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510 lines
21 KiB
C++
510 lines
21 KiB
C++
// normal_misc_examples.cpp
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// Copyright Paul A. Bristow 2007, 2010.
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// Use, modification and distribution are subject to the
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// Boost Software License, Version 1.0.
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// (See accompanying file LICENSE_1_0.txt
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// or copy at http://www.boost.org/LICENSE_1_0.txt)
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// Example of using normal distribution.
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// Note that this file contains Quickbook mark-up as well as code
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// and comments, don't change any of the special comment mark-ups!
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//[normal_basic1
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/*`
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First we need some includes to access the normal distribution
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(and some std output of course).
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*/
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#include <boost/math/distributions/normal.hpp> // for normal_distribution
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using boost::math::normal; // typedef provides default type is double.
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#include <iostream>
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using std::cout; using std::endl; using std::left; using std::showpoint; using std::noshowpoint;
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#include <iomanip>
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using std::setw; using std::setprecision;
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#include <limits>
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using std::numeric_limits;
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int main()
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{
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cout << "Example: Normal distribution, Miscellaneous Applications.";
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try
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{
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{ // Traditional tables and values.
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/*`Let's start by printing some traditional tables.
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*/
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double step = 1.; // in z
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double range = 4; // min and max z = -range to +range.
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int precision = 17; // traditional tables are only computed to much lower precision.
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// but std::numeric_limits<double>::max_digits10; on new Standard Libraries gives
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// 17, the maximum number of digits that can possibly be significant.
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// std::numeric_limits<double>::digits10; == 15 is number of guaranteed digits,
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// the other two digits being 'noisy'.
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// Construct a standard normal distribution s
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normal s; // (default mean = zero, and standard deviation = unity)
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cout << "Standard normal distribution, mean = "<< s.mean()
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<< ", standard deviation = " << s.standard_deviation() << endl;
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/*` First the probability distribution function (pdf).
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*/
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cout << "Probability distribution function values" << endl;
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cout << " z " " pdf " << endl;
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cout.precision(5);
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for (double z = -range; z < range + step; z += step)
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{
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cout << left << setprecision(3) << setw(6) << z << " "
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<< setprecision(precision) << setw(12) << pdf(s, z) << endl;
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}
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cout.precision(6); // default
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/*`And the area under the normal curve from -[infin] up to z,
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the cumulative distribution function (cdf).
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*/
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// For a standard normal distribution
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cout << "Standard normal mean = "<< s.mean()
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<< ", standard deviation = " << s.standard_deviation() << endl;
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cout << "Integral (area under the curve) from - infinity up to z " << endl;
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cout << " z " " cdf " << endl;
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for (double z = -range; z < range + step; z += step)
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{
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cout << left << setprecision(3) << setw(6) << z << " "
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<< setprecision(precision) << setw(12) << cdf(s, z) << endl;
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}
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cout.precision(6); // default
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/*`And all this you can do with a nanoscopic amount of work compared to
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the team of *human computers* toiling with Milton Abramovitz and Irene Stegen
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at the US National Bureau of Standards (now [@http://www.nist.gov NIST]).
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Starting in 1938, their "Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables",
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was eventually published in 1964, and has been reprinted numerous times since.
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(A major replacement is planned at [@http://dlmf.nist.gov Digital Library of Mathematical Functions]).
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Pretty-printing a traditional 2-dimensional table is left as an exercise for the student,
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but why bother now that the Math Toolkit lets you write
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*/
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double z = 2.;
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cout << "Area for z = " << z << " is " << cdf(s, z) << endl; // to get the area for z.
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/*`
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Correspondingly, we can obtain the traditional 'critical' values for significance levels.
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For the 95% confidence level, the significance level usually called alpha,
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is 0.05 = 1 - 0.95 (for a one-sided test), so we can write
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*/
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cout << "95% of area has a z below " << quantile(s, 0.95) << endl;
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// 95% of area has a z below 1.64485
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/*`and a two-sided test (a comparison between two levels, rather than a one-sided test)
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*/
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cout << "95% of area has a z between " << quantile(s, 0.975)
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<< " and " << -quantile(s, 0.975) << endl;
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// 95% of area has a z between 1.95996 and -1.95996
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/*`
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First, define a table of significance levels: these are the probabilities
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that the true occurrence frequency lies outside the calculated interval.
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It is convenient to have an alpha level for the probability that z lies outside just one standard deviation.
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This will not be some nice neat number like 0.05, but we can easily calculate it,
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*/
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double alpha1 = cdf(s, -1) * 2; // 0.3173105078629142
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cout << setprecision(17) << "Significance level for z == 1 is " << alpha1 << endl;
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/*`
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and place in our array of favorite alpha values.
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*/
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double alpha[] = {0.3173105078629142, // z for 1 standard deviation.
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0.20, 0.1, 0.05, 0.01, 0.001, 0.0001, 0.00001 };
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/*`
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Confidence value as % is (1 - alpha) * 100 (so alpha 0.05 == 95% confidence)
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that the true occurrence frequency lies *inside* the calculated interval.
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*/
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cout << "level of significance (alpha)" << setprecision(4) << endl;
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cout << "2-sided 1 -sided z(alpha) " << endl;
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for (int i = 0; i < sizeof(alpha)/sizeof(alpha[0]); ++i)
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{
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cout << setw(15) << alpha[i] << setw(15) << alpha[i] /2 << setw(10) << quantile(complement(s, alpha[i]/2)) << endl;
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// Use quantile(complement(s, alpha[i]/2)) to avoid potential loss of accuracy from quantile(s, 1 - alpha[i]/2)
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}
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cout << endl;
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/*`Notice the distinction between one-sided (also called one-tailed)
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where we are using a > *or* < test (and not both)
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and considering the area of the tail (integral) from z up to +[infin],
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and a two-sided test where we are using two > *and* < tests, and thus considering two tails,
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from -[infin] up to z low and z high up to +[infin].
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So the 2-sided values alpha[i] are calculated using alpha[i]/2.
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If we consider a simple example of alpha = 0.05, then for a two-sided test,
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the lower tail area from -[infin] up to -1.96 is 0.025 (alpha/2)
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and the upper tail area from +z up to +1.96 is also 0.025 (alpha/2),
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and the area between -1.96 up to 12.96 is alpha = 0.95.
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and the sum of the two tails is 0.025 + 0.025 = 0.05,
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*/
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//] [/[normal_basic1]
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//[normal_basic2
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/*`Armed with the cumulative distribution function, we can easily calculate the
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easy to remember proportion of values that lie within 1, 2 and 3 standard deviations from the mean.
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*/
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cout.precision(3);
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cout << showpoint << "cdf(s, s.standard_deviation()) = "
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<< cdf(s, s.standard_deviation()) << endl; // from -infinity to 1 sd
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cout << "cdf(complement(s, s.standard_deviation())) = "
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<< cdf(complement(s, s.standard_deviation())) << endl;
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cout << "Fraction 1 standard deviation within either side of mean is "
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<< 1 - cdf(complement(s, s.standard_deviation())) * 2 << endl;
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cout << "Fraction 2 standard deviations within either side of mean is "
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<< 1 - cdf(complement(s, 2 * s.standard_deviation())) * 2 << endl;
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cout << "Fraction 3 standard deviations within either side of mean is "
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<< 1 - cdf(complement(s, 3 * s.standard_deviation())) * 2 << endl;
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/*`
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To a useful precision, the 1, 2 & 3 percentages are 68, 95 and 99.7,
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and these are worth memorising as useful 'rules of thumb', as, for example, in
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[@http://en.wikipedia.org/wiki/Standard_deviation standard deviation]:
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[pre
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Fraction 1 standard deviation within either side of mean is 0.683
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Fraction 2 standard deviations within either side of mean is 0.954
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Fraction 3 standard deviations within either side of mean is 0.997
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]
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We could of course get some really accurate values for these
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[@http://en.wikipedia.org/wiki/Confidence_interval confidence intervals]
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by using cout.precision(15);
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[pre
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Fraction 1 standard deviation within either side of mean is 0.682689492137086
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Fraction 2 standard deviations within either side of mean is 0.954499736103642
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Fraction 3 standard deviations within either side of mean is 0.997300203936740
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]
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But before you get too excited about this impressive precision,
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don't forget that the *confidence intervals of the standard deviation* are surprisingly wide,
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especially if you have estimated the standard deviation from only a few measurements.
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*/
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//] [/[normal_basic2]
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//[normal_bulbs_example1
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/*`
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Examples from K. Krishnamoorthy, Handbook of Statistical Distributions with Applications,
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ISBN 1 58488 635 8, page 125... implemented using the Math Toolkit library.
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A few very simple examples are shown here:
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*/
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// K. Krishnamoorthy, Handbook of Statistical Distributions with Applications,
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// ISBN 1 58488 635 8, page 125, example 10.3.5
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/*`Mean lifespan of 100 W bulbs is 1100 h with standard deviation of 100 h.
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Assuming, perhaps with little evidence and much faith, that the distribution is normal,
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we construct a normal distribution called /bulbs/ with these values:
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*/
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double mean_life = 1100.;
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double life_standard_deviation = 100.;
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normal bulbs(mean_life, life_standard_deviation);
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double expected_life = 1000.;
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/*`The we can use the Cumulative distribution function to predict fractions
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(or percentages, if * 100) that will last various lifetimes.
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*/
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cout << "Fraction of bulbs that will last at best (<=) " // P(X <= 1000)
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<< expected_life << " is "<< cdf(bulbs, expected_life) << endl;
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cout << "Fraction of bulbs that will last at least (>) " // P(X > 1000)
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<< expected_life << " is "<< cdf(complement(bulbs, expected_life)) << endl;
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double min_life = 900;
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double max_life = 1200;
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cout << "Fraction of bulbs that will last between "
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<< min_life << " and " << max_life << " is "
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<< cdf(bulbs, max_life) // P(X <= 1200)
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- cdf(bulbs, min_life) << endl; // P(X <= 900)
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/*`
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[note Real-life failures are often very ab-normal,
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with a significant number that 'dead-on-arrival' or suffer failure very early in their life:
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the lifetime of the survivors of 'early mortality' may be well described by the normal distribution.]
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*/
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//] [/normal_bulbs_example1 Quickbook end]
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}
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{
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// K. Krishnamoorthy, Handbook of Statistical Distributions with Applications,
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// ISBN 1 58488 635 8, page 125, Example 10.3.6
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//[normal_bulbs_example3
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/*`Weekly demand for 5 lb sacks of onions at a store is normally distributed with mean 140 sacks and standard deviation 10.
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*/
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double mean = 140.; // sacks per week.
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double standard_deviation = 10;
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normal sacks(mean, standard_deviation);
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double stock = 160.; // per week.
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cout << "Percentage of weeks overstocked "
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<< cdf(sacks, stock) * 100. << endl; // P(X <=160)
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// Percentage of weeks overstocked 97.7
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/*`So there will be lots of mouldy onions!
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So we should be able to say what stock level will meet demand 95% of the weeks.
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*/
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double stock_95 = quantile(sacks, 0.95);
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cout << "Store should stock " << int(stock_95) << " sacks to meet 95% of demands." << endl;
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/*`And it is easy to estimate how to meet 80% of demand, and waste even less.
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*/
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double stock_80 = quantile(sacks, 0.80);
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cout << "Store should stock " << int(stock_80) << " sacks to meet 8 out of 10 demands." << endl;
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//] [/normal_bulbs_example3 Quickbook end]
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}
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{ // K. Krishnamoorthy, Handbook of Statistical Distributions with Applications,
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// ISBN 1 58488 635 8, page 125, Example 10.3.7
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//[normal_bulbs_example4
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/*`A machine is set to pack 3 kg of ground beef per pack.
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Over a long period of time it is found that the average packed was 3 kg
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with a standard deviation of 0.1 kg.
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Assuming the packing is normally distributed,
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we can find the fraction (or %) of packages that weigh more than 3.1 kg.
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*/
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double mean = 3.; // kg
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double standard_deviation = 0.1; // kg
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normal packs(mean, standard_deviation);
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double max_weight = 3.1; // kg
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cout << "Percentage of packs > " << max_weight << " is "
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<< cdf(complement(packs, max_weight)) << endl; // P(X > 3.1)
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double under_weight = 2.9;
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cout <<"fraction of packs <= " << under_weight << " with a mean of " << mean
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<< " is " << cdf(complement(packs, under_weight)) << endl;
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// fraction of packs <= 2.9 with a mean of 3 is 0.841345
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// This is 0.84 - more than the target 0.95
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// Want 95% to be over this weight, so what should we set the mean weight to be?
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// KK StatCalc says:
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double over_mean = 3.0664;
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normal xpacks(over_mean, standard_deviation);
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cout << "fraction of packs >= " << under_weight
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<< " with a mean of " << xpacks.mean()
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<< " is " << cdf(complement(xpacks, under_weight)) << endl;
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// fraction of packs >= 2.9 with a mean of 3.06449 is 0.950005
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double under_fraction = 0.05; // so 95% are above the minimum weight mean - sd = 2.9
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double low_limit = standard_deviation;
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double offset = mean - low_limit - quantile(packs, under_fraction);
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double nominal_mean = mean + offset;
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normal nominal_packs(nominal_mean, standard_deviation);
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cout << "Setting the packer to " << nominal_mean << " will mean that "
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<< "fraction of packs >= " << under_weight
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<< " is " << cdf(complement(nominal_packs, under_weight)) << endl;
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/*`
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Setting the packer to 3.06449 will mean that fraction of packs >= 2.9 is 0.95.
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Setting the packer to 3.13263 will mean that fraction of packs >= 2.9 is 0.99,
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but will more than double the mean loss from 0.0644 to 0.133.
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Alternatively, we could invest in a better (more precise) packer with a lower standard deviation.
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To estimate how much better (how much smaller standard deviation) it would have to be,
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we need to get the 5% quantile to be located at the under_weight limit, 2.9
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*/
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double p = 0.05; // wanted p th quantile.
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cout << "Quantile of " << p << " = " << quantile(packs, p)
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<< ", mean = " << packs.mean() << ", sd = " << packs.standard_deviation() << endl; //
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/*`
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Quantile of 0.05 = 2.83551, mean = 3, sd = 0.1
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With the current packer (mean = 3, sd = 0.1), the 5% quantile is at 2.8551 kg,
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a little below our target of 2.9 kg.
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So we know that the standard deviation is going to have to be smaller.
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Let's start by guessing that it (now 0.1) needs to be halved, to a standard deviation of 0.05
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*/
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normal pack05(mean, 0.05);
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cout << "Quantile of " << p << " = " << quantile(pack05, p)
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<< ", mean = " << pack05.mean() << ", sd = " << pack05.standard_deviation() << endl;
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cout <<"Fraction of packs >= " << under_weight << " with a mean of " << mean
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<< " and standard deviation of " << pack05.standard_deviation()
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<< " is " << cdf(complement(pack05, under_weight)) << endl;
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//
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/*`
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Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.05 is 0.9772
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So 0.05 was quite a good guess, but we are a little over the 2.9 target,
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so the standard deviation could be a tiny bit more. So we could do some
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more guessing to get closer, say by increasing to 0.06
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*/
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normal pack06(mean, 0.06);
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cout << "Quantile of " << p << " = " << quantile(pack06, p)
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<< ", mean = " << pack06.mean() << ", sd = " << pack06.standard_deviation() << endl;
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cout <<"Fraction of packs >= " << under_weight << " with a mean of " << mean
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<< " and standard deviation of " << pack06.standard_deviation()
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<< " is " << cdf(complement(pack06, under_weight)) << endl;
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/*`
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Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.06 is 0.9522
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Now we are getting really close, but to do the job properly,
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we could use root finding method, for example the tools provided, and used elsewhere,
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in the Math Toolkit, see __root_finding_without_derivatives.
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But in this normal distribution case, we could be even smarter and make a direct calculation.
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*/
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normal s; // For standard normal distribution,
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double sd = 0.1;
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double x = 2.9; // Our required limit.
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// then probability p = N((x - mean) / sd)
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// So if we want to find the standard deviation that would be required to meet this limit,
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// so that the p th quantile is located at x,
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// in this case the 0.95 (95%) quantile at 2.9 kg pack weight, when the mean is 3 kg.
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double prob = pdf(s, (x - mean) / sd);
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double qp = quantile(s, 0.95);
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cout << "prob = " << prob << ", quantile(p) " << qp << endl; // p = 0.241971, quantile(p) 1.64485
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// Rearranging, we can directly calculate the required standard deviation:
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double sd95 = std::abs((x - mean)) / qp;
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cout << "If we want the "<< p << " th quantile to be located at "
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<< x << ", would need a standard deviation of " << sd95 << endl;
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normal pack95(mean, sd95); // Distribution of the 'ideal better' packer.
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cout <<"Fraction of packs >= " << under_weight << " with a mean of " << mean
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<< " and standard deviation of " << pack95.standard_deviation()
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<< " is " << cdf(complement(pack95, under_weight)) << endl;
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// Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.0608 is 0.95
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/*`Notice that these two deceptively simple questions
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(do we over-fill or measure better) are actually very common.
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The weight of beef might be replaced by a measurement of more or less anything.
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But the calculations rely on the accuracy of the standard deviation - something
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that is almost always less good than we might wish,
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especially if based on a few measurements.
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*/
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//] [/normal_bulbs_example4 Quickbook end]
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}
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{ // K. Krishnamoorthy, Handbook of Statistical Distributions with Applications,
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// ISBN 1 58488 635 8, page 125, example 10.3.8
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//[normal_bulbs_example5
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/*`A bolt is usable if between 3.9 and 4.1 long.
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From a large batch of bolts, a sample of 50 show a
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mean length of 3.95 with standard deviation 0.1.
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Assuming a normal distribution, what proportion is usable?
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The true sample mean is unknown,
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but we can use the sample mean and standard deviation to find approximate solutions.
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*/
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normal bolts(3.95, 0.1);
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double top = 4.1;
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double bottom = 3.9;
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cout << "Fraction long enough [ P(X <= " << top << ") ] is " << cdf(bolts, top) << endl;
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cout << "Fraction too short [ P(X <= " << bottom << ") ] is " << cdf(bolts, bottom) << endl;
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cout << "Fraction OK -between " << bottom << " and " << top
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<< "[ P(X <= " << top << ") - P(X<= " << bottom << " ) ] is "
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<< cdf(bolts, top) - cdf(bolts, bottom) << endl;
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cout << "Fraction too long [ P(X > " << top << ") ] is "
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<< cdf(complement(bolts, top)) << endl;
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cout << "95% of bolts are shorter than " << quantile(bolts, 0.95) << endl;
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//] [/normal_bulbs_example5 Quickbook end]
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}
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}
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catch(const std::exception& e)
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{ // Always useful to include try & catch blocks because default policies
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// are to throw exceptions on arguments that cause errors like underflow, overflow.
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// Lacking try & catch blocks, the program will abort without a message below,
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// which may give some helpful clues as to the cause of the exception.
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|
std::cout <<
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"\n""Message from thrown exception was:\n " << e.what() << std::endl;
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}
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return 0;
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} // int main()
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|
|
|
|
|
/*
|
|
|
|
Output is:
|
|
|
|
Autorun "i:\boost-06-05-03-1300\libs\math\test\Math_test\debug\normal_misc_examples.exe"
|
|
Example: Normal distribution, Miscellaneous Applications.Standard normal distribution, mean = 0, standard deviation = 1
|
|
Probability distribution function values
|
|
z pdf
|
|
-4 0.00013383022576488537
|
|
-3 0.0044318484119380075
|
|
-2 0.053990966513188063
|
|
-1 0.24197072451914337
|
|
0 0.3989422804014327
|
|
1 0.24197072451914337
|
|
2 0.053990966513188063
|
|
3 0.0044318484119380075
|
|
4 0.00013383022576488537
|
|
Standard normal mean = 0, standard deviation = 1
|
|
Integral (area under the curve) from - infinity up to z
|
|
z cdf
|
|
-4 3.1671241833119979e-005
|
|
-3 0.0013498980316300959
|
|
-2 0.022750131948179219
|
|
-1 0.1586552539314571
|
|
0 0.5
|
|
1 0.84134474606854293
|
|
2 0.97724986805182079
|
|
3 0.9986501019683699
|
|
4 0.99996832875816688
|
|
Area for z = 2 is 0.97725
|
|
95% of area has a z below 1.64485
|
|
95% of area has a z between 1.95996 and -1.95996
|
|
Significance level for z == 1 is 0.3173105078629142
|
|
level of significance (alpha)
|
|
2-sided 1 -sided z(alpha)
|
|
0.3173 0.1587 1
|
|
0.2 0.1 1.282
|
|
0.1 0.05 1.645
|
|
0.05 0.025 1.96
|
|
0.01 0.005 2.576
|
|
0.001 0.0005 3.291
|
|
0.0001 5e-005 3.891
|
|
1e-005 5e-006 4.417
|
|
cdf(s, s.standard_deviation()) = 0.841
|
|
cdf(complement(s, s.standard_deviation())) = 0.159
|
|
Fraction 1 standard deviation within either side of mean is 0.683
|
|
Fraction 2 standard deviations within either side of mean is 0.954
|
|
Fraction 3 standard deviations within either side of mean is 0.997
|
|
Fraction of bulbs that will last at best (<=) 1.00e+003 is 0.159
|
|
Fraction of bulbs that will last at least (>) 1.00e+003 is 0.841
|
|
Fraction of bulbs that will last between 900. and 1.20e+003 is 0.819
|
|
Percentage of weeks overstocked 97.7
|
|
Store should stock 156 sacks to meet 95% of demands.
|
|
Store should stock 148 sacks to meet 8 out of 10 demands.
|
|
Percentage of packs > 3.10 is 0.159
|
|
fraction of packs <= 2.90 with a mean of 3.00 is 0.841
|
|
fraction of packs >= 2.90 with a mean of 3.07 is 0.952
|
|
Setting the packer to 3.06 will mean that fraction of packs >= 2.90 is 0.950
|
|
Quantile of 0.0500 = 2.84, mean = 3.00, sd = 0.100
|
|
Quantile of 0.0500 = 2.92, mean = 3.00, sd = 0.0500
|
|
Fraction of packs >= 2.90 with a mean of 3.00 and standard deviation of 0.0500 is 0.977
|
|
Quantile of 0.0500 = 2.90, mean = 3.00, sd = 0.0600
|
|
Fraction of packs >= 2.90 with a mean of 3.00 and standard deviation of 0.0600 is 0.952
|
|
prob = 0.242, quantile(p) 1.64
|
|
If we want the 0.0500 th quantile to be located at 2.90, would need a standard deviation of 0.0608
|
|
Fraction of packs >= 2.90 with a mean of 3.00 and standard deviation of 0.0608 is 0.950
|
|
Fraction long enough [ P(X <= 4.10) ] is 0.933
|
|
Fraction too short [ P(X <= 3.90) ] is 0.309
|
|
Fraction OK -between 3.90 and 4.10[ P(X <= 4.10) - P(X<= 3.90 ) ] is 0.625
|
|
Fraction too long [ P(X > 4.10) ] is 0.0668
|
|
95% of bolts are shorter than 4.11
|
|
|
|
*/
|