Hello guys, i need help for estimating FAR from this formula can review my code please

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Ted Erdenekhuyag
Ted Erdenekhuyag 2021-12-16
回答： Yazan 2021-12-16
mu = 3;
sd = 1;
x= linspace(0,10)
y1 = 1/(2*pi*sd)*exp(-(x-mu).^2/(2*sd^2))
plot(x1,y1);
hold on
mu = 5;
sd = 1;
x2 = linspace(0,10);
y2 = 1/(2*pi*sd)*exp(-(x2-mu).^2/(2*sd^2));
plot(x2,y2);
hold on
plot([1 1]*4, ylim, '--r') % Draw A Red Vertical Line At ‘x=5’
hold off
ylabel('q(x)and p(x)')
xlabel('x')
hold on
x1=linspace(4,10)
trapz(x1,y1)

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回答（1 个）

Yazan
Yazan 2021-12-16
Ran in:
Have you noticed that the distributions defined in your codes are not Gaussian? See the demo below
mu = 3;
sigma = 1;
x = mu-6*sigma:0.05:mu+6*sigma;
dist = 1/sqrt(2*pi*sigma)*exp(-(x-mu).^2/(2*sigma^2));
figure, subplot(1,2,1), plot(x, dist);
x0 = 4;
hold on, xline(x0, 'LineStyle', '--', 'Color', 'r'); hold off
% If you have 'Statistics and Machine Learning Toolbox'
if license('test', 'Statistics_Toolbox')
y1 = 1 - cdf('Normal', 4, mu, sigma);
end
% Otherwise:
x2 = 4:0.05:mu+6*sigma;
dist2 = 1/sqrt(2*pi*sigma)*exp(-(x2-mu).^2/(2*sigma^2));
subplot(1,2,2), plot(x2, dist2);
y2 = trapz(x2, dist2);
if license('test', 'Statistics_Toolbox')
fprintf('Result using the function "cdf" is %g, and using "trapz" is %g', y1, y2)
else
fprintf('Result using the function "trapz" is %g', y2)
end
Result using the function "cdf" is 0.158655, and using "trapz" is 0.158706

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