Supplementary Exercise 1.127 of IPS7e
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X = cholesterol levels of women aged 20-34, measured in mg/dL.
Assume X to follow N(185,39).
Denote also (as usual) by Z a random variable from N(0,1).

(a) compute P(X>240)

In the extended notation of the textbooks we would write

             X > 240
or  (X-185)/39 > (240-185)/39
or           Z > 1.41026

which gives us the z-score z=1.41026, and tells us we need to calculate
P(Z>1.41026). In a more condensed notation, we may write

P(X>240) = P((X-185)/39) > (240-185)/39) 
         = P(Z>1.41026) 
         = P(Z<-1.41026) 
         = 0.0792 (computer), or 0.0793 (Table A).

We conclude that about 8% of the women have cholesterol levels that
require medical attention.


(b) compute P(200<X<240)

In the extended notation we would write

              200 <      X     < 240
or   (200-185)/39 < (X-185)/39 < (240-185)/39
or         0.3846 <      Z     < 1.41026

In the condensed notation, it looks like this:

P(200<X<240) = P( (200-185)/39 < (X-185)/39 < (240-185)/39 ) 
             = P(0.3846 < Z < 1.41026)
             = P(Z<1.41026) - P(Z<=0.3846)
             = (1-0.0792) - 0.6497 = 0.271 (computer)

Therefore, about 27% of the women have borderline high cholesterol 
levels.

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Minitab commands and output to display the two cumulative probabilities:

 CDF 240;
  Normal 185 39.

Normal with mean = 185 and standard deviation = 39

  x  P( X <= x )
240     0.920768

 CDF 200;
  Normal 185 39.

Normal with mean = 185 and standard deviation = 39

  x  P( X <= x )
200     0.649739

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Minitab commands and listing for all calculations (using constants k1,k2,...):

 CDF 240 k1;
  Normal 185 39.
 Let k2 = 1-k1
 Print  K1 K2.

Data Display 

K1    0.920768
K2    0.0792320

Note: K2 is the probability for (a).

 CDF 200 k3;
  Normal 185 39.
 Let k4=k1-k3
 Print k3 k4

Data Display 

K3    0.649739
K4    0.271029

Note: K4 is the probability for (b).