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Transformation to Standard Normal Distribution
Question: What does the transformation accomplish
?
Answer:
It takes a Normal r.v. with any mean and SD and rescales it to the
same location (m=0) and gives it the same "spread"
(s=1).
It rescales the pdf curve "correctly." Suppose X~N(m
,s2). Then Z=(X-m
)/s Þ X=Zs+m
and
P(X£x)=P[Zs+m
£ x]= P[Z £ (x-m)/s
] = F[(x-m )/s].
That is, P(X£ x) = F(x)
= F(z), where z=(x-m)/s
Values of F(z) have been accurately computed
and tabulated for a wide range of values for z. Here is a NORMAL
TABLE from which you can read off the value of F(z)
for any z, or conversely, find the value of z that gives rise to a particular
cumulative probability F(z).
If we need to find P(X£ x) = F(x) for
X~N(m ,s2),
we use 2) and 3) above:
first compute z=(x-m)/s
look up the value of F(z) from tables; this
is the desired probability
Conversely, suppose we are given P(X£
x) = F(x) for X~N(m ,s2)
and we wish to find x.
since P(X£ x)=F(z)
find the value of z corresponding to F (z) from
tables
then use 2) to compute x=zs +m
P(a£X£b)
= F(b)-F(a) = F[(b-m)/s]
- F[(a-m)/s]
The (100a )th percentile of X~N(m
,s2) = m+s
Z1-a
Next: Normal Distribution as an Approximation
Previous: Standard Normal Distribution
Jayant Rajgopal
