Question A.SE.4: The percent carbon in a sugar is measured four times: 42.01%...
The percent carbon in a sugar is measured four times: 42.01%, 42.28%, 41.79%, and 42.25%. Calculate (a) the average and (b) the standard deviation for these measurements.
(a) The average is found by adding the quantities and dividing by the number of measurements:
\bar{x}=\frac{42.01+42.28+41.79+42.25}{4}=\frac{168.33}{4}=42.08
(b) The standard deviation is found using the preceding equation:
s=\sqrt{\frac{\sum_{i=1}^N\left(x_i-\bar{x}\right)^2}{N-1}}
Let’s tabulate the data so the calculation of \sum_{i=1}^N\left(x_i-\bar{x}\right)^2 can be seen clearly.
\begin{array}{clr}\hline\text{Percent C}& \begin{array}{l}\text{Difference between}\\\text{Measurement}\\\text{and Average, }(x_i – x)\\\end{array}& \begin{array}{l}\text{Square of}\\\text{Difference,}\left(x_{i}-x\right)^2\end{array}\\\hline 42.01 & 42.01-42.08=-0.07 & (-0.07)^2=0.005 \\42.28 & 42.28-42.08=0.20 & (0.20)^2=0.040 \\41.79 & 41.79-42.08=-0.29 & (-0.29)^2=0.084 \\42.25 & 42.25-42.08=0.17 & (0.17)^2=0.029 \\\hline\end{array}
The sum of the quantities in the last column is
\sum_{i=1}^N\left(x_i-\bar{x}\right)^2=0.005+0.040+0.084+0.029=0.16
Thus, the standard deviation is
s=\sqrt{\frac{\sum_{i=1}^N\left(x_i-\bar{x}\right)^2}{N-1}}=\sqrt{\frac{0.16}{4-1}}=\sqrt{\frac{0.16}{3}}=\sqrt{0.053}=0.23
Based on these measurements, it would be appropriate to represent the measured percent carbon as 42.08 ± 0.23.