Suppose X ~ N(100, 36). Use a
standard normal table to determine the .15-quantile
−1(.15)
of X.
Solution
Let Z ~ N(0,1) such
that:
[s1]
[s2]
From the standard normal table, the .15-quantile of Z
is –1.04, so
[s3]
[s4]
[s5]
Accordingly, the .15-quantile
−1(.15)
of X is 93.76.
A more direct and intuitive solution is
to recognize that the .15-quantile of any normal distribution occurs 1.04
standard deviations below its mean. Accordingly, the .15-quantile of X
is
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−1(.15)
of X.
