Ninth Grade Mathematics: Measures of Dispersion: Range & Deviations
Measures of central tendency (mean, median, and mode) describe the center of a distribution, but they tell us nothing about how the data is scattered around that center. Two completely different datasets can have the exact same mean yet exhibit vastly different levels of stability or volatility. In this chapter, you will master measures of dispersion—exploring the range, mean deviation, and the conceptual foundation of variance and standard deviation.
Why Central Tendency Alone is Insufficient
Consider the test scores of two students across five exams:
Student A: 70, 70, 70, 70, 70 ===> Mean = 70
Student B: 40, 55, 70, 85, 100 ===> Mean = 70
Both students have an identical mean score of 70. However, Student A is perfectly consistent, while Student B exhibits wide, volatile swings in performance. To distinguish between these two scenarios, statisticians use measures of dispersion (or measures of spread).
Measures of Dispersion
- Low Dispersion: Data values cluster tightly around the mean (Consistent, Predictable)
- High Dispersion: Data values scatter widely away from the mean (Volatile, Variable)
The Range: The Simplest Measure of Spread
The range is the simplest measure of dispersion, defined as the difference between the maximum and minimum observations in a dataset:
Range Formula:
Range = Maximum Value - Minimum Value = X_max - X_min
Coefficient of Range (Relative Dispersion):
X_max - X_min
Coeff of Range = -------------------
X_max + X_min
For Student A: Range = 70 - 70 = 0
For Student B: Range = 100 - 40 = 60
While the range is simple to compute, it depends entirely on the two extreme values and tells us nothing about how the remaining intermediate data points are distributed.
Mean Deviation (Mean Absolute Deviation)
To measure how far all data points deviate from the center, we compute deviations from the mean: d_i = x_i - x-bar. However, a fundamental mathematical property of the arithmetic mean is that the sum of signed deviations is always zero:
Algebraic Invariance:
Sigma (x_i - x-bar) = 0 (Positive and negative deviations cancel completely!)
To prevent positive and negative deviations from cancelling each other out, we take the absolute value of each deviation:
Mean Absolute Deviation (MAD) about the Mean:
Sigma |x_i - x-bar|
MAD(x-bar) = -------------------
n
Example: Compute MAD for Student B (40, 55, 70, 85, 100 with Mean = 70):
|40 - 70| = |-30| = 30
|55 - 70| = |-15| = 15
|70 - 70| = |0| = 0
|85 - 70| = |15| = 15
|100 - 70| = |30| = 30
Sum of absolute deviations = 30 + 15 + 0 + 15 + 30 = 90.
MAD = 90 / 5 = 18.
On average, Student B's scores deviate by 18 points from the mean.
Introduction to Variance and Standard Deviation
An alternative mathematical way to eliminate negative signs without using absolute value bars is to square the deviations:
The Variance (sigma^2):
The arithmetic mean of the squared deviations from the mean:
Sigma (x_i - x-bar)^2
sigma^2 = ---------------------
n
Because squaring numbers changes the physical unit (e.g., turning centimeters into square centimeters), we take the positive square root of variance to return to the original physical units:
The Standard Deviation (sigma):
/ Sigma (x_i - x-bar)^2 sigma = | ----------------------- |
\ n /
The standard deviation is the universal gold standard of statistical dispersion used in physics, quality control, finance, and machine learning.
Chapter Exercises
Exercise 1: Find the range and coefficient of range for the following set of temperatures (in degrees Celsius): 14, 22, 19, 28, 35, 12, 25, 30.
Exercise 2: Prove algebraically that the sum of the signed deviations of any set of n observations from their arithmetic mean is always equal to zero: Sigma (x_i - x-bar) = 0.
Exercise 3: Compute the Mean Absolute Deviation (MAD) about the mean for the dataset: 6, 8, 10, 12, 14.
Solutions and Explanations
Solution 1: Identify the minimum and maximum values in the temperature dataset: X_min = 12 deg C, and X_max = 35 deg C. The range is Range = X_max - X_min = 35 - 12 = 23 degrees Celsius. The coefficient of range is (X_max - X_min) / (X_max + X_min) = (35 - 12) / (35 + 12) = 23 / 47 approx 0.489.
Solution 2: Expand the summation of signed deviations across all n observations: Sigma (x_i - x-bar) = (x_1 - x-bar) + (x_2 - x-bar) + ... + (x_n - x-bar). Group all data values together and all mean terms together: Sigma (x_i - x-bar) = (x_1 + x_2 + ... + x_n) - (x-bar + x-bar + ... + x-bar). The first group is Sigma x_i. In the second group, x-bar is added n times, giving n * x-bar. By definition of the arithmetic mean, x-bar = (Sigma x_i) / n, which means Sigma x_i = n * x-bar. Substitute this into the expansion: Sigma (x_i - x-bar) = (n * x-bar) - (n * x-bar) = 0. The sum of signed deviations from the mean is identically zero.
Solution 3: First compute the arithmetic mean x-bar for the dataset 6, 8, 10, 12, 14: n = 5; Sum = 6 + 8 + 10 + 12 + 14 = 50. Mean x-bar = 50 / 5 = 10. Next, compute the absolute deviation |x_i - x-bar| for each value: |6 - 10| = |-4| = 4; |8 - 10| = |-2| = 2; |10 - 10| = |0| = 0; |12 - 10| = |2| = 2; |14 - 10| = |4| = 4. Sum the absolute deviations: 4 + 2 + 0 + 2 + 4 = 12. Compute the Mean Absolute Deviation: MAD = 12 / 5 = 2.4. On average, the data points deviate from the mean by 2.4 units.