MATLAB HELP STANDARD DEVIATION, MEAN, HISTOGRAMS

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BEN MILLER
BEN MILLER 2020-10-31
回答： Meg Noah 2025-8-5，23:55
PLEASE LEAVE NOTES SO I MAY UNDERSTAND THE STEPS ON HOW TO FIND THE STANDARD DEVIATION AND MEAN

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回答（2 个）

Walter Roberson
Walter Roberson 2020-10-31
mean is mean()
Standard deviation is std()
Meg Noah
Meg Noah 2025-8-5，23:55
Ran in:
Try this:
force_lbs = [243,236,389,628,143,417,205,404,464,605,137,123,372,439,...
497,500,535,577,441,231,675,132,196,217,660,569,865,725,547,347];
mean_lbs = mean(force_lbs);
std_lbs = std(force_lbs);
fprintf(1,'Mean force = %f [lbs]\nStandard Deviation force=%f [lbs]\n' ,...
mean_lbs,std_lbs);
Mean force = 417.300000 [lbs] Standard Deviation force=199.789743 [lbs]
edges_lbs = linspace(-3*std_lbs+mean_lbs,3*std_lbs+mean_lbs,13);
histogram(force_lbs,edges_lbs);
% 68% of the population is approx within 1 standard deviation of the mean
x = norminv([(1-0.68)/2 (1-0.68)/2+0.68]);
upper_limit_68 = mean_lbs + x(2)*std_lbs;
lower_limit_68 = mean_lbs + x(1)*std_lbs;
percentage_in_limit_68 = 100* ...
sum(lower_limit_68 <= force_lbs & force_lbs <= upper_limit_68)/numel(force_lbs);
fprintf(1,'%.4f%s are within the normal 68%s limits [%.4f,%.4f] lbs\n', ...
percentage_in_limit_68, ...
'%','%',lower_limit_68,upper_limit_68);
60.0000% are within the normal 68% limits [218.6175,615.9825] lbs
% 96% of the population is approx within 2.1 standard deviation of the mean
x = norminv([(1-0.96)/2 (1-0.96)/2+0.96]);
upper_limit_96 = mean_lbs + x(2)*std_lbs;
lower_limit_96 = mean_lbs + x(1)*std_lbs;
percentage_in_limit_96 = 100* ...
sum(lower_limit_96 <= force_lbs & force_lbs <= upper_limit_96)/numel(force_lbs);
fprintf(1,'%.4f%s are within the normal 96%s limits [%.4f,%.4f] lbs\n', ...
percentage_in_limit_96, ...
'%','%',lower_limit_96,upper_limit_96);
96.6667% are within the normal 96% limits [6.9820,827.6180] lbs

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