Supplementary Exercise 5.64 of IPS7e
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Student acceptance of admission to a college.
Let X = number of students who accept, and assume X ~ B(1500,0.7).

(a) Use the formulae for mean and standard deviation for a binomial distribution.
EX (mean) = 1500*0.7 = 1050
sdX (std.dev.) = sqrt(1500*0.7*0.3) = 17.7482


(b) Before we use the normal approximation, we check whether the rule of thumb
for its use is satisfied (from slide 5L-7):
  1500*0.7*0.3 = 315 >> 10 
We should definitely be ok to use the normal approximation.

Now use the normal approximation with continuity correction (5L-7)

P(X>=1000) = 1-P(X<1000) ~= 1-P(Z<=(1000-0.5-1050)/17.7482) = 1-P(Z<=-2.845)
                          = 1-0.0022 = 0.9978 (using table)

Direct computation from the binomial distribution:

MTB > CDF 999 k1;
SUBC>   Binomial 1500 0.7.
MTB > let k2=1-k1
MTB > print k1 k2
Data Display 

K1    0.00239749
K2    0.997603

So the approximation error is about 0.0002.

A cautionary note:
Using the normal approximation without continuity correction often
gives less accurate results, even though in this case we would get:
  P(X>=1000) ~= P(Z>=(1000-1050)/17.7482) = P(Z>=-2.817) = 0.9976 (Table A)
We could however equally have used the approximation for the lower
probability, as follows
  P(X>=1000) = 1-P(X<=999) ~= 1-P(Z<=(999-1050)/17.7482) = 1-P(Z>=-2.874)
                            = 1-0.0021 = 0.9979
There is no way to know which of these two approximations are going to
be best, and sometimes (see e.g. question (d) below) both of them are
clearly worse than the approximation with the continuity correction.
Therefore it is generally safer to use the continuity correction.


(c) Same procedure with 1200 instead of 1000 but now 1200 is not
included in the event:

P(X>1200) = 1-P(X<=1200) ~= 1-P(Z<=(1200+0.5-1050)/17.7482) = 1-P(Z<=8.48)
                          = 0.0000 (using table)

It is completely safe to assume that at most 1200 students will apply,
provided that the normal distribution model is correct.


(d) Same procedure with 1700 instead of 1500, so now X ~ B(1700,0.7) and
EX=1700*0.7=1190,
sdX=sqrt(1700*0.7*0.3)=18.8944,

P(X>1200) = 1-P(X<=1200) ~= 1-P(Z<=(1200+0.5-1190)/18.8944) = 1-P(Z<=0.5557)
                          = 1-0.7108 = 0.2892 (using Minitab)

The exact probability computed in Minitab is

MTB > CDF 1200 k1;
SUBC>   Binomial 1700 0.7.
MTB > let k2=1-k1
MTB > print k1 k2
Data Display 

K1    0.709957
K2    0.290043

In this case, the normal approximations without continuity correction
give 0.280 and 0.298, and are much less precise.