Math Problem Statement

For the following set of data identify any outlier, if there is one. 840, 296, 157, 750, 5,580, 400.

Solution

To identify outliers in the data set {840,296,157,750,5,580,400}\{840, 296, 157, 750, 5,580, 400\}, we will use the Interquartile Range (IQR) method.

Steps to Identify Outliers:

  1. Sort the Data: Arrange the data in ascending order. {5,157,296,400,750,840}\{5, 157, 296, 400, 750, 840\}

  2. Calculate the Quartiles:

    • Median (Q2): The middle value of the sorted data: Q2=296+4002=348Q2 = \frac{296 + 400}{2} = 348
    • Lower Quartile (Q1): Median of the lower half: Q1=5+1572=81Q1 = \frac{5 + 157}{2} = 81
    • Upper Quartile (Q3): Median of the upper half: Q3=750+8402=795Q3 = \frac{750 + 840}{2} = 795
  3. Calculate the IQR: IQR=Q3Q1=79581=714\text{IQR} = Q3 - Q1 = 795 - 81 = 714

  4. Determine Outlier Bounds:

    • Lower Bound: Q11.5×IQR=811.5×714=990Q1 - 1.5 \times \text{IQR} = 81 - 1.5 \times 714 = -990 (no lower outliers as data can't go below 0).
    • Upper Bound: Q3+1.5×IQR=795+1.5×714=1866Q3 + 1.5 \times \text{IQR} = 795 + 1.5 \times 714 = 1866.
  5. Identify Outliers:

    • Any data point less than the lower bound or greater than the upper bound is an outlier.
    • Observing the data: {840,296,157,750,5,580,400}\{840, 296, 157, 750, 5, 580, 400\} All values except 5,5805,580 are within the range [0,1866][0, 1866]. Therefore, 5,5805,580 is an outlier.

Conclusion:

The only outlier in the data is 5,5805,580.

Would you like further details or have any questions?


Related Questions:

  1. What is the significance of IQR in outlier detection?
  2. How would the presence of multiple outliers affect statistical measures like mean and variance?
  3. Could the Z-score method detect 5,5805,580 as an outlier in this data set?
  4. How can removing outliers impact the analysis of a data set?
  5. What are other methods for identifying outliers besides IQR and Z-scores?

Tip:

Outlier detection is vital for improving data quality and ensuring statistical analyses are not skewed by extreme values.

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Math Problem Analysis

Mathematical Concepts

Statistics
Outlier Detection
Interquartile Range (IQR)

Formulas

Median (Q2): (n/2th value if n is odd or average of (n/2)th and (n/2 + 1)th values if n is even)
Lower Quartile (Q1): Median of lower half of the data
Upper Quartile (Q3): Median of upper half of the data
IQR = Q3 - Q1
Outlier Bounds: Lower Bound = Q1 - 1.5 * IQR, Upper Bound = Q3 + 1.5 * IQR

Theorems

IQR Method for Outlier Detection

Suitable Grade Level

Grades 8-10