which set of data does not contain any outliers?\n113, 115, 103, 154, 109, 111, 119\n141, 151, 111, 142…

Explanation:

Step1: Arrange data in ascending order

For the first set: $103, 109, 111, 113, 115, 154$. For the second set: $111, 140, 141, 142, 149, 150, 151$. For the third set: $91, 97, 99, 103, 104, 107, 109$. For the fourth set: $99, 131, 135, 136, 138, 140$.

Step2: Calculate the inter - quartile range (IQR) and check for outliers

For the first set:

• First, find the median ($Q2$). Since $n = 6$, the median is the average of the 3rd and 4th values. $Q2=\frac{111 + 113}{2}=112$.
• The lower half is $103, 109, 111$, so $Q1 = 109$.
• The upper half is $113, 115, 154$, so $Q3=115$.
• $IQR=Q3 - Q1=115 - 109 = 6$.
• Lower fence: $Q1-1.5\times IQR=109-1.5\times6=109 - 9 = 100$.
• Upper fence: $Q3 + 1.5\times IQR=115+1.5\times6=115 + 9 = 124$. $154$ is an outlier.

For the second set:

• Since $n = 7$, $Q2 = 142$.
• The lower half is $111, 140, 141$, so $Q1 = 140$.
• The upper half is $149, 150, 151$, so $Q3 = 150$.
• $IQR=Q3 - Q1=150 - 140 = 10$.
• Lower fence: $Q1-1.5\times IQR=140-1.5\times10=140 - 15 = 125$.
• Upper fence: $Q3 + 1.5\times IQR=150+1.5\times10=150 + 15 = 165$. There are no outliers.

For the third set:

• Since $n = 7$, $Q2 = 103$.
• The lower half is $91, 97, 99$, so $Q1 = 97$.
• The upper half is $104, 107, 109$, so $Q3 = 107$.
• $IQR=Q3 - Q1=107 - 97 = 10$.
• Lower fence: $Q1-1.5\times IQR=97-1.5\times10=97 - 15 = 82$.
• Upper fence: $Q3 + 1.5\times IQR=107+1.5\times10=107 + 15 = 122$. There are no outliers.

For the fourth set:

• Since $n = 6$, $Q2=\frac{135+136}{2}=135.5$.
• The lower half is $99, 131, 135$, so $Q1 = 131$.
• The upper half is $136, 138, 140$, so $Q3 = 138$.
• $IQR=Q3 - Q1=138 - 131 = 7$.
• Lower fence: $Q1-1.5\times IQR=131-1.5\times7=131 - 10.5 = 120.5$.
• Upper fence: $Q3 + 1.5\times IQR=138+1.5\times7=138 + 10.5 = 148.5$. $99$ is an outlier.

Answer:

$141, 151, 111, 142, 149, 140, 150$; $99, 103, 91, 104, 109, 107, 97$