Mean of grouped data

A class interval such as $10$–$20$ tells you how many observations fall between $10$ and $20$, but not their individual values. The standard convention is to assume every observation in that class behaves like the interval's midpoint, the class mark:

where: the lower and upper limits are the boundaries of class $i$, so for $10$–$20$ the class mark is $x_i = (10+20)/2 = 15$. Once every class has a single representative value $x_i$ and a frequency $f_i$ (the count of observations in that class), the mean is found the same way as an ordinary weighted average — each class mark contributes to the total in proportion to how many observations sit in it.

Direct method. Multiply each class mark by its frequency, add these products, and divide by the total frequency:

where: $f_i$ is the frequency of class $i$, $x_i$ is its class mark, $\sum f_i$ is the total number of observations, and $\sum f_ix_i$ is the sum of every class mark weighted by its frequency.

Assumed-mean method. Pick a convenient class mark as a reference point, an assumed mean $a$ (usually a round number near the middle of the data), and work with deviations $d_i = x_i - a$ instead of raw class marks:

where: $d_i = x_i - a$ is how far class mark $i$ sits from the assumed mean, so deviations for classes below $a$ are negative and those above are positive.

Step-deviation method. When every class has the same width $h$, divide the deviations by $h$ to get smaller whole-number values $u_i = (x_i - a)/h$:

where: $h$ is the common class width and $u_i$ is the step-deviation of class $i$, an integer such as $-2, -1, 0, 1, 2$ when classes are evenly spaced around $a$. The figure below shows the same five-class distribution processed by all three methods side by side, landing on one shared mean.

This grounded derivation shows that the assumed-mean and step-deviation formulas are not new rules to memorise — they are the direct-method formula rearranged for smaller arithmetic, so if two of your three answers disagree on the same data, the error is arithmetic, not conceptual.

Step 1. Start from the direct-method definition, $\bar{x} = \dfrac{\sum f_ix_i}{\sum f_i}$, and rewrite each class mark as the assumed mean plus its deviation, $x_i = a + d_i$. Substituting gives $\bar{x} = \dfrac{\sum f_i(a + d_i)}{\sum f_i}$.

Step 2. Split the sum in the numerator, since $a$ is a constant that multiplies every $f_i$: $\sum f_i(a+d_i) = a\sum f_i + \sum f_id_i$. Dividing through by $\sum f_i$ gives $\bar{x} = a + \dfrac{\sum f_id_i}{\sum f_i}$ — exactly the assumed-mean formula. Nothing new was assumed; this is the direct method with $x_i$ replaced by $a + d_i$.

Step 3. Now suppose every class has the same width $h$, so the deviations are evenly spaced multiples of $h$: $d_i = h\,u_i$ for some integer $u_i$. Substitute into the assumed-mean result: $\bar{x} = a + \dfrac{\sum f_i(hu_i)}{\sum f_i} = a + h\left(\dfrac{\sum f_iu_i}{\sum f_i}\right)$, since $h$ is a constant and factors out of the sum. This is the step-deviation formula.

Step 4. Because each formula was obtained from the previous one by pure algebraic substitution — no approximation, no rounding — all three must give identical results on the same data, provided the arithmetic is carried out correctly. The only thing that changes between methods is the size of the numbers you add: raw class marks (can be large), deviations from a central value (smaller), or whole-number step-deviations (smallest, and easiest to sum by hand).

Find the mean of the distribution below using all three methods, taking assumed mean $a = 25$, and confirm they agree.

Step 1. Find the class marks: $x_i = 5, 15, 25, 35, 45$, and note the class width is $h = 10$ throughout.

Step 2 — direct method. Compute $f_ix_i$: $5(5)=25$, $8(15)=120$, $15(25)=375$, $16(35)=560$, $6(45)=270$. Totals: $\sum f_i = 5+8+15+16+6 = 50$ and $\sum f_ix_i = 25+120+375+560+270 = 1350$. So $\bar{x} = 1350/50 = 27$.

Step 3 — assumed-mean method. With $a = 25$, deviations are $d_i = x_i - 25$: that is $-20, -10, 0, 10, 20$. Then $f_id_i$: $5(-20)=-100$, $8(-10)=-80$, $15(0)=0$, $16(10)=160$, $6(20)=120$. Summing, $\sum f_id_i = -100-80+0+160+120 = 100$. So $\bar{x} = 25 + 100/50 = 25 + 2 = 27$.

Step 4 — step-deviation method. Divide each deviation by $h=10$: $u_i = -2, -1, 0, 1, 2$. Then $f_iu_i$: $5(-2)=-10$, $8(-1)=-8$, $15(0)=0$, $16(1)=16$, $6(2)=12$. Summing, $\sum f_iu_i = -10-8+0+16+12 = 10$. So $\bar{x} = 25 + 10(10/50) = 25 + 2 = 27$. All three methods land on 27, exactly as the algebra of the previous section guarantees.

Competitive stretch

For the distribution with $a=25$ and $\sum f_i = 50$, if $\sum f_iu_i$ had instead come out to $-3$ (as in a differently-weighted version of this table), the mean would shift to $\bar{x} = 25 + 10(-3/50) = 25 - 0.6 = 24.4$ — a reminder that the step-deviation total can be negative when more weight sits in classes below the assumed mean, and the final mean can land below $a$ even though $a$ was chosen "near the middle."

• Using a class boundary instead of the class mark. The formula needs $x_i$, the midpoint, never the lower or upper limit of the interval.
• Applying step-deviation with unequal class widths. $u_i = (x_i-a)/h$ only makes sense when every class shares one common width $h$; with unequal widths, use the direct or assumed-mean method instead.
• Dropping a sign while summing $f_id_i$ or $f_iu_i$. Deviations below the assumed mean are negative — losing one minus sign shifts the whole mean.
• Picking an assumed mean far from the data's centre. The method still gives the correct answer, but $d_i$ or $u_i$ become large and the arithmetic advantage is lost.

Ask the tutor: Want to try the step-deviation method on a distribution with six classes and check whether your answer matches the direct method?