Supplementary Exercise 5.40 of IPS7e
-----------------------------------

Simple random sample (SRS) of 200 households in the area of a restaurant
to determine the proportion of adults committed to eating nutritious food when 
dining out. Denote by X the number of respondents answering yes to the 
question (whether they are committed to eating nutritious food when dining 
out). The National Restaurant Association estimate 40% of adults to have 
this commitment.

(a) 
It is probably quite reasonable to use a binomial distribution 
B(200,0.4) for X if the area where the survey is conducted has a large
number of households.

The sampling is without replacement from a finite population of unknown size,
but it seems reasonable to assume the size to be large enough to make the
binomial a good approximation. (Depending of course on the location of the
restaurant; if it is in a small village, the sample may even exhaust the 
population completely). The rule of thumb is that the population should be at
least 20 times as large as the sample; that is, the number of households in the
area should be at least 4000.

(b)
The mean is EX = 200*0.4 = 80, by the formula for the mean of a binomial
distribution.

The question asks for the probability that X lies between 75 and 85. We will get
different answers if we interpret that to mean all values in the range 75-85 
including 75 and 85, or all values in the range excluding 75 and 85. This is
because the binomial distribution is discrete and has non-zero probability of
taking the values 75 and 85 exactly. The way the question is worded, I would
consider either of these interpretations as valid. 

For the purpose of this solution we have chosen the latter interpretation. The 
probability P(75<X<85) can be computed in different ways. The table
in IPS only has n up to 20, so we need to use either the software or the
normal approximation.

Normal approximation:
The condition for using the normal approximation is that n*p*(1-p)>10.
We compute 200*0.4*(1-0.4)=48>10, so we can indeed use the approximation.
In order to use the formula on slide 5L-7, we need to rephrase the event
is question: P(75<X<85) = P(76<=X<=84). To ease the notation, we
also compute VarX = n*p*(1-p) = 48. Then we have
  P(76<=X<=84) ~= P(Z<=(84+0.5-80)/sqrt(48)) - P(Z<=(76-0.5-80)/sqrt(48))
                = P(Z<=0.6495)-P(Z<=-0.6495) = 0.7420-0.2580 = 0.4840

Direct calculation in a B(200,0.4) using Minitab:
  P(76<=X<=84) = P(X<=84) - P(X<=75) = 0.742849-0.258956 = 0.483893

MTB > CDF 84;
SUBC>   Binomial 200 .4.
Cumulative Distribution Function 
Binomial with n = 200 and p = 0.4

 x  P( X <= x )
84     0.742849

MTB > CDF 75;
SUBC>   Binomial 200 .4.
Cumulative Distribution Function 
Binomial with n = 200 and p = 0.4

 x  P( X <= x )
75     0.258956

(c)
As under (b), the probability P(X>=100) can be computed in different
ways.

Normal approximation:
In order to use the formula on slide 5L-7, we need to rephrase the event in
question: P(X>=100) = 1-P(X<=99). As above, VarX = n*p*(1-p) = 48. Then we have
  P(X>=100) = 1 - P(X<=99) ~= 1 - P(Z<=(99+0.5-80)/sqrt(48))
            = 1 - P(Z<=2.8146) = P(Z<=-2.8146) = 0.00244

Direct calculation in a B(200,0.4) using Minitab:
  P(X>=100) = 1 - P(X<=99) = 1 - 0.997365 = 0.00264

MTB > CDF 99;
SUBC>   Binomial 200 .4.
Cumulative Distribution Function 
Binomial with n = 200 and p = 0.4

 x  P( X <= x )
99     0.997365

The probability is low, so observing 100 out of 200 respondents
being committed to nutritious food seems to indicate that the local
commitment is greater than the value reported by the National Restaurant 
Association. Stated more precisely, the probability of observing 100 or 
more committed respondents by chance when the true p is 0.4, is about 0.25%.