Question

which data set has an outlier? 6, 13, 13, 15, 15, 18, 18, 22 4, 4, 4, 8, 9, 9, 11, 18 2, 3, 5, 7, 8, 8, 9, 10, 12, 17 3, 6, 7, 7, 8, 8, 9, 9, 9, 10

Explanation:

Step1: Find the quartiles for each data set

For the data set \(6,13,13,15,15,18,18,22\):

• First, order the data (already ordered).
• \(n = 8\) (number of data points).
• \(Q_1=\frac{13 + 13}{2}=13\) (the average of the 2nd and 3rd values).
• \(Q_3=\frac{18+18}{2} = 18\) (the average of the 6th and 7th values).
• \(IQR=Q_3 - Q_1=18 - 13 = 5\).
• Lower - bound \(=Q_1-1.5\times IQR=13-1.5\times5=13 - 7.5 = 5.5\).
• Upper - bound \(=Q_3 + 1.5\times IQR=18+1.5\times5=18 + 7.5=25.5\). No outliers.

For the data set \(4,4,4,8,9,9,11,18\):

• \(n = 8\).
• \(Q_1=\frac{4 + 4}{2}=4\) (average of 2nd and 3rd values).
• \(Q_3=\frac{9+11}{2}=10\) (average of 6th and 7th values).
• \(IQR=Q_3 - Q_1=10 - 4 = 6\).
• Lower - bound \(=Q_1-1.5\times IQR=4-1.5\times6=4 - 9=-5\).
• Upper - bound \(=Q_3+1.5\times IQR=10 + 9=19\). \(18<19\), no outliers.

For the data set \(2,3,5,7,8,8,9,10,12,17\):

• \(n = 10\).
• \(Q_1\) is the 3rd value (\(n = 10\), \(Q_1\) is at position \(\frac{n + 1}{4}=\frac{10+1}{4}=2.75\), so \(Q_1 = 5\)).
• \(Q_3\) is the 8th value (at position \(3\times\frac{n + 1}{4}=3\times\frac{10 + 1}{4}=8.25\), so \(Q_3=10\)).
• \(IQR=Q_3 - Q_1=10 - 5 = 5\).
• Lower - bound \(=Q_1-1.5\times IQR=5-1.5\times5=5 - 7.5=-2.5\).
• Upper - bound \(=Q_3+1.5\times IQR=10+7.5 = 17.5\). \(17<17.5\), no outliers.

For the data set \(3,6,7,7,8,8,9,9,9,10\):

• \(n = 10\).
• \(Q_1\) is the 3rd value (\(n = 10\), \(Q_1\) is at position \(\frac{n + 1}{4}=\frac{10+1}{4}=2.75\), so \(Q_1 = 7\)).
• \(Q_3\) is the 8th value (at position \(3\times\frac{n + 1}{4}=3\times\frac{10 + 1}{4}=8.25\), so \(Q_3=9\)).
• \(IQR=Q_3 - Q_1=9 - 7 = 2\).
• Lower - bound \(=Q_1-1.5\times IQR=7-1.5\times2=7 - 3 = 4\).
• Upper - bound \(=Q_3+1.5\times IQR=9+3=12\). \(3<4\), so \(3\) is an outlier.

Answer:

\(3,6,7,7,8,8,9,9,9,10\)