To determine which description best fits the data set, we need to evaluate both its accuracy and precision using the given trials and correct value. Let's go through the steps:
1. Calculate the Mean of the Trials:
The first step is to find the mean (average) of the given trials.
[tex]\[
\text{Mean} = \frac{58.7 + 59.3 + 60.0 + 58.9 + 59.2}{5} = \frac{296.1}{5} = 59.22
\][/tex]
2. Calculate the Standard Deviation:
Next, we calculate the standard deviation to measure the spread of the trials from the mean. The formula for the standard deviation [tex]\(\sigma\)[/tex] of a sample is:
[tex]\[
\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2}
\][/tex]
where [tex]\(\bar{x}\)[/tex] is the mean of the trials and [tex]\(x_i\)[/tex] is each individual trial value.
Let's denote the mean as [tex]\(\bar{x} = 59.22\)[/tex].
Calculate the variance:
[tex]\[
\text{Variance} = \frac{(58.7 - 59.22)^2 + (59.3 - 59.22)^2 + (60.0 - 59.22)^2 + (58.9 - 59.22)^2 + (59.2 - 59.22)^2}{5}
\][/tex]
[tex]\[
= \frac{(-0.52)^2 + (0.08)^2 + (0.78)^2 + (-0.32)^2 + (-0.02)^2}{5}
\][/tex]
[tex]\[
= \frac{0.2704 + 0.0064 + 0.6084 + 0.1024 + 0.0004}{5}
\][/tex]
[tex]\[
= \frac{0.988}{5} = 0.1976
\][/tex]
Standard deviation:
[tex]\[
\sigma = \sqrt{0.1976} \approx 0.4445
\][/tex]
3. Evaluate Accuracy:
To check if the data is accurate, compare the mean of the trials to the correct value (59.2). If the difference is within an acceptable threshold (0.5), the data is considered accurate.
[tex]\[
\left| 59.22 - 59.2 \right| = 0.02 \leq 0.5
\][/tex]
Since 0.02 is less than or equal to the threshold of 0.5, the data is accurate.
4. Evaluate Precision:
For precision, compare the standard deviation to a predetermined threshold (0.3). If the standard deviation is within this threshold, the data is considered precise.
[tex]\[
0.4445 > 0.3
\][/tex]
Since 0.4445 is greater than the threshold of 0.3, the data is not precise.
Given these evaluations, we can conclude:
- The data set is accurate (mean is close to the correct value).
- The data set is not precise (standard deviation is too high).
Therefore, the data set is accurate but not precise.
