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

In this Hackerrank Day 6: The Central Limit Theorem III 10 Days of Statistics problem You have a sample of 100 values from a population with a mean of 500 and with a standard deviation of 80. Compute the interval that covers the middle 95% of the distribution of the sample mean; in other words, compute A and B such that P(A<x<B)=0.95. Use the value of z=1.96.

## Problem solution in Python programming.

```# Enter your code here. Read input from STDIN. Print output to STDOUT
# true population distribution
mu, sigma = 500, 80

# sample mean distribution
muS, sigmaS = mu, sigma/(100**0.5)

# confidence intervals of sample mean dist
A = mu - (1.96*sigmaS)
B = mu + (1.96*sigmaS)

print(round(A,2))
print(round(B,2))```

## Problem solution in Java Programming.

```import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
float mean = scan.nextFloat();
float sigma = scan.nextFloat();
float goal = scan.nextFloat();
float z = scan.nextFloat();

float sampleSigma = sigma/(float)Math.sqrt(n);

float A = mean - z*sampleSigma;
float B = mean + z*sampleSigma;
System.out.printf("%.2f\n",A);
System.out.printf("%.2f\n",B);
}

}```

## Problem solution in C++ programming.

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

int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
double x1,x2;
x1=-8*1.96+500;
x2=8*1.96+500;
printf("%.2f\n%.2f",x1,x2);
return 0;
}```

## Problem solution in C programming.

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

int main() {

double mean   = 500;
double std    = 80;
int    n      = 100;
double zScore = 1.96;

double marginOfError =(double) (zScore * std) / (sqrt(n));

/* Print output */
printf("%.2f\n", mean - marginOfError);
printf("%.2f\n", mean + marginOfError);
return 0;
}
```

## Problem solution in JavaScript programming.

```function processData(input) {
//Enter your code here
input = input.split('\n')

let mx = parseFloat(input[0]),
m = parseFloat(input[1]),
sd = parseFloat(input[2]),
i = parseFloat(input[3]),
z = parseFloat(input[4])

function confidenceInterval(z, sd, mx) {
return z * (sd / Math.sqrt(mx))
}

console.log((m - confidenceInterval(z, sd, mx)).toFixed(2))
console.log((m + confidenceInterval(z, sd, mx)).toFixed(2))
}

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

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