Supplementary Exercise 1.110 of IPS7e
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We denote by Z a random variable from N(0,1), the standard normal
distribution.

(a) P(Z<=-2) = 0.0228

(b) P(Z>=-2) = 1-P(Z<=-2) = 1-0.0228 = 0.9772
(or directly as P(Z<2))

(c) P(Z>1.67) = 1-P(Z<1.67) = 1-0.9525 = 0.0475
(or directly as P(Z<=-1.67))

(d) P(-2<Z<1.67) = P(Z<1.67) - P(Z<-2)
                 = 0.9525-0.0228 = 0.9297

Note that for all these calculations, it does not matter whether the
endpoints of the intervals are included or not; this is because the
probability of any single value (such as the interval endpoints) is zero
in a continuous probability distribution (such as the normal distribution). 

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Minitab commands and listing to compute the cumulative probabilities:

MTB > CDF -2;
SUBC>   Normal 0.0 1.0.
Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

 x  P( X <= x )
-2    0.0227501


MTB > CDF 2;
SUBC>   Normal 0.0 1.0.
Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

x  P( X <= x )
2     0.977250


MTB > CDF 1.67;
SUBC>   Normal 0.0 1.0.
Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

   x  P( X <= x )
1.67     0.952540


MTB > CDF -1.67;
SUBC>   Normal 0.0 1.0.
Cumulative Distribution Function 

Normal with mean = 0 and standard deviation = 1

    x  P( X <= x )
-1.67    0.0474597