Mean = Σfx / Σf. Calculate midpoints, multiply by frequency, sum and divide. Class 10 method.

Find Mean of Grouped Data Using Direct Method — Step by Step

Question

The following table shows the number of plants in 40 houses in a locality. Find the mean number of plants per house.

Number of Plants	Number of Houses (Frequency)
0 – 2	1
2 – 4	2
4 – 6	1
6 – 8	5
8 – 10	6
10 – 12	2
12 – 14	3

Total houses = 20 (standard NCERT version uses 20 houses)

Solution — Step by Step

For any class interval, the midpoint is $x_i = \dfrac{\text{lower limit} + \text{upper limit}}{2}$ .

This midpoint represents the “average” value of all data points in that class — we don’t know individual values, so we assume everyone in the class is at the centre.

Class	Midpoint $x_i$	Frequency $f_i$
0–2	1	1
2–4	3	2
4–6	5	1
6–8	7	5
8–10	9	6
10–12	11	2
12–14	13	3

Multiply each frequency by its class midpoint. This gives the “total plants contributed” by each group.

Class	$x_i$	$f_i$	$f_i x_i$
0–2	1	1	1
2–4	3	2	6
4–6	5	1	5
6–8	7	5	35
8–10	9	6	54
10–12	11	2	22
12–14	13	3	39

\Sigma f_i = 1 + 2 + 1 + 5 + 6 + 2 + 3 = 20

\Sigma f_i x_i = 1 + 6 + 5 + 35 + 54 + 22 + 39 = 162

Always verify $\Sigma f_i$ matches the total number of observations given in the problem. If it doesn’t, you’ve made an arithmetic error — catch it here, not after the final answer.

\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{162}{20} = 8.1

Mean = 8.1 plants per house.

Why This Works

In raw data, mean = sum of all values ÷ count. Grouped data hides individual values inside classes, so we use midpoints as “stand-ins.” It’s an approximation — we assume data is uniformly spread within each class.

The formula $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$ is just a weighted average. A class with frequency 6 pulls the mean toward its midpoint six times harder than a class with frequency 1. That’s why the 8–10 class (frequency 6, midpoint 9) has strong influence on our answer of 8.1.

This method works cleanly when numbers are small. For messier data with large class midpoints (like ages 0–80 in 10-year groups), the Assumed Mean Method cuts down arithmetic — but the answer is identical.

Alternative Method

Shortcut check: Once you have $\Sigma f_i x_i = 162$ and $\Sigma f_i = 20$ , notice $162 \div 20 = 8.1$ directly. No need to simplify a fraction — just divide.

We could also use the Assumed Mean Method to cross-verify. Take assumed mean $A = 7$ (midpoint of the 6–8 class, roughly central):

Class	$x_i$	$d_i = x_i - 7$	$f_i$	$f_i d_i$
0–2	1	–6	1	–6
2–4	3	–4	2	–8
4–6	5	–2	1	–2
6–8	7	0	5	0
8–10	9	+2	6	+12
10–12	11	+4	2	+8
12–14	13	+6	3	+18

\Sigma f_i d_i = -6 - 8 - 2 + 0 + 12 + 8 + 18 = 22

\bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 7 + \frac{22}{20} = 7 + 1.1 = 8.1 \checkmark

Same answer. Use Assumed Mean when midpoints are large numbers (reduces calculation errors significantly).

Common Mistake

Using class limits instead of midpoints. A very common error: students write $f_i x_i$ using the lower class limit (0, 2, 4, 6…) instead of the midpoint (1, 3, 5, 7…).

The lower limit is just the starting boundary — it doesn’t represent the class. If you use lower limits, you’ll get $\Sigma f_i x_i = 128$ and mean $= 6.4$ , which is wrong. Always compute midpoint as $\dfrac{\text{lower} + \text{upper}}{2}$ first, before touching frequencies.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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