%Here is a script to help you with Exercise 2.1
%p=0.1, n=5,r=0, page 16 of Davis, probability
% of an exploration program of 5 holes with no discoveries
clear
p=0.1;
n=5;
r=0;
nmk=n-r;
%cjw's little script to calculate the combination
%of two numbers, nCr (Eqn. 1.52, GG313 lecture notes)
%note that it might go wrong for large n
temp1=max(r,nmk);
temp2=min(r,nmk);
stop=temp1+0.1;
value=1;
for i=n:-1:stop;
value=value*i;
end
for i=1:temp2
value=value/i;
end
%value is the value of nCr
value
P=value*p^r*(1-p)^(nmk)

clear;
p=0.1;
n=5;
r=1;
nmk=n-r;
temp1=max(r,nmk);
temp2=min(r,nmk);
stop=temp1+0.1;
value=1;
for i=n:-1:stop;
value=value*i;
end
for i=1:temp2
value=value/i;
end
%value is the value of nCr
value
P=value*p^r*(1-p)^(nmk)

%This will help you with Exercise 2.2
clear
load balance.dat
whos
b1=balance(:,1)
b2=balance(:,2)
subplot(2,1,1)
hist(b1);
hold on
subplot(2,1,2)
hist(b2)
pause
hold off
mean1=mean(b1)
mean2=mean(b2)
%or
mean1=sum(b1)/length(b1)
mean2=sum(b2)/length(b2)
%calculate standard deviation--see help std
std(b1)
std(b2)

%Excerise 2.3
clear
load WLYONS.dat
whos
xcor=WLYONS(:,1);
ycor=WLYONS(:,2);
por=WLYONS(:,3);
h20=WLYONS(:,4);
thick=WLYONS(:,5);
%you will need the function normcdf for this problem (1-4)
clf
hist(por,20)
pause
%next part, while you could use a chi-2 test, do this problem instead using
%equation 2.42 for a 
%population with mean 15% and standard deviation (of the population)
%of 5%, get the p value

%the next part, test the equality of the means
hist(thick,20)
i1=0;
i2=0;
for i=1:length(por)
if (thick(i)< 30); 
i1=i1+1
por1(i1)=por(i);
else
i2=i2+1
por2(i2)=por(i);
end
end
subplot(2,1,1)
hist(por1,20);
hold on
subplot(2,1,2)
hist(por2,20);
pause;
clf