Average (Mean) of Data

Mathematical Foundation

The mean, or arithmetic average, is a fundamental measure of central tendency that represents the typical value in a dataset. It provides a single representative value that summarizes the entire collection of data points by distributing the total sum equally across all observations. The mean is sensitive to extreme values and serves as the foundation for many statistical calculations and data analysis techniques.

Arithmetic Mean Formula:
  • For individual data values: $\overline{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{x_1 + x_2 + \cdots + x_n}{n}$
  • For grouped data: $\overline{x} = \frac{\sum (f_i \cdot x_i)}{\sum f_i}$ where $f_i$ is frequency
  • For data with weights: $\overline{x} = \frac{\sum (w_i \cdot x_i)}{\sum w_i}$ where $w_i$ is weight
Where:
  • $\overline{x}$ = sample mean
  • $n$ = number of data points
  • $x_i$ = individual data values
  • $\sum$ = summation symbol

Interactive Mean Calculator

Data Visualization

6.83
Mean
41
Total Sum
6
Data Points
4.07
Variance

Worked Example 1: Calculating Mean of Test Scores

A student scored 85, 92, 78, 96, and 88 on five math tests. Find the mean score.

Solution:

Sum = 85 + 92 + 78 + 96 + 88 = 439

Mean = 439 ÷ 5 = 87.8

The student's average test score is 87.8

Worked Example 2: Mean with Frequencies (Grouped Data)

The table shows the number of hours students spent studying:

Hours Frequency
2 5
4 8
6 6
8 3

Solution:

Mean = (2×5 + 4×8 + 6×6 + 8×3) ÷ (5+8+6+3) = (10 + 32 + 36 + 24) ÷ 22 = 102 ÷ 22 = 4.64

The average study time is 4.64 hours

Worked Example 3: Weighted Mean

A student's grades are: Quiz (20% weight): 85, Midterm (30%): 78, Final (50%): 92. Find the weighted mean.

Solution:

Weighted Mean = (85×0.2 + 78×0.3 + 92×0.5) ÷ (0.2 + 0.3 + 0.5) = (17 + 23.4 + 46) ÷ 1 = 86.4

The weighted average grade is 86.4

Properties of the Mean
Comparison with Other Measures

Mean (Average)

  • Uses all data points
  • Affected by outliers
  • Mathematically useful
  • Balance point of data

Median

  • Middle value when ordered
  • Resistant to outliers
  • Position-based measure
  • Not affected by extreme values

Mode

  • Most frequent value
  • Can have multiple modes
  • Resistant to outliers
  • May not exist

Practice Problems

Problem 1: Find the mean of: 12, 15, 18, 22, 25, 30
Solution:
Sum = 12 + 15 + 18 + 22 + 25 + 30 = 122
Mean = 122 ÷ 6 = 20.33
Problem 2: The mean of 8 numbers is 25. If one number is removed, the mean becomes 22. Find the removed number.
Solution:
Original sum = 8 × 25 = 200
New sum = 7 × 22 = 154
Removed number = 200 - 154 = 46
Problem 3: Calculate the mean of the following frequency distribution:
Score Frequency
60 3
70 5
80 4
90 2

Solution:
Sum = (60×3) + (70×5) + (80×4) + (90×2) = 180 + 350 + 320 + 180 = 1030
Total frequency = 3 + 5 + 4 + 2 = 14
Mean = 1030 ÷ 14 = 73.57
Problem 4: The mean of three numbers is 15. When a fourth number is added, the mean becomes 17. Find the fourth number.
Solution:
Original sum = 3 × 15 = 45
New sum = 4 × 17 = 68
Fourth number = 68 - 45 = 23
Problem 5: Find the missing value if the mean of 5, 8, 12, x, 15 is 10.
Solution:
Sum = 5 + 8 + 12 + x + 15 = 40 + x
Mean = (40 + x) ÷ 5 = 10
40 + x = 50
x = 10

Key Takeaways