Worked exercise — Mean and standard deviation

Problem

Five measurements of the period of a pendulum (in seconds): $2{,}01;\ 2{,}03;\ 1{,}99;\ 2{,}02;\ 2{,}00$. Find the mean and the standard deviation of the mean.

Solution

Mean. $\bar{T} = \frac{2{,}01 + 2{,}03 + 1{,}99 + 2{,}02 + 2{,}00}{5} = \frac{10{,}05}{5} = 2{,}010\ \text{s}$

Deviations from the mean. $0;\quad +0{,}02;\quad -0{,}02;\quad +0{,}01;\quad -0{,}01$

Standard deviation of the measurements. From the sum of the squared deviations $\sum (\Delta T)^2 = 0 + 0{,}0004 + 0{,}0004 + 0{,}0001 + 0{,}0001 = 0{,}0010\ \text{s}^2$: $\sigma = \sqrt{\frac{\sum(\Delta T)^2}{N-1}} = \sqrt{\frac{0{,}0010}{4}} \approx 0{,}016\ \text{s}$

Standard deviation of the mean. $\sigma_{\bar{T}} = \frac{\sigma}{\sqrt{N}} = \frac{0{,}016}{\sqrt{5}} \approx 0{,}007\ \text{s}$

$\ev{T = (2{,}010 \pm 0{,}007)\ \text{s}}$

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Topics: Metodo sperimentale Concepts: Cifre significative e incertezza Skills: Uso delle cifre significative