Question

the data set in the table forms a cluster and does not include outliers. which point, if added to the table, would be an outlier? (0.5,42) (1.25,35) (2.75,41) (3.25,28)

Explanation:

Step1: Analyze the data range

Looking at the \(x\) - values: \(0.5,1,1.5,1.5,2,2.25,2.75\). The \(x\) - values are mostly in the range \(0.5\leq x\leq2.75\). Looking at the \(y\) - values: \(41,35,37,38,35,30,31\). The \(y\) - values are mostly in the range \(30\leq y\leq41\).

Step2: Check each option

• For the point \((0.5,42)\): The \(x = 0.5\) is within the \(x\) - value range. The \(y=42\) is only slightly larger than the maximum \(y\) - value of \(41\) in the original data.
• For the point \((1.25,35)\): The \(x = 1.25\) is within the \(x\) - value range (\(0.5\leq x\leq2.75\)) and \(y = 35\) is also within the \(y\) - value range (\(30\leq y\leq41\)).
• For the point \((2.75,41)\): The \(x = 2.75\) is the maximum \(x\) - value in the original data and \(y = 41\) is the maximum \(y\) - value in the original data.
• For the point \((3.25,28)\): The \(x=3.25\) is outside the range of \(x\) - values (\(0.5\leq x\leq2.75\)) in the original data. And while \(y = 28\) is close to the lower end of the \(y\) - value range (\(30\leq y\leq41\)), the \(x\) - value deviation is more significant.

Answer:

\((3.25,28)\)