Mathematics · Ch 13 — Statistics

Shortcut Method to Find Variance and Standard Deviation

13.5.4

Shortcut Method to Find Variance and Standard Deviation

Why a Shortcut Method is Needed

When the values of $x_i$ in a discrete distribution, or the mid-points of classes in a continuous distribution, are large numbers, the direct calculation of mean and variance becomes tedious and time-consuming. The step-deviation method simplifies this by shifting the origin to an assumed mean and reducing the scale by the class width.

The core idea is to transform the original variable $x$ into a new variable $y$ that has much smaller, simpler values. Once we compute the mean and variance for $y$ , we can easily convert them back to the original scale.

Defining the Step-Deviation Variable

Let the assumed mean be $A$ and let the class width (the size of each class interval) be $h$ . For each observation $x_i$ (or each class mid-point in a continuous distribution), define the step-deviation $y_i$ as:

y_i = \frac{x_i - A}{h}

This can be rearranged to express $x_i$ in terms of $y_i$ :

x_i = A + h y_i \qquad(1)

The variable $y_i$ is simply the number of steps (of width $h$ ) that $x_i$ is away from the assumed mean $A$ . A negative $y_i$ means $x_i$ is below $A$ , a positive $y_i$ means it is above $A$ , and $y_i = 0$ when $x_i = A$ .

Deriving the Mean Using Step-Deviation

We know the formula for the mean of a grouped frequency distribution (where $f_i$ are the frequencies and $N = \sum f_i$ ):

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

Substitute $x_i = A + h y_i$ from (1) into (2):

\bar{x} = \frac{1}{N} \sum_{i=1}^{n} f_i (A + h y_i)

\bar{x} = \frac{1}{N} \left( \sum_{i=1}^{n} f_i A + \sum_{i=1}^{n} f_i h y_i \right)

Since $A$ and $h$ are constants, they can be taken out of the sums:

\bar{x} = \frac{1}{N} \left( A \sum_{i=1}^{n} f_i + h \sum_{i=1}^{n} f_i y_i \right)

We know that $\sum_{i=1}^{n} f_i = N$ . Therefore:

\bar{x} = \frac{1}{N} \left( A N + h \sum_{i=1}^{n} f_i y_i \right)

\bar{x} = A + h \left( \frac{1}{N} \sum_{i=1}^{n} f_i y_i \right)

The term in brackets is simply the mean of the $y$ variable, which we denote as $\bar{y}$ . So we arrive at the key result:

\boxed{\bar{x} = A + h \bar{y}} \qquad(3)

Important

The mean of the original variable $x$ is the assumed mean $A$ plus the product of the class width $h$ and the mean of the step-deviation variable $y$ .

Deriving the Variance Using Step-Deviation

The variance of $x$ is defined as:

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

Substitute $x_i = A + h y_i$ from (1) and $\bar{x} = A + h \bar{y}$ from (3):

\sigma_x^2 = \frac{1}{N} \sum_{i=1}^{n} f_i \left[ (A + h y_i) - (A + h \bar{y}) \right]^2

The $A$ terms cancel:

\sigma_x^2 = \frac{1}{N} \sum_{i=1}^{n} f_i \left( h y_i - h \bar{y} \right)^2

\sigma_x^2 = \frac{1}{N} \sum_{i=1}^{n} f_i \left[ h (y_i - \bar{y}) \right]^2

\sigma_x^2 = \frac{1}{N} \sum_{i=1}^{n} f_i h^2 (y_i - \bar{y})^2

Since $h^2$ is a constant, it can be taken out of the sum:

\sigma_x^2 = h^2 \left[ \frac{1}{N} \sum_{i=1}^{n} f_i (y_i - \bar{y})^2 \right]

The term in brackets is precisely the variance of the $y$ variable, $\sigma_y^2$ . Therefore:

\boxed{\sigma_x^2 = h^2 \sigma_y^2} \qquad(4)

Taking the square root gives the relationship for standard deviation:

\boxed{\sigma_x = h \sigma_y} \qquad(4)

Watch out

When converting variance back from the $y$ scale to the $x$ scale, you multiply by $h^2$ , not by $h$ . The standard deviation is multiplied by $h$ because it is the square root of the variance.

The Direct Formula for Standard Deviation

We can combine the results above into a single formula that does not require computing $\bar{y}$ separately. Recall that the variance of $y$ can be written as:

\sigma_y^2 = \frac{1}{N} \sum_{i=1}^{n} f_i y_i^2 - \bar{y}^2

Substituting this into $\sigma_x^2 = h^2 \sigma_y^2$ :

\sigma_x^2 = h^2 \left( \frac{1}{N} \sum_{i=1}^{n} f_i y_i^2 - \bar{y}^2 \right)

Since $\bar{y} = \frac{1}{N} \sum f_i y_i$ , we have:

\sigma_x^2 = h^2 \left( \frac{1}{N} \sum_{i=1}^{n} f_i y_i^2 - \left( \frac{1}{N} \sum_{i=1}^{n} f_i y_i \right)^2 \right)

Multiplying through by $h^2$ :

\boxed{\sigma_x^2 = \frac{h^2}{N^2} \left[ N \sum_{i=1}^{n} f_i y_i^2 - \left( \sum_{i=1}^{n} f_i y_i \right)^2 \right]} \qquad(5)

And for standard deviation:

\boxed{\sigma_x = \frac{h}{N} \sqrt{ N \sum_{i=1}^{n} f_i y_i^2 - \left( \sum_{i=1}^{n} f_i y_i \right)^2 }} \qquad(5)

\sigma_x = \frac{h}{N} \sqrt{ N \sum f_i y_i^2 - \left( \sum f_i y_i \right)^2 }

Worked Example: Full Solution

Let us apply the shortcut method to the following distribution:

Classes	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	3	7	12	15	8	3	2

Step 1: Choose the assumed mean and identify the class width.

Choose $A = 65$ (a value near the centre of the distribution). The class width is $h = 10$ (each interval spans 10 units).

Step 2: Compute the mid-points and step-deviations.

For each class, find the mid-point $x_i$ and then compute $y_i = \frac{x_i - 65}{10}$ .

Class	Frequency $f_i$	Mid-point $x_i$	$y_i = \frac{x_i - 65}{10}$	$y_i^2$	$f_i y_i$	$f_i y_i^2$
30-40	3	35	$-3$	9	$-9$	27
40-50	7	45	$-2$	4	$-14$	28
50-60	12	55	$-1$	1	$-12$	12
60-70	15	65	0	0	0	0
70-80	8	75	1	1	8	8
80-90	3	85	2	4	6	12
90-100	2	95	3	9	6	18
Total	$N = 50$				$\sum f_i y_i = -15$	$\sum f_i y_i^2 = 105$

Step 3: Compute the mean using $\bar{x} = A + h \bar{y}$ . …

Table 13.11Mean, variance and standard deviation by the shortcut method

Class	Frequency $f_i$	Mid-point $x_i$	$y_i=\dfrac{x_i-65}{10}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
30-40	3	35	$-3$	9	$-9$	27
40-50	7	45	$-2$	4	$-14$	28
50-60	12	55	$-1$	1	$-12$	12
60-70	15	65	0	0	0	0
70-80	8	75	1	1	8	8
80-90	3	85	2	4	6	12

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