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C# Program to Reverse a NumberC# Program to Find the Standard Deviation

C# Program to Find the Standard Deviation

 

 

This is a C# Program to find the standard deviation of a set of given numbers.

Problem Description

This C# Program Finds the Standard Deviation of a Set of Given Numbers.

Problem Solution

Here the mean, variance and standard deviation are calculated and displayed.

Program/Source Code

Here is source code of the C# Program to Find the Standard Deviation of a Set of Given Numbers. The C# program is successfully compiled and executed with Microsoft Visual Studio. The program output is also shown below.

/*
 * C# Program to Find the Standard Deviation of a Set of Given Numbers
 */
using System;
using System.Collections.Generic;
namespace SampleApp
{
    internal class Program
    {
        private static void Main()
        {
            List<double> number = new List<double> { 1, 2, 3, 4, 5, 6 };
 
            double mean = number.Mean();
            double variance = number.Variance();
            double sd = number.StandardDeviation();
 
            Console.WriteLine("Mean: {0}  , Variance: {1}  , SD: {2}  ", 
                               mean, variance, sd);
            Console.ReadKey();
        }
    }
    public static class list
    {
        public static double Mean(this List<double> values)
        {
            return values.Count == 0 ? 0 : values.Mean(0, values.Count);
        }
 
        public static double Mean(this List<double> values, 
                                  int start, int end)
        {
            double s = 0;
 
            for (int i = start; i < end; i++)
            {
                s += values[i];
            }
 
            return s / (end - start);
        }
 
        public static double Variance(this List<double> values)
        {
            return values.Variance(values.Mean(), 0, values.Count);
        }
 
        public static double Variance(this List<double> values, double mean)
        {
            return values.Variance(mean, 0, values.Count);
        }
 
        public static double Variance(this List<double> values, double mean, 
                                      int start, int end)
        {
            double variance = 0;
 
            for (int i = start; i < end; i++)
            {
                variance += Math.Pow((values[i] - mean), 2);
            }
 
            int n = end - start;
            if (start > 0) n -= 1;
 
            return variance / (n);
        }
        public static double StandardDeviation(this List<double> values)
        {
            return values.Count == 0 ? 0 : values.StandardDeviation(0, values.Count);
        }
        public static double StandardDeviation(this List<double> values, 
                                               int start, int end)
        {
            double mean = values.Mean(start, end);
            double variance = values.Variance(mean, start, end);
            return Math.Sqrt(variance);
        }
    }
}
Program Explanation

This C# program is used to find the standard deviation of a set of given numbers. Create a list using ‘number’ variable. For loop is used to compute the mean from the start and up to the end of the list.

 

Compute the summation of all the values and assigning the value to‘s’ variable. Divide the value of ‘s’ variable value by the difference of end and start variable value. For loop is used to compute the mean from the start and up to the end of the list. Compute the power value by 2 to the difference between the values in the list by the mean variable value.

If condition statement is used to check the value of ‘start’ variable is greater than 0. If the condition is true then execute the statement and assign negative value of one to the n variable. Divide the value of ‘variance’ variable by the value of ‘n’ variable.

Compute the square to the value of ‘variance’ variable. Print the computed values of the mean, variance and standard deviation.

 
Runtime Test Cases
 
Mean : 3.5  Variance : 2.916666666667  S.D = 1.7078251256993

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