Question

which set of data does not contain any outliers?
113, 115, 103, 154, 109, 111, 119
141, 151, 111, 142, 149, 140, 150
99, 103, 91, 104, 109, 107, 97
131, 135, 131, 99, 138, 136, 140

Explanation:

Step1: Sort the data sets

• First set: \(103,109,111,113,115,119,154\)
• Second set: \(111,140,141,142,149,150,151\)
• Third set: \(91,97,99,103,104,107,109\)
• Fourth set: \(99,131,131,135,136,138,140\)

Step2: Find the median (\(Q_2\)), lower quartile (\(Q_1\)) and upper quartile (\(Q_3\))

• For the second set (\(n = 7\)):
• \(Q_2=\) the \(4^{th}\) value \(=142\)
• \(Q_1=\) the \(2^{nd}\) value \(=140\)
• \(Q_3=\) the \(6^{th}\) value \(=150\)
• Calculate the inter - quartile range (\(IQR=Q_3 - Q_1\)): \(IQR = 150-140 = 10\)
• Find the lower and upper bounds for outliers:
• Lower bound \(=Q_1-1.5\times IQR=140 - 1.5\times10=140 - 15=125\)
• Upper bound \(=Q_3 + 1.5\times IQR=150+1.5\times10=150 + 15=165\)
• Check each data point in the second set: \(111<125\) (outlier in the second set)
• For the third set (\(n = 7\)):
• \(Q_2=\) the \(4^{th}\) value \(=103\)
• \(Q_1=\) the \(2^{nd}\) value \(=97\)
• \(Q_3=\) the \(6^{th}\) value \(=107\)
• \(IQR=107 - 97=10\)
• Lower bound \(=97-1.5\times10=97 - 15 = 82\)
• Upper bound \(=107+1.5\times10=107 + 15=122\)
• All data points \(91,97,99,103,104,107,109\) lie within \([82,122]\)
• For the first set (\(n = 7\)):
• \(Q_2=\) the \(4^{th}\) value \(=113\)
• \(Q_1=\) the \(2^{nd}\) value \(=109\)
• \(Q_3=\) the \(6^{th}\) value \(=119\)
• \(IQR=119 - 109=10\)
• Lower bound \(=109-1.5\times10=109 - 15 = 94\)
• Upper bound \(=119+1.5\times10=119 + 15=134\)
• \(154>134\) (outlier in the first set)
• For the fourth set (\(n = 7\)):
• \(Q_2=\) the \(4^{th}\) value \(=135\)
• \(Q_1=\) the \(2^{nd}\) value \(=131\)
• \(Q_3=\) the \(6^{th}\) value \(=138\)
• \(IQR=138 - 131=7\)
• Lower bound \(=131-1.5\times7=131 - 10.5 = 120.5\)
• Upper bound \(=138+1.5\times7=138+10.5 = 148.5\)
• \(99<120.5\) (outlier in the fourth set)

Answer:

The set \(99,103,91,104,109,107,97\)