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: Recall Outlier Formula

To find outliers, we use the interquartile range (IQR) method. The formula for outliers is: values less than \( Q_1 - 1.5 \times \text{IQR} \) or greater than \( Q_3 + 1.5 \times \text{IQR} \) are outliers, where \( Q_1 \) is the first quartile and \( Q_3 \) is the third quartile.

Step2: Analyze First Data Set (113, 115, 103, 154, 109, 111, 119)

- Sort: 103, 109, 111, 113, 115, 119, 154
- \( Q_1 = 109 \), \( Q_3 = 119 \)
- \( \text{IQR} = 119 - 109 = 10 \)
- Lower bound: \( 109 - 15 = 94 \), Upper bound: \( 119 + 15 = 134 \)
- 154 > 134, so outlier exists.

Step3: Analyze Second Data Set (141, 151, 111, 142, 149, 140, 150)

- Sort: 111, 140, 141, 142, 149, 150, 151
- \( Q_1 = 140 \), \( Q_3 = 150 \)
- \( \text{IQR} = 150 - 140 = 10 \)
- Lower bound: \( 140 - 15 = 125 \), Upper bound: \( 150 + 15 = 165 \)
- All values (111,140,141,142,149,150,151) are between 125 and 165? Wait, 111 < 125. Wait, maybe miscalculation. Wait, re-sort: 111, 140, 141, 142, 149, 150, 151. Wait, median is 142 (4th term). \( Q_1 \) is median of first half (111,140,141) → 140. \( Q_3 \) is median of second half (149,150,151) → 150. IQR=10. Lower bound=140 -15=125. 111 <125, so 111 is outlier? Wait, maybe I made a mistake. Wait, let's check third data set.

Step4: Analyze Third Data Set (99, 103, 91, 104, 109, 107, 97)

- Sort: 91, 97, 99, 103, 104, 107, 109
- \( Q_1 = 97 \), \( Q_3 = 107 \)
- \( \text{IQR} = 107 - 97 = 10 \)
- Lower bound: \( 97 - 15 = 82 \), Upper bound: \( 107 + 15 = 122 \)
- All values (91,97,99,103,104,107,109) are between 82 and 122? 91 >82, 109 <122. Wait, no outlier? Wait, no, wait 91 is in the data. Wait, maybe I messed up. Wait, let's check fourth data set.

Step5: Analyze Fourth Data Set (131, 135, 131, 99, 138, 136, 140)

- Sort: 99, 131, 131, 135, 136, 138, 140
- \( Q_1 = 131 \), \( Q_3 = 138 \)
- \( \text{IQR} = 138 - 131 = 7 \)
- Lower bound: \( 131 - 10.5 = 120.5 \), Upper bound: \( 138 + 10.5 = 148.5 \)
- 99 < 120.5, so 99 is outlier.

Wait, maybe I made a mistake in second data set. Wait, let's re-examine the second data set: 141, 151, 111, 142, 149, 140, 150. Sort: 111, 140, 141, 142, 149, 150, 151. Median (Q2) is 142. \( Q_1 \) is median of [111,140,141] → 140. \( Q_3 \) is median of [149,150,151] → 150. IQR=10. Lower bound=140 -15=125. 111 <125, so 111 is outlier. So second data set has outlier.

Third data set: 99,103,91,104,109,107,97. Sort: 91,97,99,103,104,107,109. Median=103. \( Q_1 \) (median of 91,97,99) → 97. \( Q_3 \) (median of 104,107,109) → 107. IQR=10. Lower bound=97-15=82. Upper bound=107+15=122. All values: 91 (≥82), 97,99,103,104,107,109 (≤122). So no outliers? Wait, but let's check fourth data set: 131,135,131,99,138,136,140. Sort:99,131,131,135,136,138,140. Median=135. \( Q_1 \) (median of 99,131,131) →131. \( Q_3 \) (median of 136,138,140) →138. IQR=7. Lower bound=131-10.5=120.5. 99 <120.5, so 99 is outlier.

First data set: 113,115,103,154,109,111,119. Sort:103,109,111,113,115,119,154. Median=113. \( Q_1 \) (103,109,111) →109. \( Q_3 \) (115,119,154) →119. IQR=10. Upper bound=119+15=134. 154>134, so 154 is outlier.

Wait, so third data set: 91,97,99,103,104,107,109. All values between 82 and 122. So no outliers. Wait, but the option selected in the image is the second one, but maybe I made a mistake. Wait, maybe I messed up the second data set. Wait, maybe the correct one is the third data set? Wait, no, let's recheck.

Wait, the third data set: 99,103,91,104,109,107,97. Sorted:91,97,99,103,104,107,109. \( Q_1 \): median of first three (91,97,99) is 97. \( Q_3 \): median of last three (104,107,109) is 107. IQR=10. Lower fence: 97 - 1.5*10 = 97 -15=82. Upper fence:107 +15=122. All data points: 91 (≥82), 97,99,103,104,107,109 (≤122). So no outliers.

Second data set: 141,151,111,142,149,140,150. Sorted:111,140,141,142,149,150,151. \( Q_1 \)=140, \( Q_3 \)=150, IQR=10. Lower fence=140-15=125. 111 <125, so outlier.

Fourth data set: 131,135,131,99,138,136,140. Sorted:99,131,131,135,136,138,140. \( Q_1 \)=131, \( Q_3 \)=138, IQR=7. Lower fence=131-10.5=120.5. 99 <120.5, outlier.

First data set: 113,115,103,154,109,111,119. Sorted:103,109,111,113,115,119,154. \( Q_1 \)=109, \( Q_3 \)=119, IQR=10. Upper fence=119+15=134. 154>134, outlier.

So the third data set (99, 103, 91, 104, 109, 107, 97) has no outliers? Wait, but the option marked in the image is the second one. Wait, maybe I made a mistake. Wait, let's recheck the second data set. Wait, 141,151,111,142,149,140,150. Sorted:111,140,141,142,149,150,151. Wait, maybe the median is 142, \( Q_1 \) is the median of the first four? No, for odd number of terms (n=7), the median is the 4th term. The first half is terms 1-3 (111,140,141), second half is terms 5-7 (149,150,151). So \( Q_1 \) is median of 111,140,141 →140. \( Q_3 \) is median of 149,150,151 →150. IQR=10. Lower fence=140-15=125. 111 is below 125, so outlier. So second data set has outlier.

Third data set: 91,97,99,103,104,107,109. All values between 82 and 122. So no outliers. So the correct answer is the third option: 99, 103, 91, 104, 109, 107, 97. Wait, but the image has the second option selected, maybe a mistake. Wait, let's check again.

Wait, maybe I messed up the second data set. Let's calculate \( Q_1 \) and \( Q_3 \) correctly. For n=7, positions: 1:111, 2:140, 3:141, 4:142, 5:149, 6:150, 7:151. \( Q_1 \) is the value at position \( \frac{n+1}{4} = 2 \) (since (7+1)/4=2). So \( Q_1 = 140 \) (2nd term). \( Q_3 \) is at position \( \frac{3(n+1)}{4} = 6 \) (3*(8)/4=6). So \( Q_3 = 150 \) (6th term). IQR=150-140=10. Lower fence=140 -1.5*10=125. Upper fence=150 +1.5*10=165. Now, check all values: 111 (1st term) <125 → outlier. 140,141,142,149,150,151 are between 125 and 165. So 111 is outlier. So second data set has outlier.

Third data set: n=7. Positions:1:91,2:97,3:99,4:103,5:104,6:107,7:109. \( Q_1 \) at position 2:97. \( Q_3 \) at position 6:107. IQR=10. Lower fence=97-15=82. Upper fence=107+15=122. All values:91 (≥82),97,99,103,104,107,109 (≤122). So no outliers. So the correct answer is the third option: 99, 103, 91, 104, 109, 107, 97.

Answer:

The set of data that does not contain any outliers is 99, 103, 91, 104, 109, 107, 97 (the third option in the list).