Mathematics · Ch 13 — Statistics

Standard Deviation of a Continuous Frequency Distribution

13.5.3

Standard Deviation of a Continuous Frequency Distribution

When you have a continuous frequency distribution — data grouped into class intervals like 30–40, 40–50, and so on — you cannot directly use the raw values because you only know how many observations fall in each interval, not their exact individual values. The standard technique is to convert the continuous distribution into a discrete one by replacing each class interval with its mid-point (the average of the lower and upper class limits). Once you have these mid-points $x_i$ and their corresponding frequencies $f_i$ , you treat the distribution exactly as you would a discrete frequency distribution.

So, for a distribution with $n$ classes, where the $i$ -th class has mid-point $x_i$ and frequency $f_i$ , the total number of observations is

N = \sum_{i=1}^{n} f_i

and the mean is

\bar{x} = \frac{1}{N} \sum_{i=1}^{n} f_i x_i.

The standard deviation $\sigma$ is then given by the same formula you used for discrete data:

\sigma = \sqrt{ \frac{1}{N} \sum_{i=1}^{n} f_i (x_i - \bar{x})^2 }.

This is the direct definition. But as you will see, there is a more convenient formula that avoids calculating each deviation $(x_i - \bar{x})$ separately.

Watch out

A common mistake is to forget that $N$ is the sum of frequencies, not the number of classes. Always compute $N = \sum f_i$ first.

Another Formula for Standard Deviation (The Computational Formula)

The direct formula $\sigma^2 = \frac{1}{N} \sum f_i (x_i - \bar{x})^2$ requires you to first find $\bar{x}$ , then compute each squared deviation. The textbook derives an algebraically equivalent formula that works directly with $x_i$ and $x_i^2$ , which is often easier for calculations — especially when the mean is not a nice round number.

Start with the definition of variance:

\sigma^2 = \frac{1}{N} \sum_{i=1}^{n} f_i (x_i - \bar{x})^2.

Expand the square:

\sigma^2 = \frac{1}{N} \sum_{i=1}^{n} f_i \left( x_i^2 - 2 x_i \bar{x} + \bar{x}^2 \right).

Now split the sum into three separate sums:

\sigma^2 = \frac{1}{N} \left[ \sum_{i=1}^{n} f_i x_i^2 - 2\bar{x} \sum_{i=1}^{n} f_i x_i + \bar{x}^2 \sum_{i=1}^{n} f_i \right].

Recall that $\sum f_i = N$ and $\sum f_i x_i = N \bar{x}$ . Substitute these in:

\sigma^2 = \frac{1}{N} \left[ \sum f_i x_i^2 - 2\bar{x} (N \bar{x}) + \bar{x}^2 (N) \right].

Simplify the terms inside the brackets:

\sigma^2 = \frac{1}{N} \left[ \sum f_i x_i^2 - 2N \bar{x}^2 + N \bar{x}^2 \right] = \frac{1}{N} \left[ \sum f_i x_i^2 - N \bar{x}^2 \right].

Now replace $\bar{x}$ with $\frac{1}{N} \sum f_i x_i$ :

\sigma^2 = \frac{1}{N} \left[ \sum f_i x_i^2 - N \left( \frac{1}{N} \sum f_i x_i \right)^2 \right] = \frac{1}{N} \left[ \sum f_i x_i^2 - \frac{(\sum f_i x_i)^2}{N} \right].

Multiply through by $\frac{1}{N}$ :

\sigma^2 = \frac{1}{N} \sum f_i x_i^2 - \left( \frac{1}{N} \sum f_i x_i \right)^2.

This is the variance. Taking the square root gives the standard deviation:

\sigma = \sqrt{ \frac{1}{N} \sum_{i=1}^{n} f_i x_i^2 - \left( \frac{1}{N} \sum_{i=1}^{n} f_i x_i \right)^2 }.

The textbook writes this in an equivalent, slightly different form that is also very common:

\sigma = \frac{1}{N}\sqrt{ N \sum f_i x_i^2 - \left( \sum f_i x_i \right)^2 }.

Both are the same — the second version just clears the denominators inside the root and then divides by $N$ . You can use whichever you find easier to remember.

Tip

This formula is often called the step-deviation formula when you also shift the origin (subtract an assumed mean) and change the scale (divide by class width). But the version above is the basic computational formula — it only uses $x_i$ , $x_i^2$ , and $f_i$ . No need to compute $\bar{x}$ separately.

Worked Example 10 (from the textbook)

Problem: Calculate the mean, variance, and standard deviation for the following distribution:

Class	30–40	40–50	50–60	60–70	70–80	80–90	90–100
Frequency	3	7	12	15	8	3	2

Step 1: Find mid-points and construct the table.

The mid-point $x_i$ of a class is $\frac{\text{lower limit} + \text{upper limit}}{2}$ . For 30–40, $x_i = 35$ ; for 40–50, $x_i = 45$ ; and so on.

Class	$f_i$	$x_i$	$f_i x_i$	$(x_i - \bar{x})^2$	$f_i (x_i - \bar{x})^2$
30–40	3	35	105	729	2187
40–50	7	45	315	289	2023
50–60	12	55	660	49	588
60–70	15	65	975	9	135
70–80	8	75	600	169	1352
80–90	3	85	255	529	1587
90–100	2	95	190	1089	2178
Total	50		3100		10050

Step 2: Compute the mean.

\bar{x} = \frac{1}{N} \sum f_i x_i = \frac{3100}{50} = 62.

Step 3: Compute variance using the direct formula.

\sigma^2 = \frac{1}{N} \sum f_i (x_i - \bar{x})^2 = \frac{1}{50} \times 10050 = 201.

Step 4: Standard deviation.

\sigma = \sqrt{201} \approx 14.18.

Note

The squared deviations $(x_i - \bar{x})^2$ in the table were computed using $\bar{x} = 62$ . For example, $(35 - 62)^2 = (-27)^2 = 729$ , $(45 - 62)^2 = (-17)^2 = 289$ , and so on.

Worked Example 11 (using the computational formula)

Problem: Find the standard deviation for the following data:

$x_i$	3	8	13	18	23
$f_i$	7	10	15	10	6

Here the data is already given as mid-points (discrete), so the same formula applies directly.

Step 1: Construct the table with $x_i^2$ and $f_i x_i^2$ . …

Table 13.9Mean, variance and standard deviation of a continuous frequency distribution

Class	Frequency $(f_i)$	Mid-point $(x_i)$	$f_ix_i$	$(x_i-\bar{x})^2$	$f_i(x_i-\bar{x})^2$
30-40	3	35	105	729	2187
40-50	7	45	315	289	2023
50-60	12	55	660	49	588
60-70	15	65	975	9	135
70-80	8	75	600	169	1352
80-90	3	85	255	529	1587

Table 13.10Standard deviation using the $\sum f_ix_i^2$ formula

$x_i$	$f_i$	$f_ix_i$	$x_i^2$	$f_ix_i^2$
3	7	21	9	63
8	10	80	64	640
13	15	195	169	2535
18	10	180	324	3240

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