Spread of the average of nine readings

Question

A lab takes 9 independent readings of the same quantity. Each reading has a standard deviation of 6 (variance 36). You report the average of the 9 readings.

What is the standard deviation of that average?

Accepted Answer

Averaging n independent, identically distributed values reduces the spread. The standard deviation of the average (the standard error) is 	ext{sd(average)} = \frac{\sigma}{\sqrt{n}}. With \sigma = 6 and n = 9: 	ext{sd(average)} = \frac{6}{\sqrt{9}} = \frac{6}{3} = 2. The variance of the average is \sigma^2/n = 36/9 = 4, whose square root is 2. Averaging 9 readings cuts the noise by a factor of \sqrt{9} = 3. Takeaway. The standard error of a mean is \sigma/\sqrt{n}; averaging 9 readings with sd 6 tightens it to sd 2.

Spread of the average of nine readings

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