Suppose X ~ N(2, .09).
Use a standard normal table to determine the .90-quantile
−1(.90)
of X.
Solution
Let Z ~ N(0,1) such
that:
[s1]
[s2]
From the standard normal table, the .90-quantile of Z
is 1.28, so
[s3]
[s4]
[s5]
Accordingly, the .90-quantile
−1(.90)
of X is 2.38.
A more direct and intuitive solution is
to recognize that the .90-quantile of any normal distribution occurs 1.28
standard deviations above its mean. Accordingly, the .90-quantile of X
is
_small.jpg)
−1(.90)
of X.
