Find the mean marks of the students using assumed mean method.

| Class Interval | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| Frequency | 3 | 2 | 5 | 6 | 4 |

Question

Find the mean marks of the students using assumed mean method.

| Class Interval | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
|---|---|---|---|---|---|
| Frequency | 3 | 2 | 5 | 6 | 4 |

Accepted Answer

> [!TLDR] The assumed-mean method takes a convenient class mark $A$ as a reference, computes deviations $d_i=x_i-A$, and corrects $A$ by the mean deviation.
>
> [!FORMULA] $\bar{x}=A+\dfrac{\sum f_id_i}{\sum f_i}, \quad d_i=x_i-A$
> where $A$ = assumed mean (a chosen class mark), $x_i$ = class mark of the $i$-th class, $f_i$ = frequency.

1. **Tabulate class marks:**

| Class Interval | $f_i$ | $x_i$ |
   |---|---|---|
   | 0–20 | 3 | 10 |
   | 20–40 | 2 | 30 |
   | 40–60 | 5 | 50 |
   | 60–80 | 6 | 70 |
   | 80–100 | 4 | 90 |

2. **Choose the assumed mean** as the class mark of the middle class: $A=50$.

3. **Compute deviations $d_i=x_i-A$ and $f_id_i$:**

| Class Interval | $f_i$ | $x_i$ | $d_i=x_i-50$ | $f_id_i$ |
   |---|---|---|---|---|
   | 0–20 | 3 | 10 | −40 | −120 |
   | 20–40 | 2 | 30 | −20 | −40 |
   | 40–60 | 5 | 50 | 0 | 0 | …

Concept understanding — Mean

What Does "Mean" Actually Mean?

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