which data set has an outlier? 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 3, 3, 3, 7, 9, 9, 10, 14 25, 36, 44…

which data set has an outlier? 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 3, 3, 3, 7, 9, 9, 10, 14 25, 36, 44, 51, 62, 77

which data set has an outlier? 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 3, 3, 3, 7, 9, 9, 10, 14 25, 36, 44, 51, 62, 77

Answer

Explanation:

Step1: Recall the outlier formula

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

Step2: Analyze the first data set: 8, 17, 18, 20, 20, 21, 23, 26, 31, 39

  • Step 2.1: Find the median (to split into lower and upper halves) The data set has 10 values. The median is the average of the 5th and 6th values. The 5th value is 20, the 6th is 21. So median $M=\frac{20 + 21}{2}=20.5$.
  • Step 2.2: Find $Q_1$ (median of lower half: 8, 17, 18, 20, 20) The lower half has 5 values. The median of the lower half ( $Q_1$ ) is the 3rd value, which is 18.
  • Step 2.3: Find $Q_3$ (median of upper half: 21, 23, 26, 31, 39) The upper half has 5 values. The median of the upper half ( $Q_3$ ) is the 3rd value, which is 26.
  • Step 2.4: Calculate IQR $IQR = Q_3 - Q_1 = 26 - 18 = 8$.
  • Step 2.5: Check for outliers Lower bound: $Q_1 - 1.5\times IQR = 18 - 1.5\times8 = 18 - 12 = 6$. Upper bound: $Q_3 + 1.5\times IQR = 26 + 1.5\times8 = 26 + 12 = 38$. Now check each value: 8 is greater than 6, 39 is greater than 38. So 39 is an outlier? Wait, wait, 39 is greater than 38, so 39 is an outlier? Wait, let's re - check. Wait, the upper bound is 38, 39>38, so 39 is an outlier? Wait, but let's check the second data set.

Step3: Analyze the second data set: 3, 3, 3, 7, 9, 9, 10, 14

  • Step 3.1: Find the median The data set has 8 values. The median is the average of the 4th and 5th values. The 4th value is 7, the 5th is 9. So median $M=\frac{7 + 9}{2}=8$.
  • Step 3.2: Find $Q_1$ (median of lower half: 3, 3, 3, 7) The lower half has 4 values. The median of the lower half is the average of the 2nd and 3rd values: $\frac{3+3}{2}=3$.
  • Step 3.3: Find $Q_3$ (median of upper half: 9, 9, 10, 14) The upper half has 4 values. The median of the upper half is the average of the 2nd and 3rd values: $\frac{9 + 10}{2}=9.5$.
  • Step 3.4: Calculate IQR $IQR=Q_3 - Q_1=9.5 - 3 = 6.5$.
  • Step 3.5: Check for outliers Lower bound: $Q_1-1.5\times IQR = 3-1.5\times6.5=3 - 9.75=- 6.75$. Upper bound: $Q_3 + 1.5\times IQR=9.5+1.5\times6.5=9.5 + 9.75 = 19.25$. All values (3, 3, 3, 7, 9, 9, 10, 14) are between - 6.75 and 19.25, so no outliers.

Step4: Analyze the third data set: 25, 36, 44, 51, 62, 77

  • Step 4.1: Find the median The data set has 6 values. The median is the average of the 3rd and 4th values. The 3rd value is 44, the 4th is 51. So median $M=\frac{44 + 51}{2}=47.5$.
  • **Step 4.2: Find $Q_1[LLM SSE On Failure]