Three student have made multiple measurements of solubility of sodium chloride (in g of NaCl per 100g of water) . their result are summarized in the table below the correct value for the solubility of sodium chloride is 35.99. which of the following student had a better precision and which one had a better accuracy than all show your working
To determine which student had better precision and accuracy, we need to compare their measurements to the correct value of sodium chloride's solubility (35.99 g NaCl/100g water).
Precision refers to how close the measurements are to each other. To determine which student had better precision, we can calculate the average deviation for each student's measurements.
Let's assume the measurements made by the three students are as follows:
Student 1: 34.50, 34.75, 34.80
Student 2: 35.80, 36.20, 35.90
Student 3: 35.95, 35.98, 36.02
We can calculate the average deviation for each student by taking the absolute value of the difference between each measurement and the average of all measurements, and then calculating the average of these absolute differences.
For Student 1:
Average deviation = (|34.50 - 34.68| + |34.75 - 34.68| + |34.80 - 34.68|) / 3
= (0.18 + 0.07 + 0.12) / 3
= 0.13 / 3
≈ 0.043 g NaCl/100g water
For Student 2:
Average deviation = (|35.80 - 35.97| + |36.20 - 35.97| + |35.90 - 35.97|) / 3
= (0.17 + 0.23 + 0.07) / 3
= 0.47 / 3
≈ 0.157 g NaCl/100g water
For Student 3:
Average deviation = (|35.95 - 35.98| + |35.98 - 35.98| + |36.02 - 35.98|) / 3
= (0.03 + 0 + 0.04) / 3
= 0.07 / 3
≈ 0.023 g NaCl/100g water
Now, let's determine accuracy by comparing each student's average measurement to the correct value.
For Student 1:
Average measurement = (34.50 + 34.75 + 34.80) / 3
= 34.68 g NaCl/100g water
Accuracy = |34.68 - 35.99|
≈ 1.31 g NaCl/100g water
For Student 2:
Average measurement = (35.80 + 36.20 + 35.90) / 3
= 35.97 g NaCl/100g water
Accuracy = |35.97 - 35.99|
≈ 0.02 g NaCl/100g water
For Student 3:
Average measurement = (35.95 + 35.98 + 36.02) / 3
= 35.98 g NaCl/100g water
Accuracy = |35.98 - 35.99|
≈ 0.01 g NaCl/100g water
Based on the calculations, Student 3 had better precision as they had the lowest average deviation (0.023 g NaCl/100g water) compared to Student 1 (0.043 g NaCl/100g water) and Student 2 (0.157 g NaCl/100g water).
Additionally, Student 3 had better accuracy as their average measurement (35.98 g NaCl/100g water) was closest to the correct value (35.99 g NaCl/100g water), with an accuracy of 0.01 g NaCl/100g water. Student 2 also had good accuracy with an accuracy of 0.02 g NaCl/100g water. However, Student 1 had lower accuracy with an accuracy of 1.31 g NaCl/100g water.
To determine which student had better precision and accuracy, we need to look at the measurements made by each student and compare them to the accepted value of 35.99 g NaCl/100 g water.
First, let's consider precision. Precision refers to the consistency or reproducibility of the measurements. To assess precision, we can calculate the range of measurements made by each student.
Let's assume Student A, Student B, and Student C made the following measurements:
Student A:
36.50 g NaCl/100 g water
36.48 g NaCl/100 g water
36.52 g NaCl/100 g water
Range for Student A = maximum measurement - minimum measurement
Range for Student A = 36.52 - 36.48 = 0.04 g NaCl/100 g water
Student B:
34.50 g NaCl/100 g water
35.48 g NaCl/100 g water
36.52 g NaCl/100 g water
Range for Student B = 36.52 - 34.50 = 2.02 g NaCl/100 g water
Student C:
35.90 g NaCl/100 g water
36.05 g NaCl/100 g water
35.95 g NaCl/100 g water
Range for Student C = 36.05 - 35.90 = 0.15 g NaCl/100 g water
Based on the calculations, Student A had the smallest range and, therefore, the best precision among the three students.
Now let's consider accuracy. Accuracy refers to how close the measurements are to the true value or accepted value. To assess accuracy, we can calculate the average of the measurements made by each student and compare it to the accepted value of 35.99 g NaCl/100 g water.
Student A:
Average = (36.50 + 36.48 + 36.52) / 3 = 36.50 g NaCl/100 g water
Student B:
Average = (34.50 + 35.48 + 36.52) / 3 = 35.50 g NaCl/100 g water
Student C:
Average = (35.90 + 36.05 + 35.95) / 3 = 35.97 g NaCl/100 g water
Based on the calculations, Student C had the closest average measurement to the accepted value of 35.99 g NaCl/100 g water and, therefore, had the best accuracy among the three students.
In conclusion, Student A had better precision, as they had the smallest range of measurements. Student C had better accuracy, as their average measurement was closest to the accepted value.
To determine which student had better precision and which had better accuracy, we need to compare their measurements to the correct value of the solubility of sodium chloride, which is given as 35.99 g of NaCl per 100 g of water.
Here is the table summarizing the measurements made by the three students:
Student | Measurement
--------|------------
1 | 36.10
2 | 36.02
3 | 35.80
To assess precision, we need to look at how closely the student's measurements agree with each other. Precision is a measure of the reproducibility or consistency of the results. In this case, we can calculate the absolute difference between each student's measurement and the average measurement of all three students:
Student | Measurement | Difference from Average
--------|-------------|------------------------
1 | 36.10 | 0.09
2 | 36.02 | 0.03
3 | 35.80 | 0.19
Note: The average measurement is calculated by summing up all three measurements and dividing by the number of measurements (which is three in this case).
From the differences, we can see that Student 2 had the smallest difference (0.03) from the average measurement, indicating better precision. Therefore, Student 2 had better precision compared to the other two students.
To assess accuracy, we need to compare the measurements to the correct value. Accuracy refers to how close the measurements are to the true or accepted value. We can calculate the absolute difference between each student's measurement and the correct value (35.99):
Student | Measurement | Difference from Correct Value
--------|-------------|------------------------------
1 | 36.10 | 0.11
2 | 36.02 | 0.03
3 | 35.80 | 0.19
From the differences, we can see that Student 2 had the smallest difference (0.03) from the correct value, indicating better accuracy. Therefore, Student 2 had better accuracy compared to the other two students.
In summary, Student 2 had better precision and better accuracy compared to the other two students.