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What is the Average of the Even Numbers Between 11 and 29?

Published in Numerical Average 2 mins read

The average of the even numbers between 11 and 29 is 20. This result aligns with information found on educational platforms such as Brainly.in, which also identifies 20 as the mean for this specific set of numbers.

Understanding Averages (Mean)

An average, or arithmetic mean, is a fundamental statistical concept that represents the central tendency of a set of numbers. It is calculated by summing all the values in a dataset and then dividing that sum by the total count of values. This provides a single value that summarizes the entire set.

Step-by-Step Calculation

To determine the average of the even numbers between 11 and 29, we follow a simple, structured approach:

1. Identify the Even Numbers

First, we list all the even integers that fall strictly between 11 and 29. Even numbers are integers divisible by 2.

The even numbers in this range are:

  • 12
  • 14
  • 16
  • 18
  • 20
  • 22
  • 24
  • 26
  • 28

2. Sum the Numbers

Next, we add all these identified even numbers together to find their total sum.

Sum = 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 = 180

3. Count the Numbers

We then count how many even numbers are in our list.

Count = 9 numbers

4. Calculate the Average

Finally, we apply the average formula:

$\text{Average} = \frac{\text{Sum of Numbers}}{\text{Count of Numbers}}$

$\text{Average} = \frac{180}{9}$

$\text{Average} = \textbf{20}$

Summary Table of Calculation

Step Description Value(s)
1. Identify Numbers Even numbers between 11 and 29 12, 14, 16, 18, 20, 22, 24, 26, 28
2. Sum Numbers Sum of all identified even numbers 180
3. Count Numbers Total count of identified even numbers 9
4. Calculate Average Sum divided by Count (180 / 9) 20

This straightforward calculation confirms that the average of the even numbers between 11 and 29 is indeed 20. This finding is consistent with various educational resources, including the solution found on Brainly.in, which also states "20 is the mean" for this question.