91Ó°ÊÓ

Organize the data in ascending order: \(32.7, 37.4, 39.6, 46.4, 46.8, 47.3, 52.1, 53.7, 57.8, 63.0, 63.2, 64.4, 66.1, 66.3, 68.6, 79.2, 89.2, 92.4\). Calculate the mean: \(\frac{32.7+37.4+\cdots+89.2+92.4}{18} \approx 59.53 \: ml\). Calculate the median: The middle values are \(57.8\) and \(63.0\), so the median is \(\frac{57.8+63.0}{2} = 60.4 \: ml\). Calculate the standard deviation: First, find the squared differences: \((32.7-59.53)^2, (37.4-59.53)^2, \cdots, (89.2-59.53)^2, (92.4-59.53)^2\). Find the average of these squared differences: \(\frac{726.43+487.89+\cdots+1078.25+1084.22}{18} \approx 337.27\), and take the square root to obtain the standard deviation, which is approximately \(18.36 \: ml\). The mean of the dataset is \(59.53 \: ml\), the median is \(60.4 \: ml\), and the standard deviation is \(18.36 \: ml\). This indicates that the individuals generally poured more than the desired \(44.3 \: ml\), and there was considerable variability in the amounts poured.