HackerRank Day 6: The Central Limit Theorem II | 10 Days of Statistics solution

In this Hackerrank Day 6: The Central Limit Theorem II 10 Days of Statistics problem The number of tickets purchased by each student for the University X vs. University Y football game follows a distribution that has a mean of 2.4 and a standard deviation of 2.0.

A few hours before the game starts, 100 eager students line up to purchase last-minute tickets. If there are only 250 tickets left, what is the probability that all 100 students will be able to purchase tickets?

Problem solution in Python programming.

import math

numTic = float(input())
numStd = float(input())
mu = numStd * float(input())
sig = math.sqrt(numStd) * float(input())

print(round(0.5*(1+math.erf((numTic - mu)/(sig * math.sqrt(2)))),4))

Problem solution in Java Programming.

import java.util.Scanner;
import java.util.stream.IntStream;

public class Solution {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int maxTicketNum = sc.nextInt();
		int n = sc.nextInt();
		double mean = sc.nextDouble();
		double std = sc.nextDouble();

		System.out.printf("%.4f\n", solve(maxTicketNum, n, mean, std));

		sc.close();
	}

	static double solve(int maxTicketNum, int n, double mean, double std) {
		return phi(n * mean, Math.sqrt(n) * std, maxTicketNum);
	}

	static double phi(double mean, double std, double x) {
		return 0.5 * (1 + erf((x - mean) / (std * Math.sqrt(2))));
	}

	// https://en.wikipedia.org/wiki/Error_function#Taylor_series
	static double erf(double z) {
		double result = 0;
		for (int n = 0;; n++) {
			double term = 2 / Math.sqrt(Math.PI) * ((n % 2 == 0) ? 1 : -1) * Math.pow(z, 2 * n + 1)
					/ (factorial(n) * (2 * n + 1));

			if (Math.abs(term) < 1e-5) {
				break;
			}

			result += term;
		}
		return result;
	}

	static double factorial(int n) {
		return IntStream.rangeClosed(1, n).mapToDouble(x -> x).reduce(1, (x, y) -> x * y);
	}
}

Problem solution in C++ programming.

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


int main() {

    int max, N;
    float mu, sigma;

    cin >> max >> N >> mu >> sigma;

    float central_mu = 1.0f * mu * N;
    float central_sigma = 1.0f * sigma * pow(N,0.5);

    float ans = ( erf( (max-central_mu)/central_sigma/pow(2,0.5) ) + 1  )*0.5;

    cout << ans;


    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int n = 100;
    double mean = n * 2.4;
    double sd = 2.0 * pow(n, 0.5);

    double res = erf((250 - mean) / (sd * sqrt(2.0)));

    printf("%.4f", (0.5) * (1 + res));

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

Problem solution in JavaScript programming.

function erf(x) {
    //jacked from the internet
    let a1 = 0.254829592,
        a2 = -0.284496736,
        a3 = 1.421413741,
        a4 = -1.453152027,
        a5 = 1.061405429,
        p = 0.3275911

    let sign = 1

    if (x < 0)
        sign = -1

    x = Math.abs(x)

    let t = 1.0 / (1.0 + p * x)
    let y = 1.0 - (((((a5 * t + a4) * t) + a3) * t + a2) * t + a1) * t * Math.exp(-x * x)

    return sign * y
}

function cumdev(m, sd, x) {
    return 1 / 2 * (1 + erf((x - m) / (sd * Math.sqrt(2))))
}

function processData(input) {
    let max = 250;
    let n = 100;
    let mu = 2.4;
    let stdev = 2.0;

    console.log(cumdev(n * mu, Math.sqrt(n) * stdev, max).toFixed(4));
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});