variance wrong calculation c#

Question

I have problem with my variance when i put 2,5,1,3 numbers result is 3,46 but correct is 2,18

double result=0;
string x1 = textBox1.Text;
string[] tab = x1.Split(',');
int n = tab.Length;
double sum = 0;
double ob = 0;

for (int i = 0; i < n; i++)
{
    sum = sum + double.Parse(tab[i]);
    result = sum / n;
    ob = (ob + Math.Pow((double.Parse(tab[i]) - result),2)) / n;

}

textBox2.Text = ob.ToString();

Shouldn't the last two lines in the for loop be outside it? – Alfie Goodacre Mar 31 '16 at 13:45 — Alfie Goodacre, Mar 31 '16 at 13:45
Why is 2,18 correct? What should your code actually do? – Patrick Hofman Mar 31 '16 at 13:46 — Patrick Hofman, Mar 31 '16 at 13:46

score 1 · Accepted Answer · answered Mar 31 '16 at 14:08

1

The formula for the standard variance is:

As you can see, you have to divide by n after you calculated the sum. But your division is in the loop. If you will perform this operation at the outsite of the loop, you will recieve the correct result.

answered Mar 31 '16 at 14:08

Leonard Klausmann

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variance wrong calculation c#

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