Question

match each set of values to its correct number of outliers. 48, 47, 34, 37, 11, 39; 12, 14, 16, 1, 15, 15, 34, 2, 17; 2, 7, 4, 0, 1, 3, 8, 5, 9, 3; 185, 190, 31, 196, 549, 177, 182; 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 then calculate 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. Median of the whole set is $\frac{37 + 39}{2}=38$. Lower half is 11, 34, 37, so $Q_1 = 34$. Upper half is 39, 47, 48, so $Q_3 = 47$. $IQR=47 - 34 = 13$. Lower bound for outliers is $34-1.5\times13=34 - 19.5 = 14.5$, upper bound is $47+1.5\times13=47 + 19.5 = 66.5$. The value 11 is an outlier. So this set 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. Median of the whole set is 15. Lower half is 1, 2, 12, 14, so $Q_1=\frac{2 + 12}{2}=7$. Upper half is 15, 16, 17, 34, so $Q_3=\frac{16+17}{2}=16.5$. $IQR = 16.5 - 7=9.5$. Lower bound for outliers is $7-1.5\times9.5=7 - 14.25=-7.25$, upper bound is $16.5+1.5\times9.5=16.5 + 14.25 = 30.75$. The value 34 is an outlier. So this set 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. Median of the whole set is $\frac{3 + 4}{2}=3.5$. Lower half is 0, 1, 2, 3, 3, so $Q_1 = 2$. Upper half is 4, 5, 7, 8, 9, so $Q_3 = 7$. $IQR=7 - 2 = 5$. Lower bound for outliers is $2-1.5\times5=2 - 7.5=-5.5$, upper bound is $7+1.5\times5=7 + 7.5 = 14.5$. There are no outliers. So this set contains no outliers.

Step5: For the fourth set: 185, 190, 31, 196, 549, 177, 182

Arrange in ascending order: 31, 177, 182, 185, 190, 196, 549. Median of the whole set is 185. Lower half is 31, 177, 182, so $Q_1 = 177$. Upper half is 190, 196, 549, so $Q_3 = 196$. $IQR=196 - 177 = 19$. Lower bound for outliers is $177-1.5\times19=177 - 28.5 = 148.5$, upper bound is $196+1.5\times19=196 + 28.5 = 224.5$. The values 31 and 549 are outliers. So this set contains two outliers.

Answer:

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