Standard Deviation in LINQ

Dynami's answer works but makes multiple passes through the data to get a result. This is a single pass method that calculates the sample standard deviation:

public static double StdDev(this IEnumerable<double> values)
{
    // ref: http://warrenseen.com/blog/2006/03/13/how-to-calculate-standard-deviation/
    double mean = 0.0;
    double sum = 0.0;
    double stdDev = 0.0;
    int n = 0;
    foreach (double val in values)
    {
        n++;
        double delta = val - mean;
        mean += delta / n;
        sum += delta * (val - mean);
    }
    if (1 < n)
        stdDev = Math.Sqrt(sum / (n - 1));

    return stdDev;
}

This is the sample standard deviation since it divides by n - 1. For the normal standard deviation you need to divide by n instead.

This uses Welford's method which has higher numerical accuracy compared to the Average(x^2)-Average(x)^2 method.

You can make your own extension calculating it

public static class Extensions
{
    public static double StdDev(this IEnumerable<double> values)
    {
       double ret = 0;
       int count = values.Count();
       if (count  > 1)
       {
          //Compute the Average
          double avg = values.Average();

          //Perform the Sum of (value-avg)^2
          double sum = values.Sum(d => (d - avg) * (d - avg));

          //Put it all together
          ret = Math.Sqrt(sum / count);
       }
       return ret;
    }
}

If you have a sample of the population rather than the whole population, then you should use ret = Math.Sqrt(sum / (count - 1));.

Transformed into extension from Adding Standard Deviation to LINQ by Chris Bennett.