Mean of Grouped Data — Direct Method

From Raw Data to Grouped Data

In Class 9 you found the mean, median and mode of ungrouped data (a plain list of numbers). Real surveys, though, usually come grouped into class intervals like $10\text{-}25$, $25\text{-}40$, and so on, with a frequency telling how many observations fall in each class.

To work with grouped data we replace every class by a single representative value — its class mark: $$x_i = \dfrac{\text{lower limit} + \text{upper limit}}{2}.$$

For example, the class $40\text{-}55$ has class mark $\dfrac{40+55}{2} = 47.5$.

Key Point: The class mark is the midpoint of the class. We treat every value in a class as if it were located at its class mark.

The Direct Method for the Mean

Once each class has a class mark $x_i$ and a frequency $f_i$, the mean of grouped data is the total of $f_i x_i$ divided by the total frequency: $$\boxed{\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}}$$

How to set it up: make a table with columns for the class, the frequency $f_i$, the class mark $x_i$, and the product $f_i x_i$. Add up the $f_i$ column (that is $n=\sum f_i$) and the $f_i x_i$ column, then divide.

Key Point: $\sum f_i = n$ is the total number of observations, and $\bar{x}=\dfrac{\sum f_i x_i}{n}$.

A Worked Table

Marks of 30 students:

Class	$f_i$	$x_i$	$f_i x_i$
10-25	2	17.5	35
25-40	3	32.5	97.5
40-55	7	47.5	332.5
55-70	6	62.5	375
70-85	6	77.5	465
85-100	6	92.5	555
Total	30		1860

$$\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{1860}{30} = 62.$$

[Board Important] So the mean mark is 62. We will reuse this same table for the mode and median later, so notice its shape.

Solved Examples

Example 1: Direct method

Find the mean of the distribution: classes $0\text{-}10, 10\text{-}20, 20\text{-}30, 30\text{-}40$ with frequencies $4, 6, 8, 2$.

Solution:

1. Class marks: $5, 15, 25, 35$.
2. $f_i x_i$: $20, 90, 200, 70$; $\sum f_i x_i = 380$; $\sum f_i = 20$.
3. $\bar{x} = \dfrac{380}{20} = 19$.

Final Answer: Mean $= 19$.

Takeaway: Class mark $=$ midpoint; multiply by frequency, add, divide.

Example 2: Class marks

Write the class marks of the classes $100\text{-}120, 120\text{-}140, 140\text{-}160$.

Solution:

1. $\dfrac{100+120}{2}=110$, $\dfrac{120+140}{2}=130$, $\dfrac{140+160}{2}=150$.

Final Answer: $110, 130, 150$.

Takeaway: The class marks of equal classes are themselves equally spaced (here by 20).

Example 3: Find a missing frequency

The mean of the following is 18. Find the missing frequency $f$: classes $11\text{-}13, 13\text{-}15, 15\text{-}17, 17\text{-}19, 19\text{-}21$ with frequencies $3, 6, 9, 13, f$; class marks $12,14,16,18,20$.

Solution:

1. $\sum f_i = 31 + f$; $\sum f_i x_i = 36 + 84 + 144 + 234 + 20f = 498 + 20f$.
2. $\bar{x}=18 \Rightarrow \dfrac{498 + 20f}{31 + f} = 18 \Rightarrow 498 + 20f = 558 + 18f$.
3. $2f = 60 \Rightarrow f = 30$.

Final Answer: $f = 30$.

Takeaway: Set $\dfrac{\sum f_ix_i}{\sum f_i}=\bar{x}$ and solve for the unknown frequency.

Example 4: Daily wages

Find the mean daily wage: $500\text{-}520, 520\text{-}540, 540\text{-}560, 560\text{-}580, 580\text{-}600$ with frequencies $12, 14, 8, 6, 10$.

Solution:

1. Class marks: $510, 530, 550, 570, 590$.
2. $f_i x_i$: $6120, 7420, 4400, 3420, 5900$; total $= 27260$; $\sum f_i = 50$.
3. $\bar{x} = \dfrac{27260}{50} = 545.2$.

Final Answer: Mean daily wage $=$ Rs 545.20.

Takeaway: The direct method works for any grouped data — the arithmetic is just larger here.