Find the standard deviation of 0.645, 0.656, 0.652.

Step-by-step solution

1. 1

   Compute the mean. Add and divide by $3$:

   $\bar x=\frac{0.645+0.656+0.652}{3}=\frac{1.953}{3}=0.651$

   The mean comes out exactly, which makes the deviations easy to read.

2. 2

   Find the deviations from the mean.

   $0.645-0.651=-0.006,\qquad 0.656-0.651=0.005,\qquad 0.652-0.651=0.001$

   Their sum is $-0.006+0.005+0.001=0$, which is always true and is a free check on the mean.

3. 3

   Square and add.

   $(-0.006)^2=3.6\times 10^{-5},\quad (0.005)^2=2.5\times 10^{-5},\quad (0.001)^2=1\times 10^{-6}$

   $\sum\left(x_i-\bar x\right)^2=6.2\times 10^{-5}$

4. 4

   Divide — and choose the right denominator. This is the decision that changes the answer. For a sample of three measurements, divide by $n-1=2$; for the whole population, divide by $n=3$:

   $s^2=\frac{6.2\times 10^{-5}}{2}=3.1\times 10^{-5},\qquad \sigma^2=\frac{6.2\times 10^{-5}}{3}=2.0667\times 10^{-5}$

   With $n$ this small the two differ noticeably — by about $22\%$ in the standard deviation.

5. 5

   Take square roots and interpret.

   $s=\sqrt{3.1\times 10^{-5}}\approx 0.005568,\qquad \sigma=\sqrt{2.0667\times 10^{-5}}\approx 0.004546$

   Three repeated measurements of one quantity are normally treated as a sample, so $s\approx 0.00557$ is usually the value to report. As a sanity check, the spread of the data is $0.656-0.645=0.011$, and a standard deviation of roughly half the range is typical for $n=3$ $\checkmark$.