Three Ways to Find the Same Average, and Why the Last One Saves the Most Time

The mean is the familiar average: the total of all values shared equally among them. It gives one number that represents the whole set.

$$\text{mean}=\frac{\text{sum of observations}}{\text{number of observations}}$$

For large grouped tables, adding every value is slow, so there are three methods — direct, short-cut and step-deviation — that give the same answer with less arithmetic. This part covers raw data, frequency tables and all three grouped methods.

**For raw or arrayed data, divide the sum by the number of values; for a frequency distribution, multiply each value by its frequency, add, and divide by the total frequency: $\bar{x}=\frac{\sum fx}{\sum f}$.

Worked example 1 — raw data.** A batsman scores $12,15,9,18,16$ in five matches.

$$\bar{x}=\frac{12+15+9+18+16}{5}=\frac{70}{5}=14$$

Arranged in order as $9,12,15,16,18$, the data is arrayed; the mean is still $14$.

Worked example 2 — discrete frequency distribution. Family sizes in a housing society:

- $x$: $2,3,4,5,6$
- $f$: $2,5,7,4,2$

$$\sum fx=4+15+28+20+12=79\sum f=20\bar{x}=\frac{79}{20}=3.95$$

An everyday example. A society secretary planning water tanks uses the mean family size to estimate daily use per flat.

The substance. The mean is pulled by extreme values: adding a score of $100$ to the batsman's five scores raises the mean from $14$ to $\frac{170}{6}\approx 28.33$.

**Replace each class by its class mark, the mid-point $x=\frac{\text{lower limit}+\text{upper limit}}{2}$, multiply by the frequency, and use $\bar{x}=\frac{\sum fx}{\sum f}$.

Worked example.** Pocket money of $40$ students in rupees:

- $0$–$10$: $4$, class mark $5$, $fx=20$
- $10$–$20$: $6$, class mark $15$, $fx=90$
- $20$–$30$: $10$, class mark $25$, $fx=250$
- $30$–$40$: $12$, class mark $35$, $fx=420$
- $40$–$50$: $8$, class mark $45$, $fx=360$

$$\sum f=40\sum fx=1140\bar{x}=\frac{1140}{40}=₹28.50$$

An everyday example. A school canteen manager estimating how much students usually spend can work from a grouped table like this instead of every individual amount.

The substance. The grouped mean is an estimate, because it assumes the values in each class are centred on the class mark.

**Choose a convenient assumed mean $A$ from the class marks, find the deviations $d=x-A$, and use $\bar{x}=A+\frac{\sum fd}{\sum f}$.

Worked example.** Same pocket-money data, with $A=25$:

- $x=5$: $d=-20$, $fd=-80$
- $x=15$: $d=-10$, $fd=-60$
- $x=25$: $d=0$, $fd=0$
- $x=35$: $d=10$, $fd=120$
- $x=45$: $d=20$, $fd=160$

$$\sum fd=140\bar{x}=25+\frac{140}{40}=25+3.5=28.5$$

Exactly the same answer as the direct method.

An everyday example. When a shopkeeper totals prices near ₹100, it is quicker to add how far each is above or below ₹100 — the same idea as using an assumed mean.

The substance. **Any value of $A$ gives the same mean**; choosing one near the middle keeps the deviations small and the arithmetic easy.

**Divide each deviation by the common class width $i$ to get $t=\frac{x-A}{i}$, then use $\bar{x}=A+\frac{\sum ft}{\sum f}\times i$.

Worked example 1.** Pocket-money data, $A=25$, $i=10$:

$$t=-2,-1,0,1,2ft=-8,-6,0,12,16\sum ft=14$$
$$\bar{x}=25+\frac{14}{40}\times 10=25+3.5=28.5$$

Worked example 2. Monthly electricity bills of $40$ households in rupees:

- $200$–$300$: $5$ $300$–$400$: $8$ $400$–$500$: $12$ $500$–$600$: $10$ $600$–$700$: $5$

With class marks $250,350,450,550,650$, $A=450$ and $i=100$:

$$ft=-10,-8,0,10,10\sum ft=2\bar{x}=450+\frac{2}{40}\times 100=₹455$$

An everyday example. A resident welfare association comparing electricity bills across a colony handles numbers in the hundreds with tiny values of $t$.

The substance. Step deviation needs equal class widths; if widths differ, use the short-cut method instead.