day5_4

// In this challenge, we go further with normal distributions. //

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;

//Credit for erf: https://introcs.cs.princeton.edu/java/21function/ErrorFunction.java.html
float errorFunction(float z) {

        double t = 1.0 / (1.0 + 0.5 * abs(z));

        // use Horner's method
        double ans = 1 - t * exp( -z*z   -   1.26551223 +
                                            t * ( 1.00002368 +
                                            t * ( 0.37409196 + 
                                            t * ( 0.09678418 + 
                                            t * (-0.18628806 + 
                                            t * ( 0.27886807 + 
                                            t * (-1.13520398 + 
                                            t * ( 1.48851587 + 
                                            t * (-0.82215223 + 
                                            t * ( 0.17087277))))))))));
        if (z >= 0) return  ans;
        else        return -ans;
}

float normal(float x, float stdDev, float mean){
    return 0.5*(1 + (errorFunction((x-mean)/(stdDev*sqrt(2)))));
}
int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT
    There are three lines of output. 
    On the first line, print the answer to question grade >= 80
    (i.e., the percentage of students having grade >= 80).
    On the second line, print the answer to question grade >= 60
    (i.e., the percentage of students having grade >= 60).
    On the third line, print the answer to question <= 60
    (i.e., the percentage of students having grade <= 60).
    */
    const int mean = 70;
    const int stdDev = 10;
    const int var = pow(stdDev, 2); 
    float probability1 = 1 - normal(80, stdDev, mean);
    float probability2 = 1 - normal(60, stdDev, mean);
    float probability3 = 1 - probability2;
    // Multiply with 100 to get percentages
    cout << fixed << setprecision(2) << 100*probability1 << endl
         << 100*probability2 << endl << 100*probability3;
    return 0;
}