Question

match each set of values to its correct number of outliers.
185, 198, 31, 196, 549, 177, 182
2, 7, 4, 0, 1, 3, 8, 5, 9, 3
12, 14, 16, 1, 15, 15, 34, 2, 17
48, 47, 34, 37, 11, 39
contains three outliers
contains two outliers
contains one outlier
contains no outliers

Explanation:

Step1: Recall outlier - finding method

To find outliers, we first need to find the first quartile ($Q_1$), the third quartile ($Q_3$), and the inter - quartile range ($IQR = Q_3 - Q_1$). Outliers are values less than $Q_1-1.5\times IQR$ or greater than $Q_3 + 1.5\times IQR$.

Step2: For the first set: 48, 47, 34, 37, 11, 39

Arrange in ascending order: 11, 34, 37, 39, 47, 48. The median of the lower half ($Q_1$) is 34, the median of the upper half ($Q_3$) is 47. $IQR=47 - 34=13$. $Q_1-1.5\times IQR=34-1.5\times13=34 - 19.5 = 14.5$, $Q_3 + 1.5\times IQR=47+1.5\times13=47 + 19.5 = 66.5$. The value 11 is an outlier. So it contains one outlier.

Step3: For the second set: 12, 14, 16, 1, 15, 15, 34, 2, 17

Arrange in ascending order: 1, 2, 12, 14, 15, 15, 16, 17, 34. The median of the lower half ($Q_1$) is 2, the median of the upper half ($Q_3$) is 16. $IQR = 16-2 = 14$. $Q_1-1.5\times IQR=2-1.5\times14=2 - 21=-19$, $Q_3 + 1.5\times IQR=16+1.5\times14=16 + 21 = 37$. The value 1 is an outlier. So it contains one outlier.

Step4: For the third set: 2, 7, 4, 0, 1, 3, 8, 5, 9, 3

Arrange in ascending order: 0, 1, 2, 3, 3, 4, 5, 7, 8, 9. The median of the lower half ($Q_1$) is 1.5, the median of the upper half ($Q_3$) is 7. $IQR=7 - 1.5 = 5.5$. $Q_1-1.5\times IQR=1.5-1.5\times5.5=1.5 - 8.25=-6.75$, $Q_3 + 1.5\times IQR=7+1.5\times5.5=7 + 8.25 = 15.25$. There are no values outside this range. So it contains no outliers.

Step5: For the fourth set: 185, 198, 31, 196, 549, 177, 182

Arrange in ascending order: 31, 177, 182, 185, 196, 198, 549. The median of the lower half ($Q_1$) is 177, the median of the upper half ($Q_3$) is 196. $IQR=196 - 177 = 19$. $Q_1-1.5\times IQR=177-1.5\times19=177 - 28.5 = 148.5$, $Q_3 + 1.5\times IQR=196+1.5\times19=196 + 28.5 = 224.5$. The values 31 and 549 are outliers. So it contains two outliers.

Answer:

Set 1: Contains one outlier
Set 2: Contains one outlier
Set 3: Contains no outliers
Set 4: Contains two outliers