Supplementary Exercises 5.51 and 5.53 of IPS7e
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(continuation of Exercise 5.49)

We use the same notation as in the solution for 5.49, where 
X is the number of word errors caught by the proofreader out of the 20
in the essay, and Y=20-X is the number of word errors missed. The
assumed distributions for X and Y are binomial B(20,0.7) and B(20,0.3),
respectively.


5.51
----
(a) By the formula for the mean of a binomial distribution
EX = 20*0.7 = 14
EY = 20*0.3 = 6

(b) By the formula for the standard deviation of a binomial distribution
sdX = sqrt(20*0.7*0.3) = sqrt(4.2) = 2.049
Note that this is also the standard devation of Y.

(c)
p=0.9:   sd(X) = sqrt(20*0.9*0.1) = sqrt(1.8) = 1.342
p=0.99:  sd(X) = sqrt(20*0.99*0.01) = sqrt(0.198) = 0.445

The standard deviation decreases (towards zero) as the probability gets
close to 1 (or close to zero).


5.53
----
In our notation, Y is the number of words missed, and Y ~ B(20,0.3). 
We are looking for the smallest number m so that P(Y>=m) is at most 0.05. 
From Exercise 5.49 we know that for m=9, that probability is 0.1133. We 
need to increase m to get a smaller probability. For m=10, the term 0.06537
will drop out of the summation (see item (b) of Exercise 5.49), so that
P(Y>=10) = 0.03082 + 0.01201 + 0.00386 + 0.00102 + 0.00022 + 0 = 0.0479.
Therefore, the desired value is m=10.

The interpretation in the question of the event Y>=10 as evidence
that the proofreader catches less than 70% of word errors comes from
statistical testing of the null hypothesis H0: p=0.7 against the
alternative Ha: p<0.7. This will be covered later in the course (and
book), and the inclusion of such an interpretation here is an attempt to
gradually introduce these ideas.