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
- 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