Day 1: Standard Deviation | 10 Days Of Statistics | HakcerRank Solution

Hello coders, today we are going to solve Day 1: Standard Deviation HackerRank Solution which is a part of 10 Days of Statistics Series.

Objective

In this challenge, we practice calculating standard deviation.

Task

Given an array, arr, of n integers, calculate and print the standard deviation. Your answer should be in decimal form, rounded to a scale of 1 decimal place (i.e., 12.3 format). An error margin of +0.1 will be tolerated for the standard deviation.

Example

arr = [2, 5, 2, 7, 4]
The sum of the array values is 20 and there are 5 elements. The mean is 4.0.
Subtract the mean from each element, square each result, and take their sum.

Their sum is 18. Take the square root of 18/5 to get 1.7, the standard deviation.

Function Description

Complete the stdDev function in the editor below.

stdDev has the following parameters:
– int arr[n]: an array of integers

Prints
– float: the standard deviation to 1 place after the decimal

Input Format

The first line contains an integer, n, denoting the size of arr.
The second line contains n space-separated integers that describe arr.

Constraints

• 5 <= n <= 100
• 0 < arr[i] <= 10⁵

Output Format

Print the standard deviation on a new line, rounded to a scale of 1 decimal place (i.e., 12.3 format).

Sample Input

STDIN           Function
-----           --------
5               arr[] size n = 5
10 40 30 50 20  arr =[10, 40, 30, 50, 20]

Sample Output

14.1

Solution – Day 1: Standard Deviation Solution

C++

#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 */ 

    int n; cin>>n; int a[n+4]; double sum=0, sum1=0;

    for(auto i=0; i<n; i++)
    {
        cin>>a[i];
        sum += a[i];
    }

    double mean = (double)(sum/n);

    for(auto i=0; i<n; i++)
    {
        sum1 += (((double)a[i]-mean)*((double)a[i]-mean));
    }

    double std = (double)sqrt(sum1/n);

    printf("%.1f\n", std);

    return 0;
}

Python

import math

# Define functionts
def mean(data):
    return sum(data) / len(data)

def stddev(data, size):
    sum = 0
    for i in range(size):
        sum = sum + (data[i] - mean(data)) ** 2
    return math.sqrt(sum / size)

# Set data
size = int(input())
numbers = list(map(int, input().split()))

# Get standard deviation
print(round(stddev(numbers, size), 1))

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