Suppose X ~
(1.05, 0.01). Use a
standard normal table to determine the .75-quantile
−1(.75) of
X.
Solution
We have the mean and standard deviation of X as
= 1.05 and
= 0.10. By
[3.91] and [3.92], we calculate m = .044 and s = .095.
Accordingly, X = exp(.095Z + .044), where Z ~
N(0,1). Denote the CDFs of X and Z as
and
. From the
standard normal table
_small.jpg)
(1.05, 0.01). Use a
standard normal table to determine the .75-quantile
−1(.75) of
X.
= 1.05 and
= 0.10. By
[3.91] and [3.92], we calculate m = .044 and s = .095.
Accordingly, X = exp(.095Z + .044), where Z ~
N(0,1). Denote the CDFs of X and Z as
and
. From the
standard normal table
