ch02 JWBK035-Vince February 12, 2007 6:50 Char Count= 0
Probability Distributions 61= 1 −. 05399093525 ∗
⎛
⎜
⎜⎜
⎜
⎝
1. 330274429 ∗. 683394433115
− 1. 821255978 ∗. 683394433114
+ 1. 781477937 ∗. 683394433113
−. 356563782 ∗. 683394433112
+. 31928153 ∗. 68339443311
⎞
⎟
⎟⎟
⎟
⎠
= 1 −. 05399093525 ∗
⎛
⎜⎜
⎝
1. 330274429 ∗. 1490587 − 1. 821255978 ∗
. 2181151 + 1. 781477937 ∗. 3191643
−. 356563782 ∗. 467028 +. 31928153 ∗
. 68339443311
⎞
⎟⎟
⎠
= 1 −. 05399093525 ∗(. 198299977 −. 3972434298
+. 5685841587 −. 16652527 +.2182635596)
= 1 −. 05399093525 ∗. 4213679955
= 1 −. 02275005216
=. 9772499478
Thus, we can say that we can expect 97.72% of the outcomes in a Nor-
mally distributed random process to fall shy of+2 standard units. This is
depicted in Figure 2.8.
If we wanted to know what the probabilities were for an event’s equal-
ing or exceeding a prescribed number of standard units (in this case+2),
we would simply amend Equation (2.21), taking out the 1−in the beginning
of the equation and doing away with the−Z provision (i.e., doing away withFIGURE 2.8 Equation (2.21) showing probability with Z=+ 2