Mean of Grouped Data for CBSE Class 10 Mathematics

In everyday language, we often talk about averages: the average rainfall during a monsoon season, the average test score of a classroom, or the average lifespan of an electric vehicle battery. In statistics, the most widely used measure of central tendency representing the central balance point of numerical data is the Arithmetic Mean (or simply the Mean, denoted by $\bar{x}$).

In Class 9, you computed the mean of raw ungrouped numbers. However, real-world data—such as census populations, industrial wages, and healthcare metrics—is organized into grouped frequency distributions. In CBSE Class 10 Mathematics, Chapter 13 (Statistics) provides three distinct mathematical techniques to compute the mean: the Direct Method, the Assumed Mean Method, and the Step-Deviation Method.

What You Will Learn

• Key terms: Class intervals, class limits, class size ($h$), and class marks ($x_i$)
• Method 1: The Direct Method ($\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$)
• Method 2: The Assumed Mean Method ($\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$)
• Method 3: The Step-Deviation Method ($\bar{x} = a + [\frac{\sum f_i u_i}{\sum f_i}] \times h$)
• When to use which method to minimize calculation time
• How to convert discontinuous class intervals into continuous intervals
• Solving for a missing frequency ($f$) when the mean is given
• Board exam presentation templates and common arithmetic traps

1. Class Intervals, Class Marks, and Definitions

A grouped continuous frequency distribution groups data into ranges called class intervals (e.g., $10 - 25, 25 - 40, \dots$):

1. Lower Class Limit & Upper Class Limit: In the interval $10 - 25$, $10$ is the lower limit and $25$ is the upper limit.
2. Class Size ($h$): The difference between the upper and lower limits of a class interval: $\mathbf{h = \text{Upper Limit} - \text{Lower Limit}}$ (For $10 - 25$, $h = 25 - 10 = 15$).
3. Class Mark ($x_i$): The midpoint or central representative value of a class interval: $\mathbf{x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2}}$ (For $10 - 25$, $x_i = \frac{10 + 25}{2} = 17.5$).

2. The Three Methods to Calculate the Mean

                            Methods to Calculate Mean
                                       |
       +-------------------------------+-------------------------------+
       |                               |                               |
Direct Method                  Assumed Mean Method             Step-Deviation Method
x̄ = (Σ fi xi) / (Σ fi)         x̄ = a + (Σ fi di) / (Σ fi)      x̄ = a + [(Σ fi ui) / (Σ fi)] × h
(Best for small numbers)       (Subtracts assumed mean a)      (Divides by class size h)

Method 1: The Direct Method

When the numerical values of frequencies ($f_i$) and class marks ($x_i$) are small integers:

$\mathbf{\bar{x} = \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}}$

• Multiply each class mark $x_i$ by its corresponding frequency $f_i$.
• Sum all the products ($\sum f_i x_i$).
• Divide by the total frequency ($N = \sum f_i$).

Method 2: The Assumed Mean Method

When $x_i$ and $f_i$ are large numbers, multiplying $f_i x_i$ directly becomes tedious and prone to arithmetic mistakes:

1. Choose an arbitrary central class mark as the Assumed Mean ($a$) (usually located in the middle row of the $x_i$ column).
2. Calculate the deviation ($d_i$) of each class mark from $a$: $\mathbf{d_i = x_i - a}$
3. Multiply each frequency $f_i$ by its deviation $d_i$ and sum them ($\sum f_i d_i$).
4. Compute the mean using: $\mathbf{\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}}$

Method 3: The Step-Deviation Method

When the class size $h$ is uniform across all intervals, the deviations $d_i$ share a common factor $h$. We can reduce the numbers even further:

1. Define the reduced step-deviation variable $u_i$: $\mathbf{u_i = \frac{x_i - a}{h} = \frac{d_i}{h}}$
2. The values of $u_i$ become tiny integers: $\dots, -2, -1, 0, +1, +2, \dots$!
3. Multiply $f_i$ by $u_i$ and sum them ($\sum f_i u_i$).
4. Compute the mean using: $\mathbf{\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h}$

Important: <u>The Step-Deviation Method is by far the fastest and most error-free method for board exams when class intervals are equal! It reduces large three-digit numbers to tiny single-digit integers $(-2, -1, 0, 1, 2)$, eliminating multi-digit multiplication entirely!</u>

3. Continuous vs. Discontinuous Class Intervals

All statistical formulas in Class 10 strictly require continuous class intervals (where the upper limit of one class equals the lower limit of the next class, e.g., $10-20, 20-30$).

• If data is given in discontinuous form (e.g., $11-20, 21-30, 31-40$):
  • Find the gap: $21 - 20 = 1$.
  • Half the gap: $1 / 2 = 0.5$.
  • Subtract $0.5$ from each lower limit, and add $0.5$ to each upper limit: $10.5 - 20.5, \quad 20.5 - 30.5, \quad 30.5 - 40.5$

4. Solved CBSE Board Examination Problems

Solved Example 1: Calculating Mean Using Step-Deviation (NCERT Classic)

Problem: Find the mean of the following distribution of daily wages of 50 workers:

Solution:

1. Class size $h = 120 - 100 = 20$.
2. Choose Assumed Mean from center of $x_i$: Let $a = 150$.

3. Calculate the Mean: $\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$ $\bar{x} = 150 + \left( \frac{-12}{50} \right) \times 20$ $\bar{x} = 150 - \frac{240}{50} = 150 - 4.8 = \mathbf{₹\,145.20}$
4. Therefore, <u>the mean daily wage of the workers is ₹$145.20$</u>.

Solved Example 2: Finding a Missing Frequency ($f$) (CBSE Classic)

Problem: The mean of the following distribution is $18$. Find the missing frequency $f$:

Solution:

1. Let Assumed Mean $a = 18$. Class size $h = 2$.

2. Apply Assumed Mean Formula: $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$ $18 = 18 + \frac{2f - 40}{44 + f}$ $18 - 18 = \frac{2f - 40}{44 + f} \implies 0 = \frac{2f - 40}{44 + f}$ $2f - 40 = 0 \implies 2f = 40 \implies \mathbf{f = 20}$
3. Therefore, <u>the missing frequency $f$ is $20$</u>.

5. Summary and Examination Tips

Exam Tip: In missing frequency questions where the given mean matches a class mark (e.g., Mean $= 18$ and one of the class marks is $18$), ALWAYS choose $a = \text{Mean}$! As seen in Solved Example 2, this makes $(18 - 18 = 0)$, collapsing the fraction immediately to $2f - 40 = 0$!

Common Mistake: Forgetting to multiply by $h$ at the end of the Step-Deviation formula. If you divide by $h$ to get $u_i$, you MUST multiply by $h$ when reconstructing the mean!