Day 6: The Central Limit Theorem I | 10 Days Of Statistics | HackerRank Solution

Hello coders, today we are going to solve Day 6: The Central Limit Theorem I HackerRank Solution which is a Part of 10 Days of Statistics Series.

Day 6: The Central Limit Theorem I



In this challenge, we practice solving problems based on the Central Limit Theorem.


A large elevator can transport a maximum of 9800 pounds. Suppose a load of cargo containing 49 boxes must be transported via the elevator. The box weight of this type of cargo follows a distribution with a mean of u = 205 pounds and a standard deviation of o = 15 pounds. Based on this information, what is the probability that all 49 boxes can be safely loaded into the freight elevator and transported?

Input Format

There are 4 lines of input (shown below):


The first line contains the maximum weight the elevator can transport. The second line contains the number of boxes in the cargo. The third line contains the mean weight of a cargo box, and the fourth line contains its standard deviation.

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

Output Format

Print the probability that the elevator can successfully transport all 49 boxes, rounded to a scale of 4 decimal places (i.e., 1.2345 format).

Solution – The Central Limit Theorem I


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

double normal_dist(double m, double sd, double x)
        p = 1/2*(1 + erf((x - m)/(sd*sqrt(2))));
    double p = 0.5*(1 + erf((x-m)/(sd*sqrt(2.0))));
    return p; 

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  
    double m = 49*205, sd = sqrt(49)*15, x = 9800;
    printf("%0.4f", normal_dist(m, sd, x));
    return 0;

Disclaimer: The above Problem (The Central Limit Theorem I) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

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