Dispersion Measures — How Spread Out Is Your Data?

import pandas as pd

# Delivery times in hours for 10 orders this week
delivery_df = pd.DataFrame({
    'order_id':        [4001, 4002, 4003, 4004, 4005, 4006, 4007, 4008, 4009, 4010],
    'customer':        ['Alice', 'Bob', 'Carol', 'Dave', 'Eve', 'Frank', 'Grace', 'Hank', 'Ivy', 'Jack'],
    'delivery_hours':  [18, 22, 19, 48, 21, 20, 23, 19, 72, 21]  # most are 18-23hrs, two outliers
})

# .max() returns the largest value in the column
# .min() returns the smallest value in the column
max_time = delivery_df['delivery_hours'].max()
min_time = delivery_df['delivery_hours'].min()

# Range = max - min — the full span of delivery times
range_val = max_time - min_time

print(f"Fastest delivery : {min_time} hours")
print(f"Slowest delivery : {max_time} hours")
print(f"Range            : {range_val} hours")
print(f"\nMean delivery    : {delivery_df['delivery_hours'].mean():.1f} hours")
print(f"Median delivery  : {delivery_df['delivery_hours'].median()} hours")

import pandas as pd

# Weekly deals closed by two sales reps over 8 weeks
sales_df = pd.DataFrame({
    'week':       [1, 2, 3, 4, 5, 6, 7, 8],
    'rep_alice':  [8, 7, 9, 8, 8, 9, 7, 8],     # very consistent — always 7-9
    'rep_bob':    [2, 15, 3, 14, 1, 16, 4, 13]   # feast or famine — wild swings
})

# .mean() — confirm both reps have the same average
print(f"Alice's mean deals/week: {sales_df['rep_alice'].mean()}")
print(f"Bob's mean deals/week  : {sales_df['rep_bob'].mean()}")

# .var() calculates variance — average of squared deviations from the mean
# ddof=1 means we use sample variance (divides by n-1, not n) — standard for real data
alice_var = sales_df['rep_alice'].var()
bob_var   = sales_df['rep_bob'].var()

print(f"\nAlice's variance: {alice_var:.2f} deals²")
print(f"Bob's variance  : {bob_var:.2f} deals²")
print(f"\nBob's variance is {bob_var/alice_var:.0f}x higher than Alice's")

import pandas as pd
import numpy as np

# Weight of 12 metal parts stamped by the machine (should be 500g each)
parts_df = pd.DataFrame({
    'part_id': range(1, 13),
    'weight_g': [499.2, 500.8, 498.7, 501.3, 499.9, 500.4,
                 500.1, 499.5, 501.0, 498.9, 500.6, 499.8]
})

# .std() calculates standard deviation — square root of variance
# This is in the same units as the data (grams), which makes it interpretable
std_dev = parts_df['weight_g'].std()
mean_wt = parts_df['weight_g'].mean()

print(f"Target weight   : 500.0g")
print(f"Mean weight     : {mean_wt:.2f}g")
print(f"Std deviation   : {std_dev:.2f}g")

# The 68% rule: most parts should fall within mean ± 1 std dev
lower_1std = mean_wt - std_dev
upper_1std = mean_wt + std_dev

print(f"\n1 std dev range : {lower_1std:.2f}g to {upper_1std:.2f}g")

# Count how many parts actually fall within that range
within_1std = parts_df[(parts_df['weight_g'] >= lower_1std) &
                        (parts_df['weight_g'] <= upper_1std)].shape[0]

total_parts = len(parts_df)
print(f"Parts within range: {within_1std} out of {total_parts} ({within_1std/total_parts*100:.0f}%)")

import pandas as pd
import numpy as np

# Customer wait times in minutes — most are normal, a few are long outliers
wait_df = pd.DataFrame({
    'customer_id': range(601, 621),   # 20 customers
    'wait_mins':   [4, 6, 5, 7, 3, 8, 5, 6, 4, 7,
                    5, 6, 45, 3, 8, 4, 7, 52, 5, 6]  # 45 and 52 are outliers
})

# .quantile() gives you any percentile — 0.25 = 25th, 0.75 = 75th
q1  = wait_df['wait_mins'].quantile(0.25)   # 25th percentile (lower quartile)
q3  = wait_df['wait_mins'].quantile(0.75)   # 75th percentile (upper quartile)
iqr = q3 - q1                               # IQR = Q3 - Q1

# Compare IQR vs standard deviation — see how outliers affect each differently
std_dev = wait_df['wait_mins'].std()
mean_wt = wait_df['wait_mins'].mean()

print(f"Q1 (25th percentile): {q1} mins")
print(f"Q3 (75th percentile): {q3} mins")
print(f"IQR                 : {iqr} mins  ← middle 50% of customers")
print(f"\nMean wait           : {mean_wt:.1f} mins  ← pulled up by outliers")
print(f"Std deviation       : {std_dev:.1f} mins  ← also inflated by outliers")
print(f"Median wait         : {wait_df['wait_mins'].median()} mins  ← honest middle")

import pandas as pd
import numpy as np

# Test scores for two schools — same mean, very different distributions
scores_df = pd.DataFrame({
    'school_a': [62, 65, 68, 70, 71, 72, 73, 74, 76, 79],   # consistent performers
    'school_b': [30, 40, 55, 68, 72, 74, 80, 88, 92, 96]    # very wide spread
})

# Build a complete dispersion summary for both schools
def dispersion_summary(series, label):
    """Returns a dict of all key dispersion measures for a given series."""
    q1  = series.quantile(0.25)
    q3  = series.quantile(0.75)
    return {
        'Mean'    : round(series.mean(), 1),
        'Median'  : series.median(),
        'Std Dev' : round(series.std(), 1),
        'Variance': round(series.var(), 1),
        'Range'   : series.max() - series.min(),
        'IQR'     : round(q3 - q1, 1),
        'Min'     : series.min(),
        'Max'     : series.max()
    }

# Generate summary for each school and display side by side
summary_a = dispersion_summary(scores_df['school_a'], 'School A')
summary_b = dispersion_summary(scores_df['school_b'], 'School B')

comparison = pd.DataFrame({'School A': summary_a, 'School B': summary_b})
print(comparison)