Question

which sets contain one or more outliers? check all that apply.
□ 187, 298, 239, 1, 984, 202, 191
□ 5, 3, 8, 6, 3, 6, 1, 2, 0
□ 15, 19, 21, 16, 25, 13, 17
□ 56, 1, 5, 72, 67, 59, 74, 60
□ 88, 7, 32, 31, 34, 39, 34, 35, 33

Explanation:

Step1: Recall Outlier Formula

To find outliers, we use the interquartile range (IQR) method. The formula for outliers is:

• Lower bound: \( Q_1 - 1.5 \times \text{IQR} \)
• Upper bound: \( Q_3 + 1.5 \times \text{IQR} \)

Values outside these bounds are outliers.

Step2: Analyze First Set (187, 298, 239, 1, 984, 202, 191)

1. Sort the data: \( 1, 187, 191, 202, 239, 298, 984 \)
2. Find \( Q_1 \) (median of lower half: \( 1, 187, 191 \)): \( Q_1 = 187 \)
3. Find \( Q_3 \) (median of upper half: \( 239, 298, 984 \)): \( Q_3 = 298 \)
4. Calculate IQR: \( 298 - 187 = 111 \)
5. Lower bound: \( 187 - 1.5 \times 111 = 187 - 166.5 = 20.5 \)
6. Upper bound: \( 298 + 1.5 \times 111 = 298 + 166.5 = 464.5 \)
7. Check values: \( 1 < 20.5 \) and \( 984 > 464.5 \) → outliers exist.

Step3: Analyze Second Set (5, 3, 8, 6, 3, 6, 1, 2, 0)

1. Sort: \( 0, 1, 2, 3, 3, 5, 6, 6, 8 \)
2. \( Q_1 \) (median of \( 0, 1, 2, 3 \)): \( \frac{1 + 2}{2} = 1.5 \)
3. \( Q_3 \) (median of \( 5, 6, 6, 8 \)): \( \frac{6 + 6}{2} = 6 \)
4. IQR: \( 6 - 1.5 = 4.5 \)
5. Lower bound: \( 1.5 - 1.5 \times 4.5 = 1.5 - 6.75 = -5.25 \)
6. Upper bound: \( 6 + 1.5 \times 4.5 = 6 + 6.75 = 12.75 \)
7. Check values: All values (0 - 8) are within bounds → no outliers.

Step4: Analyze Third Set (15, 19, 21, 16, 25, 13, 17)

1. Sort: \( 13, 15, 16, 17, 19, 21, 25 \)
2. \( Q_1 \) (median of \( 13, 15, 16 \)): \( 15 \)
3. \( Q_3 \) (median of \( 19, 21, 25 \)): \( 21 \)
4. IQR: \( 21 - 15 = 6 \)
5. Lower bound: \( 15 - 1.5 \times 6 = 15 - 9 = 6 \)
6. Upper bound: \( 21 + 1.5 \times 6 = 21 + 9 = 30 \)
7. Check values: All (13 - 25) are within bounds → no outliers.

Step5: Analyze Fourth Set (56, 1, 5, 72, 67, 59, 74, 60)

1. Sort: \( 1, 5, 56, 59, 60, 67, 72, 74 \)
2. \( Q_1 \) (median of \( 1, 5, 56, 59 \)): \( \frac{5 + 56}{2} = 30.5 \)
3. \( Q_3 \) (median of \( 60, 67, 72, 74 \)): \( \frac{67 + 72}{2} = 69.5 \)
4. IQR: \( 69.5 - 30.5 = 39 \)
5. Lower bound: \( 30.5 - 1.5 \times 39 = 30.5 - 58.5 = -28 \)
6. Upper bound: \( 69.5 + 1.5 \times 39 = 69.5 + 58.5 = 128 \)
7. Check values: \( 1 < 30.5 \) and \( 5 < 30.5 \) → outliers exist.

Step6: Analyze Fifth Set (88, 7, 32, 31, 34, 39, 34, 35, 33)

1. Sort: \( 7, 31, 32, 33, 34, 34, 35, 39, 88 \)
2. \( Q_1 \) (median of \( 7, 31, 32, 33 \)): \( \frac{31 + 32}{2} = 31.5 \)
3. \( Q_3 \) (median of \( 34, 35, 39, 88 \)): \( \frac{35 + 39}{2} = 37 \)
4. IQR: \( 37 - 31.5 = 5.5 \)
5. Lower bound: \( 31.5 - 1.5 \times 5.5 = 31.5 - 8.25 = 23.25 \)
6. Upper bound: \( 37 + 1.5 \times 5.5 = 37 + 8.25 = 45.25 \)
7. Check values: \( 7 < 23.25 \) and \( 88 > 45.25 \) → outliers exist.

Answer:

• 187, 298, 239, 1, 984, 202, 191
• 56, 1, 5, 72, 67, 59, 74, 60
• 88, 7, 32, 31, 34, 39, 34, 35, 33

(These sets contain outliers based on the IQR method.)