Supplementary Exercise 1.111 of IPS7e
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We denote by Z a random variable from N(0,1), the standard normal
distribution.


(a) Looking for z such that P(Z<z)=0.20 (that is, z is the 20% percentile
in the standard normal distribution).

Using a normal distribution table, we get z=-0.84 (closest with P(Z<z)=0.2005);
using Minitab/Stata, we get z= -0.8416.

(b) Looking for z such that P(Z>z)=0.30 (that is, z is the (100-30)% = 70% 
percentile in the standard normal distribution).

Using a normal distribution table, we get z=0.52 (closest with P(Z<z)=0.6985);
using Minitab/Stata, we get z=0.5244.


Note: in both cases, it would be valid to approximate the z-value by
interpolation from the table. For example, as P(Z<0.52)=0.6985 and
P(Z<0.53)=0.7019, it is obvious that a value between 0.52 and 0.53
achieves exactly P(Z<z)=0.70. As 0.6985 and 0.7019 are almost equally
far away from 0.70, the midpoint (that is, (0.52+0.53)/2=0.525)
should be a good approximation, and the more exact calculation 
in statistical software confirms this.

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Minitab commands and listing:

MTB > InvCDF 0.20;
SUBC>   Normal 0.0 1.0.
Inverse Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

P( X <= x )          x
        0.2  -0.841621


MTB > InvCDF 0.70;
SUBC>   Normal 0.0 1.0.
Inverse Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

P( X <= x )         x
        0.7  0.524401