Day 5: Normal Distribution II HackerRank Solution

Day 5: Normal Distribution II HackerRank Solution
Source : https://www.hackerrank.com/challenges/s10-normal-distribution-2

Objective
In this challenge, we go further with normal distributions. We recommend reviewing the previous challenge's Tutorial before attempting this problem.

Task
The final grades for a Physics exam taken by a large group of students have a mean of and a standard deviation of . If we can approximate the distribution of these grades by a normal distribution, what percentage of the students:

1. Scored higher than (i.e., have a )?
2. Passed the test (i.e., have a )?
3. Failed the test (i.e., have a )?

Find and print the answer to each question on a new line, rounded to a scale of decimal places.

Input Format

There are lines of input (shown below):

70 108060

The first line contains space-separated values denoting the respective mean and standard deviation for the exam. The second line contains the number associated with question . The third line contains the pass/fail threshold number associated with questions and .

If you do not wish to read this information from stdin, you can hard-code it into your program.

Output Format

There are three lines of output. Your answers must be rounded to a scale of decimal places (i.e., format):

1. On the first line, print the answer to question (i.e., the percentage of students having ).
2. On the second line, print the answer to question (i.e., the percentage of students having ).
3. On the third line, print the answer to question (i.e., the percentage of students having ).

Source : https://www.hackerrank.com/challenges/s10-normal-distribution-2

Solution

	// github.com/RodneyShag
	public class Solution {
	public static void main(String[] args) {
	double mean = 70;
	double std = 10;
	double result_1 = 100 * (1 - cumulative(mean, std, 80));
	double result_2 = 100 * (1 - cumulative(mean, std, 60));
	double result_3 = 100 * cumulative(mean, std, 60);
	System.out.format("%.2f%n", result_1);
	System.out.format("%.2f%n", result_2);
	System.out.format("%.2f%n", result_3);
	}
	/* Calculates cumulative probability */
	public static double cumulative(double mean, double std, double x) {
	double parameter = (x - mean) / (std * Math.sqrt(2));
	return (0.5) * (1 + erf(parameter));
	}
	/* Source: http://introcs.cs.princeton.edu/java/21function/ErrorFunction.java.html */
	// fractional error in math formula less than 1.2 * 10 ^ -7.
	// although subject to catastrophic cancellation when z in very close to 0
	// from Chebyshev fitting formula for erf(z) from Numerical Recipes, 6.2
	public static double erf(double z) {
	double t = 1.0 / (1.0 + 0.5 * Math.abs(z));
	// use Horner's method
	double ans = 1 - t * Math.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;
	}
	}