Question

A laboratory instructor gives a sample of amino-acid powder to each of four students, I, II, III, and IV, and they weigh the samples. The true value is 8.72 g. Their results for three trials are

I: 8.72 g, 8.74 g, 8.70 g            II: 8.56 g, 8.77 g, 8.83 g

III: 8.50 g, 8.48 g, 8.51 g          IV: 8.41 g, 8.72 g, 8.55 g

(a) Calculate the average mass from each set of data, and tell which set is the most accurate.

(b) Precision is a measure of the average of the deviations of each piece of data from the average value. Which set of data is the most precise? Is this set also the most accurate?

(c) Which set of data is both the most accurate and the most precise?

(d) Which set of data is both the least accurate and the least precise?

Answer 1

Solution

$\text{Precision}$ is described as a small deviation between the measurements.

$\text{Accuracy}$ is described as the proximity of the measured result and the actual result.

a) Calculate the average mass for the first set of data:

$$\begin{array}{rl}A{V}^I& =\genfrac{}{}{0ex}{}{8.72+8.74+8.70}{3}\\ & =8.72\text{ g}\end{array}$$

Calculate the average mass for the second set of data:

$$\begin{array}{rl}A{V}^{II}& =\genfrac{}{}{0ex}{}{8.56+8.77+8.83}{3}\\ & =8.72\text{ g}\end{array}$$

Calculate the average mass for the third set of data:

$$\begin{array}{rl}A{V}^{III}& =\genfrac{}{}{0ex}{}{8.5+8.48+8.51}{3}\\ & =8.50\text{ g}\end{array}$$

Calculate the average mass for the fourth set of data:

$$\begin{array}{rl}A{V}^{IV}& =\genfrac{}{}{0ex}{}{8.41+8.72+8.55}{3}\\ & =8.56\text{ g}\end{array}$$

We can clearly see that the first and the second sets are the most accurate (as the true value is $8.72\text{ g}$).

b)

Let's calculate the precision for the first set, considering $A{V}^I=8.72\text{ g}$

$$\begin{array}{rl}\Delta 1& =(8.72-8.72)\text{ g}=0\text{ g}\\ \Delta 2& =(8.74-8.72)\text{ g}=0.02\text{ g}\\ \Delta 3& =|(8.70-8.72)|\text{ g}=0.02\text{ g}\end{array}$$

Therefore:

$${\Delta }^I=\genfrac{}{}{0ex}{}{\Delta 1+\Delta 2+\Delta 3}{3}=0.013\text{g}$$

Let's calculate the precision for the third set, considering $A{V}^{III}=8.5\text{ g}$

$$\begin{array}{rl}\Delta 1& =(8.5-8.5)\text{ g}=0\text{ g}\\ \Delta 2& =|(8.48-8.5)|\text{ g}=0.02\text{ g}\\ \Delta 3& =(8.51-8.5)\text{ g}=0.01\text{ g}\end{array}$$

Therefore:

 

$${\Delta }^{IV}=\genfrac{}{}{0ex}{}{\Delta 1+\Delta 2+\Delta 3}{3}=0.01\text{g}$$

Let's calculate the precision for the fourth set, considering $A{V}^{IV}=8.56\text{ g}$

$$\begin{array}{rl}\Delta 1& =|(8.41-8.56)|\text{ g}=0.15\text{ g}\\ \Delta 2& =(8.72-8.56)\text{ g}=0.16\text{ g}\\ \Delta 3& =|(8.55-8.56)|\text{ g}=0.01\text{ g}\end{array}$$

Therefore:

 

$${\Delta }^I=\genfrac{}{}{0ex}{}{\Delta 1+\Delta 2+\Delta 3}{3}=0.107\text{g}$$

 

 

$$c)According\; to\; the\; calculation\; above,\; set\; I\; is\; both\; the\; most\; accurate\; and\; most\; precise.$$

$$d)According\; to\; the\; calculation\; above,\; set\; IV\; is\; the\; least\; accurate\; and\; the\; least\; precise.$$