Find Standard Deviation of numbers in Array

Given an array of n elements and we have to find the standard deviation of numbers.

Examples:

Input : arr[] = {9, 2, 5, 4, 12, 7, 8, 11, 9, 3, 7, 4, 12, 5, 4, 10, 9, 6, 9, 4}
Output : 2.983

Input : arr[] = {10, 5, 4, 7, 9, 34, 16, 23, 28}
Output : 10.224

Formula:
Standard deviation of numbers is calculated by the formula
Standard Deviation = sqrt(sum of(arr[i] – mean)²))/n)
Where i = 0, 1, 2, . . . , n.
mean = sum of all elements / number of elements.

Example:

arr[5] = {4, 7, 3, 9, 6}

mean = (4 + 7 + 3 + 9 + 6) / 5
     = 5.8

sd = sqrt(((4 - 5.8)2 + (7 - 5.8)2 + (3 - 5.8)2 + (9 - 5.8)2 + (6 - 5.8)2) / n)
   = 2.135

Implementation:

//Program to calculate standard deviation	
//of the given numbers. 
#include <bits/stdc++.h > 
using namespace std;	

//Function that returns standard deviation. 
float standardDeviation(int arr[], int n) { 
    
    float mean; 
    float sum = 0;	
    
    //loop for sum of all the elements in array. 
    for(int i = 0; i < n; i++) 
    sum = sum + arr[i];	
    
    //find arithmetic mean using formula sum of all	
    //numbers divided by numbers of elements. 
    mean = (float)sum / n; 
    sum = 0;	
    
    //loop for sum of square of mean deviation. 
    for(int i = 0; i < n; i++) 
    sum = sum + (arr[i] - mean) *(arr[i] - mean); 
    
    float sd;	
    //sd is the standard deviation of numbers. 
    sd = sqrt(sum / n); 
    
    return sd;
} 

int main() { 	
    
    int arr[] = {9, 2, 5, 4, 12, 7, 8, 11, 9, 3, 7, 4, 12, 5, 4, 10, 9, 6, 9, 4};
    int n = sizeof(arr)/sizeof(arr[0]); 
    
    cout<<standardDeviation(arr, n); 
    
    return 0;
}

Output:

2.98329

Time complexity: O(n)
Where n is the number of elements in the array.
Each loop runs n times.

1. First loop for summing the elements of the array. It takes O(n) time.
2. Second loop for summing the square of mean deviation. It takes O(n) time.
3. Square root function call takes constant time.

Therefore, the time complexity is O(n).