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    %% Compute Geometric Distribution icdf
%

% Copyright 2015 The MathWorks, Inc.


%%
% Suppose the probability of a five-year-old car battery not starting in
% cold weather is 0.03. If we want no more than a ten percent chance that the
% car does not start, what is the maximum number of days in a row
% that we should try to start the car?

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% To solve, compute the inverse cdf of the geometric distribution. In this
% example, a "success" means the car does not start, while a "failure"
% means the car does start. The probability of success for each trial
% _p_ equals 0.03, while the probability of observing _x_ failures in a row
% before observing a success _y_ equals 0.1.
y = 0.1;
p = 0.03;
x = geoinv(y,p)

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% The returned result indicates that if we start the car three
% times, there is at least a ten percent chance that it will not start on one of
% those tries. Therefore, if we want no greater than a ten percent chance that the
% car will not start, we should only attempt to start it for a maximum
% of two days in a row.

%%
% We can confirm this result by evaluating the cdf at values of _x_ equal
% to 2 and 3, given the probability of success for each trial _p_ equal to
% 0.03.
y2 = geocdf(2,p)  % cdf for x = 2
y3 = geocdf(3,p)  % cdf for x = 3

%%
% The returned results indicate an 8.7% chance of the car not starting if
% we try two days in a row, and an 11.5% chance of not starting if we try
% three days in a row.