Variance calculation

Suppose XVehicle is an array of the pick points of the histogram and the corresponding frequencies of each pick point are given in another array, YVehicle.

Then the variance of the histogram can be calculated using the following C# code:

double sumXY = 0.0;

double sumY = 0.0;

for(int i = 0; i < XVehicle.Length; i++)

{

sumXY += XVehicle[i] * YVehicle[i];

sumY += YVehicle[i];

}

double mean = sumXY/sumY;

double variance = 0;

for(int i = 0; i < XVehicle.Length; i++)

{

double deviation = XVehicle[i] - mean;

variance += deviation * deviation * YVehicle[i];

}

variance /= sumY;

For example if we use the data in this link:

http://phys.columbia.edu/~tutorial/estimation/tut_e_2_4.html

double[] XVehicle = {3.4, 3.5, 3.6, 3.7, 3.8, 3.9};

double[] YVehicle = {1, 2, 3, 2, 1, 1};

we get:

variance = 0.0201

and the standard deviation is then the square root of this which is:

0.141774468787578 (approximated to 0.14 in the link).

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Vulpes11y ago
Suppose XVehicle is an array of the pick points of the histogram and the corresponding frequencies of each pick point are given in another array, YVehicle.

Then the variance of the histogram can be calculated using the following C# code:

double sumXY = 0.0;
double sumY = 0.0;

for(int i = 0; i < XVehicle.Length; i++)
{
sumXY += XVehicle[i] * YVehicle[i];
sumY += YVehicle[i];
}

double mean = sumXY/sumY;
double variance = 0;

for(int i = 0; i < XVehicle.Length; i++)
{
double deviation = XVehicle[i] - mean;
variance += deviation * deviation * YVehicle[i];
}

variance /= sumY;

For example if we use the data in this link:

http://phys.columbia.edu/~tutorial/estimation/tut_e_2_4.html

double[] XVehicle = {3.4, 3.5, 3.6, 3.7, 3.8, 3.9};
double[] YVehicle = {1, 2, 3, 2, 1, 1};

we get:

variance = 0.0201

and the standard deviation is then the square root of this which is:

0.141774468787578 (approximated to 0.14 in the link).