Answer :
Let's analyze the students' measurements and determine the accuracy and precision of the data.
1. Converting Measurements:
- Student 1: 1.6 kg = 1,600 g
- Student 2: 2,300 g
- Student 3: Between 1,000 and 1,500 g → We'll use the midpoint, so (1000+1500)/2 = 1,375 g
- Student 4: 1,750 g
The measurements are:
[tex]\[ [1600, 2300, 1375, 1750] \][/tex]
2. Calculate the Mean of Measurements:
The mean is calculated by summing all measurements and dividing by the number of measurements:
[tex]\[ \text{Mean} = \frac{1600 + 2300 + 1375 + 1750}{4} = \frac{7025}{4} = 1756.25 \text{ g} \][/tex]
3. Calculate the Variance:
Variance is a measure of how much the measurements spread out from the mean. It’s calculated by averaging the squared differences from the mean:
[tex]\[ \text{Variance} = \frac{(1600 - 1756.25)^2 + (2300 - 1756.25)^2 + (1375 - 1756.25)^2 + (1750 - 1756.25)^2}{4} \][/tex]
[tex]\[ = \frac{(-156.25)^2 + 543.75^2 + (-381.25)^2 + (-6.25)^2}{4} = \frac{24390.625 + 295662.1875 + 145351.5625 + 39.0625}{4} = \frac{465443.4375}{4} = 116367.1875 \text{ g}^2 \][/tex]
4. Calculate the Standard Deviation:
The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation (Precision)} = \sqrt{116367.1875} \approx 341.13 \text{ g} \][/tex]
5. Calculate the Accuracy:
Accuracy is determined by the absolute difference between the mean of the measurements and the actual mass:
[tex]\[ \text{Accuracy} = |\text{Mean} - \text{Actual Mass}| = |1756.25 - 1864.3| = 108.05 \text{ g} \][/tex]
6. Analyzing Accuracy and Precision:
- Precision: The standard deviation of 341.13 g is quite large, indicating the measurements are not very close to each other.
- Accuracy: The accuracy of 108.05 g is moderately large, indicating the mean of the measurements is not very close to the actual mass.
Given these calculations,
- The accuracy (108.05 g) is greater than 100 g.
- The precision (341.13 g) is also greater than 100 g.
Thus, the best fit description for the measurements is:
[tex]\[ \boxed{D. \text{Neither accurate nor precise}} \][/tex]
1. Converting Measurements:
- Student 1: 1.6 kg = 1,600 g
- Student 2: 2,300 g
- Student 3: Between 1,000 and 1,500 g → We'll use the midpoint, so (1000+1500)/2 = 1,375 g
- Student 4: 1,750 g
The measurements are:
[tex]\[ [1600, 2300, 1375, 1750] \][/tex]
2. Calculate the Mean of Measurements:
The mean is calculated by summing all measurements and dividing by the number of measurements:
[tex]\[ \text{Mean} = \frac{1600 + 2300 + 1375 + 1750}{4} = \frac{7025}{4} = 1756.25 \text{ g} \][/tex]
3. Calculate the Variance:
Variance is a measure of how much the measurements spread out from the mean. It’s calculated by averaging the squared differences from the mean:
[tex]\[ \text{Variance} = \frac{(1600 - 1756.25)^2 + (2300 - 1756.25)^2 + (1375 - 1756.25)^2 + (1750 - 1756.25)^2}{4} \][/tex]
[tex]\[ = \frac{(-156.25)^2 + 543.75^2 + (-381.25)^2 + (-6.25)^2}{4} = \frac{24390.625 + 295662.1875 + 145351.5625 + 39.0625}{4} = \frac{465443.4375}{4} = 116367.1875 \text{ g}^2 \][/tex]
4. Calculate the Standard Deviation:
The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation (Precision)} = \sqrt{116367.1875} \approx 341.13 \text{ g} \][/tex]
5. Calculate the Accuracy:
Accuracy is determined by the absolute difference between the mean of the measurements and the actual mass:
[tex]\[ \text{Accuracy} = |\text{Mean} - \text{Actual Mass}| = |1756.25 - 1864.3| = 108.05 \text{ g} \][/tex]
6. Analyzing Accuracy and Precision:
- Precision: The standard deviation of 341.13 g is quite large, indicating the measurements are not very close to each other.
- Accuracy: The accuracy of 108.05 g is moderately large, indicating the mean of the measurements is not very close to the actual mass.
Given these calculations,
- The accuracy (108.05 g) is greater than 100 g.
- The precision (341.13 g) is also greater than 100 g.
Thus, the best fit description for the measurements is:
[tex]\[ \boxed{D. \text{Neither accurate nor precise}} \][/tex]