% Class 5 - Statistics
% Coded by Nigel Reuel on 9.10.19
% 
% Load the data
D = readmatrix('Bulls1997.xlsx')
Smean = mean(D(:,3));
fprintf('Mean salary is $%.2f\n',Smean)
Smedian = median(D(:,3));
fprintf('Median salary is $%.2f\n',Smedian)
% Also measure the spread
% Using the standard deviation
S_std = std(D(:,3))
% You can measure the interquartile range
S_iqr = iqr(D(:,3))
% Example of box and whisker plot
boxplot(D(:,3))
% Histogram
%histogram(D(2:end,3))
% With the class 2017 data, 
% report the average commute,
% also report 95% confidence range (+/- 2*std)
% Column 2 has the commute time
CT = readmatrix('ClassData.xlsx');
Avg_CT = mean(CT(:,2))
CT_std = std(CT(:,2));
fprintf('We are 95 percent confident that commute time is between %.2f and %.2f min\n',Avg_CT-2*CT_std,Avg_CT+2*CT_std);
% This data is a large sample, let's look
% at a histogram
histogram(CT(:,2))
xlabel('Commute Time (min)')
ylabel('Frequency')
% Example of histogram w/ more input
histogram(CT(:,2),10) %number of bins
xlabel('Commute Time (min)')
ylabel('Frequency')
% If you need to extract information,
% you can use an output variable
h1 = histogram(CT(:,2),10) %number of bins
xlabel('Commute Time (min)')
ylabel('Frequency')
% To pull off the count information
Counts = h1.Values
% Example of plotting histogram
% Determining type of distribution
% Plotting histogram w/ distribution
% You measure the lifetime of a product (months)
life = [ 6.2 16.1 16.3 19.0 12.2  8.1  8.8  5.9  7.3  8.2 ...
        16.1 12.8  9.8 11.3  5.1 10.8  6.7  1.2  8.3  2.3 ...
         4.3  2.9 14.8  4.6  3.1 13.6 14.5  5.2  5.7  6.5 ...
         5.3  6.4  3.5 11.4  9.3 12.4 18.3 15.9  4.0 10.4 ...
         8.7  3.0 12.1  3.9  6.5  3.4  8.5  0.9  9.9  7.9];
histogram(life,15)
xlabel('Product Lifetime (months)')
ylabel('Frequency')
% From inspection, you see skew,
% it is not a normal distribution
% Poisson distribution can be used for failure analysis
histfit(life,15,'Poisson')
Param = fitdist(life','Poisson') % this function needs a column vector input
% Post example of use of icdf 
%
% AFTER CLASS EXAMPLE
% I forgot to show two important things,
% generating random data for a distribution
% and using icdf
%
% 1) Assume a distribution of athlete heights
mu = 1.75; %m
sigma = 0.13;%m
% Generate 100 athlete hights from this normal distrubution
H_vec = random('Normal',mu,sigma,100,1);
% Make a histogram from this data, 20 bins
histogram(H_vec,20)
% Make a distribution from this data
pd1 = fitdist(H_vec,'Normal');
% Make a distrubution from the pdf parameters
pdf2 = makedist('Normal','mu',mu,'sigma',sigma);
% Use the pdf to make predictions
% Q1) How many athletes would you need to screen
% to find one that is > 2 meters tall?
% (NOTE: base if off of pdf2, it is more exact.  WHY?)
prob = cdf(pdf2,2)
n_screen = 1/(1-prob);
fprintf('You need to screen %d athletes to find one > 2m tall!\n',ceil(n_screen))
%
% NOW, inverse cdf is useful if you want
% to ask the opposite type of question, where
% we know the probability but want to determine
% the x-value
%
% Q2:How tall are the top 0.01% of athletes
% according to this distribution
x_top = icdf(pdf2,1-0.0001);
fprintf('The top 0.01 percent athletes are greater than %.2f m tall!\n',x_top)