Supplementary Exercises 1.113, 1.117, 1.121 of IPS7e
----------------------------------------------------

SAT scores are assumed to follow N(1026, 209)
ACT scores are assumed to follow N(20.8, 4.8)


1.113
-----
Tonya scores 1318 on SAT; Jermaine scores 27 on ACT.

To compare the scores on different tests, we compute the corresponding
z-scores. These give us the relative value taking mean and standard
deviation of the distributions into account. A higher z-score corresponds
to a higher percentile.

Tonya: z = (1318-1026)/209 = 1.397
Jermaine: z = (27-20.8)/4.8 = 1.292

Tonya has a somewhat higher score.


1.117
-----
In Exercise 1.113 we saw that Tonya's SAT-score of 1318 corresponds to a 
z-score of (1318-1026)/209 = 1.397. As P(Z<1.40)=0.9192 (using table/software), 
Tonya's value corresponds approximately to the 91.9% percentile. 

Note that the wording in the question is somewhat misleading: a percentile 
is not a proportion expressed in percent: a percentile is the value in the 
distribution corresponding to a certain percentage of the distribution 
being below it.


Minitab commands and output, in two versions - corresponding to using 
the z-scores or the N(1026, 209) distribution:

 CDF 1.397;
  Normal 0.0 1.0.
--
Normal with mean = 0 and standard deviation = 1

    x  P( X <= x )
1.397     0.918793
--

 CDF 1318;
  Normal 1026 209.
--
Normal with mean = 1026 and standard deviation = 209

   x  P( X <= x )
1318     0.91881
--


1.121
-----
The z-score for the 90% percentile is 1.28 (using a normal distribution table; 
software gives 1.28155), so the top 10% SAT scores are 1026+1.28*209 = 1294 and higher.

Again, two versions of Minitab output, corresponding to using the z-scores
or the N(1026, 209) distribution:

 InvCDF 0.9;
  Normal 0 1.
--
Normal with mean = 0 and standard deviation = 1

P( X <= x )        x
        0.9  1.28155
--

 InvCDF 0.9;
  Normal 1026 209.
--
Normal with mean = 1026 and standard deviation = 209

P( X <= x )        x
        0.9  1293.84
--