Mean of grouped data

The mean (arithmetic average) of a list of numbers is the sum divided by the count. For grouped data, we treat each class as a single value (the class mark $x_i$) with the corresponding frequency $f_i$. There are three methods to compute the mean. All three give the same answer; the choice depends on how convenient the arithmetic is.

Method 1: Direct method

Multiply each class mark by its frequency, sum, and divide by the total frequency:

$\bar x = \frac{\sum f_i x_i}{\sum f_i}.$

This is the most straightforward method but can involve large multiplications if the class marks are big.

Example.

$\bar x = 670/28 = 23.93$.

Method 2: Assumed-mean method

Pick a convenient class mark $a$ (the "assumed mean"). Compute deviations $d_i = x_i - a$. Then

$\bar x = a + \frac{\sum f_i d_i}{\sum f_i}.$

The advantage: the deviations $d_i$ are smaller than the original class marks, so the arithmetic is easier.

Example (same data, $a = 25$).

$\bar x = 25 + (-30/28) = 25 - 1.07 = 23.93$. (Matches Method 1.)

Method 3: Step-deviation method

When the class size $h$ is constant, take $u_i = (x_i - a)/h$, where $a$ is the assumed mean. Then

$\bar x = a + h \cdot \frac{\sum f_i u_i}{\sum f_i}.$

The advantage: the $u_i$ are small integers like $-2, -1, 0, 1, 2$, even easier to compute.

Example (same data, $a = 25, h = 10$).

$\bar x = 25 + 10 \cdot (-3/28) = 25 - 30/28 = 25 - 1.07 = 23.93$.

When to use which method

• Direct: simplest formula, but heavy arithmetic with large $x_i$.
• Assumed-mean: good for medium-large $x_i$; reduces multiplications.
• Step-deviation: ideal when class size $h$ is constant and $x_i$ are large; smallest arithmetic.

For board exam, all three are acceptable. The textbook usually asks for one specific method by name. Use what is asked; if no method is specified, step-deviation is usually the cleanest.

Worked examples

Example 1. Find the mean of the data: classes $0-10, 10-20, 20-30, 30-40, 40-50$ with frequencies $7, 10, 15, 8, 10$, using the direct method.

Class marks: $5, 15, 25, 35, 45$. Total $f = 50$. Sum $f x = 35 + 150 + 375 + 280 + 450 = 1290$. Mean $= 1290/50 = 25.8$.

Example 2. The mean of $30$ observations is $20$. If one of the observations is wrongly written as $25$ instead of $15$, find the correct mean.

Original sum (wrong) $= 30 \cdot 20 = 600$. Correction: subtract $25$, add $15$, so new sum = $590$. Correct mean = $590/30 = 19.67$.

Example 3. Find the mean of the data using assumed-mean method, with classes $50-60, 60-70, 70-80, 80-90, 90-100$ and frequencies $5, 10, 15, 12, 8$. Take $a = 75$.

Class marks: $55, 65, 75, 85, 95$. Deviations: $-20, -10, 0, 10, 20$. $f d$: $-100, -100, 0, 120, 160$. Sum $fd = 80$. $n = 50$.

Mean $= 75 + 80/50 = 75 + 1.6 = 76.6$.

Example 4. Use step-deviation method for the same data.

$u_i = d_i/10$: $-2, -1, 0, 1, 2$. $f u$: $-10, -10, 0, 12, 16$. Sum $= 8$.

Mean $= 75 + 10 \cdot 8/50 = 75 + 1.6 = 76.6$.

Example 5. The mean of $20$ observations is $30$. If an observation $40$ is replaced by $50$, find the new mean.

Original sum $= 600$. New sum $= 600 - 40 + 50 = 610$. New mean $= 30.5$.

Try it yourself

1. Find the mean of the data: classes $0-5, 5-10, 10-15, 15-20$ with frequencies $4, 7, 8, 1$ using direct method.
2. Same data, use assumed-mean with $a = 12.5$.
3. Same data, use step-deviation with $a = 12.5, h = 5$.
4. Find the mean of classes $20-30, 30-40, 40-50, 50-60$ with frequencies $5, 8, 12, 5$.
5. The mean of $50$ observations is $40$. Find the sum.
6. A class of $30$ students has mean weight $45$ kg. Two new students of weights $50$ and $55$ kg join. Find the new mean.
7. The mean of $5$ observations is $20$. Three are $15, 18, 25$. The other two have ratio $3:4$. Find them.
8. Find the mean of classes $5-15, 15-25, 25-35, 35-45, 45-55$ with frequencies $3, 5, 8, 6, 3$.
9. Same data, with assumed mean $30$.
10. Same data, step-deviation with $h = 10$.
11. The mean of a data set is $50$. If each value is increased by $5$, what is the new mean?
12. The mean of a data set is $50$. If each value is multiplied by $2$, what is the new mean?

Pitfalls / Insight

(1) Always compute the class marks first. A common error is using the lower or upper limits instead.

(2) For step-deviation, $h$ must be the same for all classes. If not, use assumed-mean instead.

(3) The signs of $d_i$ and $u_i$ matter: deviations below $a$ are negative.

(4) The formula gives a grand mean weighted by frequency, not a simple average of class marks.

(5) The mean has the same units as the data. If marks are in percentage points, the mean is in percentage points.