Chapter 10, Problem 1DE

SO Two students deterrmne the percen.ge of lead in a sample as a laboratory exercise Thelrue percentage is 22 52% The s.dents^r resul. ler three determinations are as fellows
1 22 52, 22 413, 22 54
220422 58,2262
a Calculate Me average percentage for each set of data and sta. which set is the more accurate based on .e average
b Precision can be Judged by examinrng the average of the devrahons from the average value for that da. (CalcAare the average value for each,. set then calculate the average aue of the absolute deviations of each measurerne. from the average ), Which.. more Precrse9

Interpretation Introduction

To determine: The average value of deviation for each set of given data; the set that is more précised.

Answer to Problem 1DE

Solution: The average value of deviation for set $\text{1}$ and set $\text{2}$ are $\text{0}\text{.02}$ and $\text{0}\text{.02}$ respectively. The set $\text{1}$ is more precise.

Explanation of Solution

Given The true average value is $\text{22}\text{.52\%}$

The values in set 1: $\text{22}\text{.52,}\text{22}\text{.48,}\text{22}\text{.54}$

The values in set 2: $\text{22}\text{.64,}\text{22}\text{.58,}\text{22}\text{.62}$

The average deviation can be determined by addition of the differences of all numbers and average of all numbers; and further dividing the outcome by the number of items present in that list.

The average deviation of set $\text{1}$ is calculated as,

  $\begin{array}{c}\frac{\left(\text{22}\text{.52}-\text{22}\text{.51}\right)+\left(\text{22}\text{.48}-\text{22}\text{.51}\right)+\left(\text{22}\text{.54}-\text{22}\text{.51}\right)}{\text{3}}=\frac{\left|+\text{0}\text{.01}\right|+\left|-\text{0}\text{.03}\right|+\left|+\text{0}\text{.03}\right|}{\text{3}}\\ =\text{0}\text{.02}\end{array}$

The average deviation of set $\text{2}$ is calculated as,

  $\begin{array}{c}\frac{\left(\text{22}\text{.64}-\text{22}\text{.61}\right)+\left(\text{22}\text{.58}-\text{22}\text{.61}\right)+\left(\text{22}\text{.62}-\text{22}\text{.61}\right)}{\text{3}}=\frac{\left|+\text{0}\text{.03}\right|+\left|-\text{0}\text{.03}\right|+\left|+\text{0}\text{.01}\right|}{\text{3}}\\ =\text{0}\text{.02}\end{array}$

Both the sets set $\text{1}$ and set $\text{2}$ have the same precision value. But set $\text{1}$ is said to be more precise as its value is more accurate than that of set $\text{2}$ .

Conclusion

Precision is the comparison of closeness of two or more measurements with each other. The set $\text{1}$ is more accurate. Hence, it is more precise.