Exercise¶
The following gas stations were fined for cheating because the delivered gasoline was less than the due quantity. Classify them in decreasing order of honesty.
\begin{array}{|ccc|}
\hline \mathrm{Brand} & \mathrm{Delivered} & \mathrm{Due}\\
\hline \mathrm{Ser} & 9.90 & 10.00\\
\mathrm{Cit} & 19.80 & 20.00\\
\mathrm{Has} & 29.10 & 30.00\\
\mathrm{Shol} & 28.90 & 30.00
\\\hline \end{array}
We will use the relative error as a criterion.
If $x$ is the actual value and $x^*$ is the approximate value:
- Absolute error $$e_a=|x-x^*|$$
- Relative error $$e_r=\frac{e_a}{|x|}$$
The relative error is expressed as a fraction of one. If we multiply it by 100 it will be expressed as a percentage.
Ser¶
- Absolute error $$e_a=|x-x^*|=|9.90-10.00|=0.10$$
- Relative error $$e_r=\frac{e_a}{|x|}=\frac{0.10}{9.90}\approx 0.01=1\%$$
Cit¶
- Absolute error $$e_a=|x-x^*|=|19.80-20.00|=0.20$$
- Relative error $$e_r=\frac{e_a}{|x|}=\frac{0.20}{19.80}\approx0.01=1\%$$
Has¶
- Absolute error $$e_a=|x-x^*|=|29.10-30.00|=0,90$$
- Relative error $$e_r=\frac{e_a}{|x|}=\frac{0.90}{29.10}\approx0.03=3\%$$
Shol¶
- Absolute error $$e_a=|x-x^*|=|28.90-30.00|=1.10$$
- Relative error $$e_r=\frac{e_a}{|x|}=\frac{1.10}{28.90}\approx0.04=4\%$$
Summing up, we have
\begin{array}{|ccc|c|c|}
\hline \mathrm{Brand} & \mathrm{Delivered} & \mathrm{Due} & e_a & e_r\\
\hline \mathrm{Ser} & 9.90 & 10.00 & 0.10 & 1\%\\
\mathrm{Cit} & 19.80 & 20.00 & 0.20 & 1\%\\
\mathrm{Has} & 29.10 & 30.00 & 0.90 & 3\%\\
\mathrm{Shol} & 28.90 & 30.00 & 1.10 & \fbox{4%}
\\\hline \end{array}
And the greatest relative difference between what is delivered and what is due is obtained for the Shol gas station and we can consider that it is the one that cheated the most.