Squaring the Gaps Gives the Most Useful Measure of Spread

Mean deviation removed negative signs with absolute values, which are hard to handle in algebra. Squaring also removes the signs, and squares can be expanded, added and averaged freely.

The average of the squared deviations is the variance, and its square root — back in the original units — is the standard deviation.

This part covers variance and standard deviation for raw data, discrete distributions and continuous distributions, and how to compare variability.

**Variance is ${\sigma }^2=\frac{1}{n}\sum (x_i-\bar{x}{)}^2$ and standard deviation is $\sigma =\sqrt{{\sigma }^2}$; the equivalent shortcut ${\sigma }^2=\frac{1}{n}\sum x_i^2-{\bar{x}}^2$ avoids working out every deviation.

Worked example — definition.** Riya's marks $60,62,64,66,68$ have $\bar{x}=64$.

$$(x_i-64{)}^2:16,4,0,4,16{\sigma }^2=\frac{40}{5}=8\sigma =\sqrt{8}\approx 2.83$$

Same data — shortcut.

$$\sum x_i^2=3600+3844+4096+4356+4624=20520{\sigma }^2=\frac{20520}{5}-64^2=4104-4096=8$$

Kabir's marks $40,50,64,78,88$ with the same mean give squared deviations $576,196,0,196,576$:

$${\sigma }^2=\frac{1544}{5}=308.8\sigma \approx 17.57$$

An everyday example. A machine filling 1 kg packets of atta is working well when the standard deviation of the packet weights is small — nearly every packet weighs the same.

The substance. Standard deviation has the same unit as the data, while variance has the unit squared.

**For values $x_i$ with frequencies $f_i$ and $N=\sum f_i$, ${\sigma }^2=\frac{1}{N}\sum f_i(x_i-\bar{x}{)}^2$, or by shortcut ${\sigma }^2=\frac{1}{N}\sum f_i{x}_i^2-{\bar{x}}^2$.

Worked example.** $x$: $10,20,30,40,50$ with $f$: $3,5,8,3,1$, so $N=20$ and $\bar{x}=\frac{540}{20}=27$.

By definition.

$$\sum f_i(x_i-27{)}^2=3(289)+5(49)+8(9)+3(169)+1(529)$$

$$=867+245+72+507+529=2220{\sigma }^2=\frac{2220}{20}=111$$

By shortcut, as a check.

$$\sum f_i{x}_i^2=300+2000+7200+4800+2500=16800{\sigma }^2=\frac{16800}{20}-27^2=840-729=111$$

$$\sigma =\sqrt{111}\approx 10.54$$

An everyday example. The number of runs scored off each ball in an over, tallied across a match, forms a discrete distribution whose standard deviation shows how steady the scoring was.

The substance. Multiply each squared deviation by its frequency before adding — each value counts as many times as it occurs.

**Using class mid-points $x_i$, an assumed mean $A$ and class width $h$, set $d_i=\frac{x_i-A}{h}$; then ${\sigma }^2=h^2\left[\frac{\sum f_i{d}_i^2}{N}-{\left(\frac{\sum f_i{d}_i}{N}\right)}^2\right]$.

Worked example.** Classes $0$-$10$ to $40$-$50$ with $f$: $5,8,15,16,6$, so $N=50$. Take $A=25$ and $h=10$.

- $d_i$: $-2,-1,0,1,2$
- $f_i{d}_i$: $-10,-8,0,16,12$, total $10$
- $f_i{d}_i^2$: $20,8,0,16,24$, total $68$

Mean.

$$\bar{x}=25+\frac{10}{50}\times 10=27$$

Variance and standard deviation.

$${\sigma }^2=100\left[\frac{68}{50}-{\left(\frac{10}{50}\right)}^2\right]=100(1.36-0.04)=132\sigma =\sqrt{132}\approx 11.49$$

An everyday example. Monthly electricity bills of the flats in a housing society, grouped in ₹500 classes, are analysed exactly this way.

The substance. **Do not forget the factor $h^2$** — the $d_i$ are measured in class widths, so the variance must be scaled back.

**When two data sets have equal means, the one with the larger standard deviation is more variable; when means or units differ, compare the coefficient of variation $\text{C.V.}=\frac{\sigma }{\bar{x}}\times 100$, and the lower C.V. is the more consistent.

Worked example 1 — equal means.** Riya and Kabir both average $64$, with $\sigma =2.83$ and $\sigma =17.57$. Riya is far more consistent.

Worked example 2 — different units. Suppose a class's heights have $\bar{x}=150$ cm and $\sigma =6$ cm, and its weights have $\bar{x}=45$ kg and $\sigma =4.5$ kg.

$$\text{C.V. (heights)}=\frac{6}{150}\times 100=4\text{C.V. (weights)}=\frac{4.5}{45}\times 100=10$$

Weights vary more, even though $4.5$ is smaller than $6$.

Worked example 3 — different means. Suppose batter P averages $50$ runs with $\sigma =10$ and batter Q averages $40$ with $\sigma =6$. C.V. is $20$ for P and $15$ for Q, so Q is more consistent while P scores more.

An everyday example. Comparing the price swings of tomatoes and rice needs C.V., because their average prices are very different.

The substance. C.V. has no unit, which is exactly why it can compare centimetres with kilograms.