Statistical Outlier Detection & Truncation Report

What you will learn

• Calculate empirical 25th and 75th quantiles using Series.quantile() in Pandas
• Establish non-parametric outlier fences using the Interquartile Range (1.5 * IQR)
• Isolate anomalous data points using vectorized boolean filter masks
• Contrast robust median-IQR statistics with sensitive mean-variance metrics

What you need first

import pandas as pd
s = pd.Series([10, 20, 30, 40])
print(float(s.quantile(0.5)))

25.0

import pandas as pd
s = pd.Series([1, 2, 3, 4, 100])
print(float(s.quantile(0.75) - s.quantile(0.25)))

2.0

Architecture and data flow

The data

A numerical transaction series containing normal commercial purchases alongside extreme card-testing outliers.

Step by step

Read each step, run its code, then check your output against the one shown.

1. 01

   Initialize transaction Series and compute summary statistics

   We initialize the raw transaction amounts Series containing 8 records: six legitimate retail transactions centered near 100.0 rupees and two hostile anomalies (500.0 and 12.0). We print the sample size and contrast the sample mean against the median. In data engineering pipelines, comparing mean and median provides an immediate diagnostic indicator of distribution skewness. Here, the mean is heavily inflated to 140.12 rupees due to the 500.0 rupee outlier, whereas the median remains stable at 101.50 rupees, reflecting the true central tendency of regular customer transactions.

   PythonStep 1Copy

   import pandas as pd
   
   s = pd.Series([100.0, 105.0, 102.0, 98.0, 101.0, 103.0, 500.0, 12.0])
   print("Sample size:", len(s))
   print(f"Mean: {s.mean():.2f}")
   print(f"Median: {s.median():.2f}")

   Show outputGuess what it prints first

   Sample size: 8
   Mean: 140.12
   Median: 101.50

   Common mistake

   Relying strictly on sample mean to assess typical transaction sizes across data distributions with extreme tail values.

   How you will know

   You will see sample size 8, mean 140.12, and median 101.50 printed to standard output.

2. 02

   Demonstrate breakdown of naive 3-sigma standard deviation filter

   Wrong turn first: this step shows the mistake before the fix

   Here is the naive approach developers often attempt first: applying three standard deviations around the mean. Because variance squares every deviation, the single 500.0 outlier inflates the standard deviation to 147.24 rupees. This pushes the three-sigma acceptance interval out to between negative 301.60 and positive 581.85 rupees. Consequently, the three-sigma check flags zero outliers, completely missing both the 500.0 fraud charge and the 12.0 micro-charge. Parametric standard deviation is mathematically contaminated by the very outliers it is supposed to catch in production workloads.

   PythonStep 2Copy

   import pandas as pd
   
   s = pd.Series([100.0, 105.0, 102.0, 98.0, 101.0, 103.0, 500.0, 12.0])
   mean_val = s.mean()
   std_val = s.std()
   lower_3sigma = mean_val - 3 * std_val
   upper_3sigma = mean_val + 3 * std_val
   naive_outliers = s[(s < lower_3sigma) | (s > upper_3sigma)]
   print(f"Mean: {mean_val:.2f}, Std: {std_val:.2f}")
   print(f"3-Sigma Bounds: [{lower_3sigma:.2f}, {upper_3sigma:.2f}]")
   print("Detected 3-Sigma Outliers:", naive_outliers.tolist())

   Show outputGuess what it prints first

   Mean: 140.12, Std: 148.76
   3-Sigma Bounds: [-306.16, 586.41]
   Detected 3-Sigma Outliers: []

   Common mistake

   Applying parametric mean and standard deviation rules to skewed or contaminated operational transaction data.

   How you will know

   Observe that 3-sigma bounds expand so wide that Detected 3-Sigma Outliers returns an empty list.

3. 03

   Compute empirical Q1 and Q3 percentiles and calculate IQR

   To build a robust filter immune to extreme values, we evaluate empirical order percentiles using the Series quantile method. We calculate the 25th percentile (Q1) and the 75th percentile (Q3), which demarcate the middle fifty percent of transactions. Subtracting Q1 from Q3 yields the Interquartile Range of exactly 4.00 rupees. Because percentiles depend strictly on data ordering rather than squared magnitude deviations, extreme values in the upper and lower tails cannot distort the calculated spread of the distribution in analytical pipelines.

   PythonStep 3Copy

   import pandas as pd
   
   s = pd.Series([100.0, 105.0, 102.0, 98.0, 101.0, 103.0, 500.0, 12.0])
   q1 = s.quantile(0.25)
   q3 = s.quantile(0.75)
   iqr = q3 - q1
   print(f"Q1 (25th percentile): {q1:.2f}")
   print(f"Q3 (75th percentile): {q3:.2f}")
   print(f"IQR: {iqr:.2f}")

   Show outputGuess what it prints first

   Q1 (25th percentile): 99.50
   Q3 (75th percentile): 103.50
   IQR: 4.00

   Common mistake

   Passing integer percentages like 25 instead of decimal fractions like 0.25 to the quantile method.

   How you will know

   Output shows Q1 as 99.50, Q3 as 103.50, and calculated IQR as 4.00.

4. 04

   Establish non-parametric 1.5 * IQR lower and upper fences

   We calculate John Tukey non-parametric outlier fences by multiplying the IQR by 1.5 and offsetting from the respective quartiles. The lower fence is established at Q1 minus 1.5 times IQR (93.50 rupees), while the upper fence is established at Q3 plus 1.5 times IQR (109.50 rupees). Any transaction falling outside these boundaries is mathematically classified as an anomaly. These thresholds remain tightly calibrated to normal retail behavior because the underlying IQR was unaffected by the presence of extreme tail records across the transaction batch.

   PythonStep 4Copy

   import pandas as pd
   
   s = pd.Series([100.0, 105.0, 102.0, 98.0, 101.0, 103.0, 500.0, 12.0])
   q1 = s.quantile(0.25)
   q3 = s.quantile(0.75)
   iqr = q3 - q1
   lower_fence = q1 - 1.5 * iqr
   upper_fence = q3 + 1.5 * iqr
   print(f"Lower Fence: {lower_fence:.2f}")
   print(f"Upper Fence: {upper_fence:.2f}")

   Show outputGuess what it prints first

   Lower Fence: 93.50
   Upper Fence: 109.50

   Common mistake

   Adding 1.5 * IQR to Q1 instead of subtracting it when computing the lower anomaly fence.

   How you will know

   Output prints Lower Fence: 93.50 and Upper Fence: 109.50.

5. 05

   Filter and isolate anomalous transaction records

   In the final step, we isolate anomalies using vectorized boolean indexing. We combine conditions using the pipe operator: values strictly less than the lower fence or strictly greater than the upper fence. This filters the Series down to exactly two anomalous values: 500.0 and 12.0 rupees. We print the sorted list of detected outliers and the outlier count. The pipeline demonstrates robust, non-parametric detection that isolates both malicious high-value probes and automated micro-charge authorizations without generating false positives across everyday commercial activity.

   PythonStep 5Copy

   import pandas as pd
   
   s = pd.Series([100.0, 105.0, 102.0, 98.0, 101.0, 103.0, 500.0, 12.0])
   q1 = s.quantile(0.25)
   q3 = s.quantile(0.75)
   iqr = q3 - q1
   lower_fence = q1 - 1.5 * iqr
   upper_fence = q3 + 1.5 * iqr
   
   outliers = s[(s < lower_fence) | (s > upper_fence)]
   print("Detected Outliers:", sorted(outliers.tolist()))
   print("Outlier Count:", len(outliers))

   Show outputGuess what it prints first

   Detected Outliers: [12.0, 500.0]
   Outlier Count: 2

   Common mistake

   Using Python logical 'or' instead of the Pandas bitwise pipe operator '|' when combining boolean Series masks.

   How you will know

   Output prints Detected Outliers: [12.0, 500.0] and Outlier Count: 2.

Build it yourself

Write your solution below, run it in your browser, and pass every check.

What the result means

Interpretation

The non-parametric Interquartile Range filter successfully isolated both anomalies ([12.0, 500.0]) while maintaining strict acceptance boundaries for normal transactions between 93.50 and 109.50 rupees. By relying on order statistics rather than squared variance, the pipeline prevented sample contamination from distorting anomaly thresholds, disproving the false alarm that would have frozen 30 lakh rupees in merchant settlements.

The message you would send

Merchant batch risk analysis complete: the overnight average transaction spike was caused by two isolated card-testing probes (12.0 and 500.0 rupees) rather than merchant delinquency. Using non-parametric IQR fences, we isolated the bad transactions without impacting normal order processing. Pending merchant settlements of 30 lakh rupees have been cleared for release on schedule.

Put it on your CV

Resume or LinkedIn bullet

Built statistical anomaly detection engines in Pandas utilizing non-parametric IQR fences to isolate fraudulent card-testing spikes.

README for GitHub

# Statistical Outlier Detection & Truncation Report in Pandas

Engineered an automated non-parametric outlier detection engine in Pandas. Replaced fragile standard deviation checks with robust Interquartile Range (IQR) fences to detect fraudulent transaction spikes.

## Technical Details
- Engine: Python 3 / Pandas
- Algorithm: Non-parametric IQR: Fences = [Q1 - 1.5*IQR, Q3 + 1.5*IQR]
- Advantage: Immune to sample skewness and extreme tail outliers.

## Results
- Cleanly isolated card testing fraud transactions (500.0, 12.0) without false-positive merchant account freezes.

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