Ralph Vince - Portfolio Mathematics

ch02 JWBK035-Vince February 12, 2007 6:50 Char Count= 0

Probability Distributions 57

FIGURE 2.7 The Normal Probability density function

Next, we need to divide the result by the standard deviation:

− 20 / 20 =− 1

We can therefore say that the number of standard units is−1, when the

data point equals−9, and the mean is 11, and the standard deviation is

20. In other words, we are 1 standard deviation away from the peak of the
    curve, the mean, and since this value is negative we know that it means we
    are 1 standard deviation to the left of the peak. To see where this places us
    on the curve itself (i.e., how high the curve is at 1 standard deviation left of
    center, or what the Y axis value of the curve is for a corresponding X axis
    value of−1), we need to now plug this into Equation (2.15a):

N′(Z)=. 398942 ∗EXP(−(Z^2 /2))

=. 398942 ∗ 2 .7182818285(−(− 12 /2))

=. 398942 ∗ 2. 7182818285

− (^1) / 2
=. 398942 ∗. 6065307
=. 2419705705