Ninth Grade Mathematics: Measures of Central Tendency: Arithmetic Mean
A measure of central tendency is a single representative value that describes the central location or typical magnitude of an entire dataset. The most widely used and mathematically versatile of these measures is the arithmetic mean—commonly referred to simply as the average. In this chapter, you will master the arithmetic mean for ungrouped data, raw frequency tables, and grouped continuous distributions using direct, assumed mean, and step-deviation methods.
The Arithmetic Mean of Ungrouped Raw Data
For an individual series of n raw numerical observations x_1, x_2, ..., x_n, the arithmetic mean (denoted by x-bar) is the sum of all values divided by the total number of observations:
Arithmetic Mean Formula (Raw Data):
x_1 + x_2 + ... + x_n Sigma x
x-bar = ----------------------- = -------
n n
Example: The heights of 5 players are 175 cm, 180 cm, 182 cm, 178 cm, 185 cm.
Sum = 175 + 180 + 182 + 178 + 185 = 900 cm.
Number of players n = 5.
Mean x-bar = 900 / 5 = 180 cm.
The arithmetic mean acts as the physical center of mass (the balance point) of the data. If the numbers were weights positioned along a ruler, the ruler would balance horizontally when supported precisely at the mean.
The Mean of Ungrouped Frequency Distributions
When values repeat with given frequencies f_1, f_2, ..., f_k, multiplying each value by its frequency computes the total sum rapidly:
Direct Method for Ungrouped Frequency Data:
f_1 * x_1 + f_2 * x_2 + ... + f_k * x_k Sigma (f * x)
x-bar = ----------------------------------------- = -------------
f_1 + f_2 + ... + f_k Sigma f
Calculation Table:
x f f * x
---------------------
10 4 40
20 6 120
30 10 300
40 5 200
---------------------
Total N = 25 Sigma(f*x) = 660
Mean x-bar = 660 / 25 = 26.4
The Mean of Grouped Continuous Distributions
In a continuous grouped distribution (such as 0-10, 10-20), individual values are lost within intervals. To compute the mean, we represent every value within an interval by its class mark (the interval midpoint x_i):
Class Mark x_i = (Lower Limit + Upper Limit) / 2
There are three algebraic methods to compute the mean of grouped data, each designed to minimize arithmetic labor:
Method 1: The Direct Method
Formula: x-bar = ( Sigma (f_i * x_i) ) / ( Sigma f_i )
Multiply each midpoint x_i by its frequency f_i, sum the products, and divide by the total frequency.
Method 2: The Assumed Mean Method (Shortcut Method)
When midpoints and frequencies are large numbers, multiplying them directly is time-consuming. We choose an arbitrary central midpoint as the Assumed Mean (A), and compute deviations d_i = x_i - A:
Assumed Mean Formula:
Sigma (f_i * d_i)
x-bar = A + ---------------------
Sigma f_i
where d_i = x_i - A
Subtracting A reduces large numbers to small, manageable positive and negative deviations.
Method 3: The Step-Deviation Method
If all class intervals have a uniform class width h, the deviations d_i = x_i - A are all divisible by h. We define the step-deviation u_i = (x_i - A) / h:
Step-Deviation Formula:
/ Sigma (f_i * u_i) x-bar = A + | ----------------- | * h
\ Sigma f_i /
where u_i = (x_i - A) / h
Step-deviation reduces numbers to single digits (-2, -1, 0, 1, 2), completely eliminating heavy arithmetic and virtually guaranteeing error-free calculations.
Chapter Exercises
Exercise 1: The mean of five numbers is 27. If one number is excluded, the mean of the remaining four numbers becomes 25. Find the value of the excluded number.
Exercise 2: Calculate the mean of the following grouped frequency distribution using the Direct Method: Class Intervals: 0 - 10 (f = 5), 10 - 20 (f = 8), 20 - 30 (f = 12), 30 - 40 (f = 10), 40 - 50 (f = 5).
Exercise 3: Compute the mean of the distribution in Exercise 2 using the Step-Deviation Method with assumed mean A = 25 and class size h = 10, verifying that both methods produce the exact same mean.
Solutions and Explanations
Solution 1: The mean of 5 numbers is 27, so the sum of all 5 numbers is Sum_5 = 5 * 27 = 135. When one number x is excluded, 4 numbers remain, and their mean is 25. The sum of the remaining 4 numbers is Sum_4 = 4 * 25 = 100. The excluded number is the difference between the two sums: x = Sum_5 - Sum_4 = 135 - 100 = 35. The excluded number is 35.
Solution 2: Find the class marks x_i for each interval: for 0-10, x_1 = 5; for 10-20, x_2 = 15; for 20-30, x_3 = 25; for 30-40, x_4 = 35; for 40-50, x_5 = 45. Multiply each midpoint by its frequency: f_1x_1 = 5 * 5 = 25; f_2x_2 = 8 * 15 = 120; f_3x_3 = 12 * 25 = 300; f_4x_4 = 10 * 35 = 350; f_5x_5 = 5 * 45 = 225. Sum of frequencies: Sigma f = 5 + 8 + 12 + 10 + 5 = 40. Sum of products: Sigma (fx) = 25 + 120 + 300 + 350 + 225 = 1,020. Apply the Direct Method formula: x-bar = Sigma (f*x) / Sigma f = 1,020 / 40 = 25.5.
Solution 3: Use the Step-Deviation Method with assumed mean A = 25 and class size h = 10. Compute u_i = (x_i - 25) / 10 for each class: for x_1 = 5, u_1 = (5 - 25)/10 = -2, and f_1u_1 = 5 * (-2) = -10. For x_2 = 15, u_2 = (15 - 25)/10 = -1, and f_2u_2 = 8 * (-1) = -8. For x_3 = 25, u_3 = (25 - 25)/10 = 0, and f_3u_3 = 12 * 0 = 0. For x_4 = 35, u_4 = (35 - 25)/10 = 1, and f_4u_4 = 10 * 1 = 10. For x_5 = 45, u_5 = (45 - 25)/10 = 2, and f_5u_5 = 5 * 2 = 10. Sum of frequencies: Sigma f = 40. Sum of products: Sigma (fu) = -10 - 8 + 0 + 10 + 10 = 2. Apply the Step-Deviation formula: x-bar = A + [ Sigma (f*u) / Sigma f ] * h = 25 + [ 2 / 40 ] * 10 = 25 + (20 / 40) = 25 + 0.5 = 25.5. Both methods produce the exact same mean of 25.5.