Comprehensive Guide to Descriptive Statistics: Deciles, Percentiles, Interquartile Range, Box Plots, and Outlier Analysis

Measures of Position: Deciles and Percentiles

  • Deciles (DkD_k):

    • Deciles divide an ordered dataset into ten equal parts.
    • Position formula for deciles:     Dk=k×N10D_k = \frac{k \times N}{10}     where kk represents the decile number (1,2,,91, 2, \dots, 9) and NN represents the total number of items in the dataset.
  • Percentiles (PkP_k):

    • Percentiles divide an ordered dataset into one hundred equal parts.
    • Position formula for percentiles:     Pk=k×N100P_k = \frac{k \times N}{100}     where kk represents the percentile rank (1,2,,991, 2, \dots, 99) and NN represents the total number of items in the dataset.
  • Example Problem: Grade 10 Math Quiz Scores:

    • Dataset: Scores of 1212 Grade 10 students in a 1515-item Math quiz:     13,20,16,14,19,15,19,11,10,18,913, 20, 16, 14, 19, 15, 19, 11, 10, 18, 9
    • Step 1: Order the Dataset (N=12N = 12):
    • 1st1\text{st} position: 99
    • 2nd2\text{nd} position: 1010
    • 3rd3\text{rd} position: 1111
    • 4th4\text{th} position: 1313
    • 5th5\text{th} position: 1414
    • 6th6\text{th} position: 1515
    • 7th7\text{th} position: 1515
    • 8th8\text{th} position: 1616
    • 9th9\text{th} position: 1818
    • 10th10\text{th} position: 1919
    • 11th11\text{th} position: 1919
    • 12th12\text{th} position: 2020
    • Part A: Calculating Deciles (D1D_1, D4D_4, and D9D_9):
    • Calculating D1D_1:       D1=1×N10D_1 = \frac{1 \times N}{10}D1=1×1210=1210=1.22nd positionD_1 = \frac{1 \times 12}{10} = \frac{12}{10} = 1.2 \rightarrow 2\text{nd}\text{ position}D1=10D_1 = 10
    • Calculating D4D_4:       D4=4×N10D_4 = \frac{4 \times N}{10}D4=4×1210=4810=4.85th positionD_4 = \frac{4 \times 12}{10} = \frac{48}{10} = 4.8 \rightarrow 5\text{th}\text{ position}D4=14D_4 = 14
    • Calculating D9D_9:       D9=9×N10D_9 = \frac{9 \times N}{10}D9=9×1210=10810=10.811th positionD_9 = \frac{9 \times 12}{10} = \frac{108}{10} = 10.8 \rightarrow 11\text{th}\text{ position}D9=19D_9 = 19
    • Part B: Proving D5=Q2D_5 = Q_2:
    • Position of D5D_5:       D5=5×N10D_5 = \frac{5 \times N}{10}D5=5×1210=6010=6th positionD_5 = \frac{5 \times 12}{10} = \frac{60}{10} = 6\text{th}\text{ position}
    • Value of D5D_5 (averaging 6th6\text{th} and 7th7\text{th} positions):       D5=6th+7th2D_5 = \frac{6\text{th} + 7\text{th}}{2}D5=15+152=302=15D_5 = \frac{15 + 15}{2} = \frac{30}{2} = 15
    • Value of Q2Q_2 (Second Quartile / Median):       Q2=15+152=302=15Q_2 = \frac{15 + 15}{2} = \frac{30}{2} = 15
    • Equivalence:       D5=Q2D_5 = Q_2
    • Part C: Calculating Percentiles (P43P_{43} and P60P_{60}):
    • Calculating P43P_{43}:       P43=43×N100P_{43} = \frac{43 \times N}{100}P43=43×12100=516100=5.166th position(round up)P_{43} = \frac{43 \times 12}{100} = \frac{516}{100} = 5.16 \rightarrow 6\text{th}\text{ position} \quad (\text{round up})P43=15P_{43} = 15
    • Calculating P60P_{60}:       P60=60×N100P_{60} = \frac{60 \times N}{100}P60=60×12100=720100=7.208th position(round up)P_{60} = \frac{60 \times 12}{100} = \frac{720}{100} = 7.20 \rightarrow 8\text{th}\text{ position} \quad (\text{round up})P60=16P_{60} = 16
    • Part D: Proving Q3=P75Q_3 = P_{75}:
    • Position of P75P_{75}:       P75=75×N100P_{75} = \frac{75 \times N}{100}P75=75×12100=900100=99th and 10th positionsP_{75} = \frac{75 \times 12}{100} = \frac{900}{100} = 9 \rightarrow 9\text{th}\text{ and } 10\text{th}\text{ positions}
    • Value of P75P_{75}:       P75=18+192=372=18.5P_{75} = \frac{18 + 19}{2} = \frac{37}{2} = 18.5
    • Calculating Q3Q_3 from upper half (15,16,18,19,19,2015, 16, 18, 19, 19, 20):       Q3=18+192=372=18.5Q_3 = \frac{18 + 19}{2} = \frac{37}{2} = 18.5
    • Equivalence:       Q3=P75Q_3 = P_{75}

Interquartile Range (IQR) and Spread

  • Interquartile Range (IQR) Definition:

    • The Interquartile Range measures the spread or dispersion of the middle 50%50\% of a dataset.
    • It quantifies the distance between the upper quartile (Q3Q_3) and lower quartile (Q1Q_1).
  • Formula:   IQR=Q3Q1\text{IQR} = Q_3 - Q_1

  • Example Problem: Finding IQR:

    • Dataset: 8,10,12,14,16,18,208, 10, 12, 14, 16, 18, 20
    • Step 1: Arrange data in ascending order (already arranged):     8,10,12,14,16,18,208, 10, 12, 14, 16, 18, 20
    • Step 2: Identify Q1Q_1 and Q3Q_3:
    • Q1=10Q_1 = 10
    • Q2=14Q_2 = 14
    • Q3=18Q_3 = 18
    • Step 3: Apply IQR formula:     IQR=Q3Q1\text{IQR} = Q_3 - Q_1IQR=1810\text{IQR} = 18 - 10IQR=8\text{IQR} = 8
    • Multiple choice question evaluation:
    • Options: A. 44, B. 66, C. 88, D. 1010
    • Correct Answer: C. 88

Graphical Interpretation: Box-and-Whisker Plot

  • Definition:

    • A Box-and-Whisker Plot is a statistical graph displaying data distribution by showing spread, median, quartiles, and potential outliers.
  • Structural Components:

    • Box:
    • Represents the interquartile range (IQR).
    • Spans the middle 50%50\% of data between Q1Q_1 (P25P_{25}) and Q3Q_3 (P75P_{75}).
    • Median Line:
    • A line drawn inside the box indicating the dataset's median (50th50\text{th} percentile).
    • Whiskers:
    • Lines extending outward from the box to the minimum and maximum values located within 1.5×IQR1.5 \times \text{IQR} from Q1Q_1 and Q3Q_3
    • Outliers:
    • Individual points plotted distinctly beyond the whiskers, representing unusually high or low values.
  • Example Problem: Assessing Outliers with Box Plot Boundaries:

    • Dataset of exam scores: {55,60,65,70,75,80,85,90,100}\{55, 60, 65, 70, 75, 80, 85, 90, 100\}
    • Step 1: Determine Q1Q_1, Median, and Q3Q_3:     Q1=60+652=62.5Q_1 = \frac{60 + 65}{2} = 62.5Median=75\text{Median} = 75Q3=85+902=87.5Q_3 = \frac{85 + 90}{2} = 87.5
    • Step 2: Compute IQR:     IQR=Q3Q1\text{IQR} = Q_3 - Q_1IQR=87.562.5=25\text{IQR} = 87.5 - 62.5 = 25
    • Step 3: Compute Whisker Limits:
    • Limit below Q1Q_1:       Q11.5×IQRQ_1 - 1.5 \times \text{IQR}62.5(1.5×25)=62.537.5=2562.5 - (1.5 \times 25) = 62.5 - 37.5 = 25
    • Limit above Q3Q_3:       Q3+1.5×IQRQ_3 + 1.5 \times \text{IQR}87.5+(1.5×25)=87.5+37.5=12587.5 + (1.5 \times 25) = 87.5 + 37.5 = 125
    • Step 4: Evaluate for Outliers:
    • Dataset minimum value = 5555
    • Dataset maximum value = 100100
    • Since all scores fall strictly within the limits 2525 and 125125, there are no outliers present in this dataset.

Outlier Detection and Analysis

  • Definition of Outliers:

    • An outlier is a value that is significantly higher or lower than the rest of the dataset.
  • Illustrative Examples:

    • In an exam dataset with typical scores ranging from 6060 to 9090, a score of 1212 or a score of 9999 would be considered an outlier.
  • Statistical Impact of Outliers:

    • Strongly skews and affects the mean.
    • Inflates standard deviation.
    • Biases linear regression models.
    • The median is resistant to outliers compared to the mean.
  • IQR Method Formulas for Outlier Detection:

    • Lower bound threshold:     Below Q1=Q11.5×IQR\text{Below } Q_1 = Q_1 - 1.5 \times \text{IQR}
    • Upper bound threshold:     Above Q3=Q3+1.5×IQR\text{Above } Q_3 = Q_3 + 1.5 \times \text{IQR}
    • Rule: Any data value falling outside these lower and upper bounds is classified as an outlier.
  • Practice Question on Outlier Criteria:

    • Question: Outliers are values that fall:
    • A. Beyond 1.5×IQR1.5 \times \text{IQR} from Q1Q_1 or Q3Q_3
    • B. Within Q1Q_1 and Q3Q_3
    • C. Equal to median
    • D. Equal to mean
    • Answer: A. Beyond 1.5×IQR1.5 \times \text{IQR} from Q1Q_1 or Q3Q_3