Answer :
Let's analyze the given data set in detail.
1. Determine the Mean of the Trials:
First, we calculate the mean (average) of the given trials:
[tex]\[ \text{Mean} = \frac{\text{Trial 1} + \text{Trial 2} + \text{Trial 3} + \text{Trial 4} + \text{Trial 5}}{5} \][/tex]
Substituting in the given values:
[tex]\[ \text{Mean} = \frac{58.7 + 59.3 + 60.0 + 58.9 + 59.2}{5} = 59.22 \][/tex]
2. Determine the Standard Deviation of the Trials:
Next, we calculate the standard deviation, which measures the dispersion or variability of the trials:
[tex]\[ \sigma = \sqrt{\frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N}} \][/tex]
Where [tex]\( N \)[/tex] is the number of trials, [tex]\( x_i \)[/tex] are the individual trial values, and [tex]\( \mu \)[/tex] is the mean.
After calculating the standard deviation using the given values, we get:
[tex]\[ \text{Standard Deviation} \approx 0.4445 \][/tex]
3. Assess Accuracy:
To determine accuracy, we check how close the mean of the trials is to the correct value (59.2). We use a threshold of [tex]\( \leq 0.5 \)[/tex]:
[tex]\[ \text{Accuracy} = \left| 59.2 - 59.22 \right| \leq 0.5 \][/tex]
[tex]\[ \text{Accuracy} = 0.02 \leq 0.5 \quad \text{(True)} \][/tex]
4. Assess Precision:
For precision, we check the standard deviation. A small standard deviation indicates high precision. Using a threshold of [tex]\( \leq 0.5 \)[/tex]:
[tex]\[ \text{Precision} = 0.4445 \leq 0.5 \quad \text{(True)} \][/tex]
Combining these assessments, we determine that the data set is both accurate and precise.
Therefore, the best description for the data set is:
- It is both accurate and precise.
1. Determine the Mean of the Trials:
First, we calculate the mean (average) of the given trials:
[tex]\[ \text{Mean} = \frac{\text{Trial 1} + \text{Trial 2} + \text{Trial 3} + \text{Trial 4} + \text{Trial 5}}{5} \][/tex]
Substituting in the given values:
[tex]\[ \text{Mean} = \frac{58.7 + 59.3 + 60.0 + 58.9 + 59.2}{5} = 59.22 \][/tex]
2. Determine the Standard Deviation of the Trials:
Next, we calculate the standard deviation, which measures the dispersion or variability of the trials:
[tex]\[ \sigma = \sqrt{\frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N}} \][/tex]
Where [tex]\( N \)[/tex] is the number of trials, [tex]\( x_i \)[/tex] are the individual trial values, and [tex]\( \mu \)[/tex] is the mean.
After calculating the standard deviation using the given values, we get:
[tex]\[ \text{Standard Deviation} \approx 0.4445 \][/tex]
3. Assess Accuracy:
To determine accuracy, we check how close the mean of the trials is to the correct value (59.2). We use a threshold of [tex]\( \leq 0.5 \)[/tex]:
[tex]\[ \text{Accuracy} = \left| 59.2 - 59.22 \right| \leq 0.5 \][/tex]
[tex]\[ \text{Accuracy} = 0.02 \leq 0.5 \quad \text{(True)} \][/tex]
4. Assess Precision:
For precision, we check the standard deviation. A small standard deviation indicates high precision. Using a threshold of [tex]\( \leq 0.5 \)[/tex]:
[tex]\[ \text{Precision} = 0.4445 \leq 0.5 \quad \text{(True)} \][/tex]
Combining these assessments, we determine that the data set is both accurate and precise.
Therefore, the best description for the data set is:
- It is both accurate and precise.