0 and so the area
under the curve between 0 and zrepresents the probability P(0 < Z ≤z) = Φ(z)−Φ(0)
and quantities like Φ(−z), by symmetry, have the value Φ(−z) = 1 −Φ(z).The standard
normal curve has the properties that Φ(−∞ ) = 0,Φ(0) = 1 / 2 and Φ(∞) = 1. The table
11.5 gives the area under the standard normal curve for values of z≥ 0 , then it is
possible to employ the symmetry of the standard normal curve to calculated specific
areas associated with probabilities. For example, to find the area from −∞ to − 1. 65
examine the table of values and find Φ(1 .65) =. 9505 so that φ(− 1 .65) = 1 −.9505 =. 0495
or the area from − 1. 65 to +∞is .9505, then one can write the probability statement
P(Z > − 1 .65) =. 9505. As another example, to find the area under the standard
normal curve between z=− 1. 65 and z= 1, first find the following values
Area from 0 to 1 = Φ(1) −Φ(0) =. 8413 − 0 .5000 =. 3413
Area from 0 to 1 .65 = Φ(1.65) −Φ(0) =. 9505 −.5000 =. 4505
Area from − 1. 65 to 0 =. 4505
Area from − 1. 65 to 1 = .4505 + .3413 = .7918 = P(− 1. 65 < Z ≤1)
Sketch a graph of the above values as areas under the normal probability density
function to get a better understanding of the values presented.
