Variance and Standard Deviation for Grouped Data

Calculating Spread for Grouped Data

A grouped frequency table can show how often phone-battery durations fall within intervals such as $6\text{-}10 \text{ hours}$ and $11\text{-}15 \text{ hours}$ . We still want to describe how widely those durations vary.

The table no longer contains each exact observation. Within $11\text{-}15 \text{ hours}$ , for example, we cannot tell whether a battery lasted $11 \text{ hours}$ , $12 \text{ hours}$ , or another duration.

To estimate the spread, we represent every observation in a class by that class's midpoint $x_i$ . This is a modeling approximation, so the resulting variance and standard deviation estimate the values we would obtain from the ungrouped observations.

Formulas for Variance and Standard Deviation of Grouped Data

We weight each midpoint $x_i$ by its class frequency $f$ . The formulas below treat the table as the complete population. A sample requires the corresponding sample-variance formula.

Variance ( $\sigma^2$ )

For grouped data, use the following computational formula:

$\sigma^2 = \frac{\sum (f \cdot x_i^2)}{\sum f} - \left( \frac{\sum (f \cdot x_i)}{\sum f} \right)^2$

The first term is the frequency-weighted mean of the squared midpoints. The second is the square of the frequency-weighted mean. Their difference is the estimated population variance.
Standard Deviation ( $\sigma$ )

Just like with ungrouped data, the standard deviation is the square root of the variance:

$\sigma = \sqrt{\sigma^2}$

Calculating Variance and Standard Deviation of Phone Battery Duration

Suppose a study records phone-battery duration in the following intervals:

Battery duration (hours)	Frequency ( $f$ )
$6\text{-}10$	$2$
$11\text{-}15$	$10$
$16\text{-}20$	$18$
$21\text{-}25$	$45$
$26\text{-}30$	$5$

We will estimate the population variance and standard deviation of these durations.

Create a Helper Table

For each class, calculate the midpoint $x_i$ , then calculate $f \cdot x_i$ and $f \cdot x_i^2$ .

Battery duration (hours)	Midpoint, $x_i$	Frequency, $f$	$f \cdot x_i$	$f \cdot x_i^2$
$6\text{-}10$	$\frac{6+10}{2}=8$	$2$	$2 \times 8 = 16$	$2 \times 8^2 = 128$
$11\text{-}15$	$\frac{11+15}{2}=13$	$10$	$10 \times 13 = 130$	$10 \times 13^2 = 1690$
$16\text{-}20$	$\frac{16+20}{2}=18$	$18$	$18 \times 18 = 324$	$18 \times 18^2 = 5832$
$21\text{-}25$	$\frac{21+25}{2}=23$	$45$	$45 \times 23 = 1035$	$45 \times 23^2 = 23805$
$26\text{-}30$	$\frac{26+30}{2}=28$	$5$	$5 \times 28 = 140$	$5 \times 28^2 = 3920$
Total		$\sum f = 80$	$\sum fx_i = 1645$	$\sum fx_i^2 = 35375$

Calculate Variance

Plug the total values from the table into the variance formula:

\sigma^2 = \frac{\sum (f \cdot x_i^2)}{\sum f} - \left( \frac{\sum (f \cdot x_i)}{\sum f} \right)^2

The estimated variance is $19.37$ square hours. Variance uses squared units because every distance from the mean is squared before averaging.

Calculate Standard Deviation

Take the square root of the variance:

\sigma = \sqrt{19.37} \approx 4.4

The estimated standard deviation is $4.4 \text{ hours}$ . Around the estimated mean of $\frac{1645}{80} \approx 20.56 \text{ hours}$ , durations differ by about $4.4 \text{ hours}$ in the root-mean-square sense.

When datasets use the same unit and a comparable scale, a smaller standard deviation indicates that values are clustered more tightly around their mean.