Each class interval is represented by its class mark (midpoint), and frequency indicates how many observations fall in that interval.
Chapter 13 · CBSE Class X
📘 Definition
🎨 SVG Diagram
Concept Visualization
📌 Note
Methods to Find Mean
- Direct Method
- Assumed Mean Method
- Step Deviation Method
📌 Note
Direct Method
\[
\overline{x} = \frac{\sum f_i x_i}{\sum f_i}
\]
This method directly uses class marks and frequencies. It is best suited when values are small and calculations are manageable.
- Find class marks \(x_i\)
- Compute \(f_i x_i\)
- Apply formula
📌 Note
Assumed Mean Method
\[
\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad d_i = x_i - a
\]
A suitable assumed mean \(a\) is taken to simplify calculations by reducing large numbers.
- Choose \(a\) near central class
- Compute deviations \(d_i\)
- Multiply with frequencies
📌 Note
Step Deviation Method
\[
\begin{aligned}
\overline{x} &= a + h \left(\frac{\sum f_i u_i}{\sum f_i}\right), \\\\ u_i &= \frac{x_i - a}{h}
\end{aligned}
\]
This method is highly efficient when class intervals are equal and values are large.
- Find class width \(h\)
- Compute \(u_i\)
- Apply simplified formula
✏️ Example
Find the mean of the following grouped data.
| Class Interval | Frequency |
|---|---|
| 10–20 | 5 |
| 20–30 | 9 |
| 30–40 | 14 |
| 40–50 | 8 |
Use Direct Method
Step by step Solution
- \[x_i = 15, 25, 35, 45\]
- \[\begin{aligned} \sum f_i x_i &= 5(15) + 9(25) + 14(35) + 8(45) \\&= 1150 \end{aligned}\]
- \[\begin{aligned}\sum f_i &=5+9+14+8\\&= 36\end{aligned}\]
- \[\begin{aligned} \overline{x} &= \frac{1150}{36} \\&\approx 31.94 \end{aligned}\]
📐 Derivation
The grouped mean formula originates from the basic definition:
\[ \overline{x} = \frac{\text{Sum of observations}}{\text{Number of observations}} \]Since exact values are unknown, each class is represented by its midpoint, leading to:
\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
⚡ Exam Tip
⚠️ Warning
Common Mistakes
📋 Case Study
A survey records marks of students in grouped form. The school wants to analyze overall performance efficiently.
Question: Which method is most efficient and why?
Answer Insight:
Step deviation method is most efficient because it simplifies calculations by scaling deviations, reducing computational complexity in large datasets.